Short story. I made a program that does addition for binary integers. I need to make it work for binary real numbers (e.g. 1010.1010(binary)=10.625(decimal)
The input is given as a binary string.
I made a lot of attempts and I couldn't find a simple way to do it. Please help create such a program.
Example: {input: 1010.1010(10.625 in decimal) 0.1(0.5 in decimal)
output: 1011.001 (11.125 in decimal)}
Code:
#include <stdio.h>
#include <string.h>
void bin_add(int c[400], int d[400])
{
int car[400]; //carry
int i = 199;
car[i] = 0;
while (i >= 0)
{
//find carry and shift it left
//find the sum
car[i - 1] = (c[i] & d[i]) | (c[i] & car[i]) | (d[i] & car[i]);
c[i] = (c[i] ^ d[i]) ^ car[i];
printf("car[i-1]=%d c[i]=%d\n", car[i - 1], c[i]);
i--;
}
// printf("\n");
}
int main()
{
int l, l1, i;//l and l1 are lengths
char a[200], b[200]; //a and b are the inputs
int c[200], d[200]; //c and d are used for processing
for (i = 0; i < 200; i++)
{
c[i] = 0;
d[i] = 0;
}
gets(a);
gets(b);
l = strlen(a);
l1 = strlen(b);
for (int i = 0; i < l; i++)
{
c[200 - l + i] = a[i] - 48;
}
////////////////////////////////////////////
for (int i = 0; i < l1; i++)
{
d[200 - l1 + i] = b[i] - 48;
}
////////////////////////////////
bin_add(c, d);
for (i = 0; i < 200; i++)
printf("%d", c[i]);
return 0;
}
What you really want to do, is handle each digit in order of increasing importance. To make that easier, you should implement the following functions:
/* Return the number of fractional bits in bs */
int bs_fractbits(const char *bs);
/* Return the number of integer bits in bs */
int bs_intbits(const char *bs);
/* Return the bit in bs corresponding to value 2**i,
0 if outside the bit string */
int bs_bit(const char *bs, int i);
/* Return -1 if bs is negative,
0 if bs is zero or NULL,
+1 if bs is positive */
int bs_sign(const char *bs);
/* Return -1 if bs1 < bs2,
0 if bs1 == bs2,
+1 if bs1 > bs2. */
int bs_cmp(const char *bs1, const char *bs2);
To support negative values, you'll need to implement both addition and subtraction (of "unsigned" bit strings):
Addition: The result has as many fractional bits as the term that has most fractional bits, and possibly one more integer bit than the term that has most integer bits. Start at the least significant bit in either term, and work your way up to the most significant bit in either term, summing each bit, and keeping the "carry bit" along, just like you'd do by hand. If the carry is nonzero at end, you'll get that one additional bit.
Subtraction: Always subtract smaller from larger. If that changes the order of the terms, negate the result. The result has at most as many fractional bits as the term that has most fractional bits, and at most as many integer bits as the term that has most integer bits. This is just like addition, except you subtract the bits, and instead of "carry bit", you use a "borrow bit". Because you subtract smaller unsigned value from larger unsigned value, the "borrow bit" will be zero at end.
Multiplication: The integer part has the number of integer bits, and the number of fractional bits, as the terms have in total (summed). You can implement the operation as if multiplying two unsigned integer values, and just insert the bit at end. (So that the result has as many fractional bits as the input terms have in total.) This usually involves a double loop, just like in long multiplication by hand.
Note that the same logic also works if you use larger radix instead of 2. The "carry"/"borrow" is a digit, between zero and one less than the radix.
Personally, I'd be very tempted to use a structure to describe each digit string:
typedef struct {
int idigits; /* Number of integral digits before point */
int fdigits; /* Number of fractional digits after point */
int size; /* Number of chars dynamically allocated at data */
char *point; /* Location of decimal point */
char *data; /* Dynamically allocated buffer */
} digitstring;
#define DIGITSTRING_INIT { 0, 0, 0, NULL, NULL }
with an additional flag if negative digit strings are to be supported.
Digit D with numerical value D×Bi, where B is the radix (number of unique digits used) and i being the position of said digit, is located at point[-i] if i < 0 (and -i <= fdigits), or at point[-i-1] if i >= 0 (and i < idigits). point[0] itself is where the decimal point is, if there is one.
For example, if we have string 0100.00, then idigits = 4, fdigits = 2, and the only nonzero digit is at position 2. (Position 0 is on the left side of the decimal point, and -1 on the right side.)
size and data fields allow reuse of the dynamically allocated buffer. Each declaration of a digitstring must be initialized, digitstring value = DIGITSTRING_INIT;, because there is no initialization function; this way you are less likely to leak memory (unless you forget to free a digitstring when no longer needed):
/* Free the specified digit string. */
static inline void digitstring_free(digitstring *ds)
{
if (ds) {
if (ds->data)
free(ds->data);
ds->idigits = 0;
ds->fdigits = 0;
ds->size = 0;
ds->point = NULL;
ds->data = NULL;
}
}
To use the digit string as a C string, you use a helper function to obtain the pointer to the most significant digit in the digit string:
/* Return a pointer to a printable version of the digit string. */
static const char *digitstring_str(const digitstring *ds, const char *none)
{
if (ds && ds->point)
return (const char *)(ds->point - ds->idigits);
else
return none;
}
I've found that rather than crash, it is often useful to pass an extra parameter that is only used for the return value when the return value is otherwise undefined. For example, if you have an initialized digit string foo without any contents, then digitstring_str(&foo, "0") returns the string literal "0".
The main point of the digit string structure is to have accessor functions that get and set each individual digit:
/* Get the value of a specific digit. */
static inline unsigned int digitstring_get(const digitstring *ds, const int position)
{
if (ds) {
if (position < 0) {
if (-position <= ds->fdigits)
return digit_to_value(ds->point[-position]);
else
return 0;
} else {
if (position < ds->idigits)
return digit_to_value(ds->point[-position-1]);
else
return 0;
}
} else
return 0;
}
/* Set the value of a specific digit. */
static inline void digitstring_set(digitstring *ds, const int position, const unsigned int value)
{
if (!ds) {
fprintf(stderr, "digitstring_set(): NULL digitstring specified.\n");
exit(EXIT_FAILURE);
} else
if (position < 0) {
if (-position > ds->fdigits) {
fprintf(stderr, "digitstring_set(): Digit position underflow (in fractional part).\n");
exit(EXIT_FAILURE);
}
ds->point[-position] = value_to_digit(value);
} else {
if (position >= ds->idigits) {
fprintf(stderr, "digitstring_set(): Digit position overflow (in integer part).\n");
exit(EXIT_FAILURE);
}
ds->point[-position-1] = value_to_digit(value);
}
}
Above, value_to_digit() is a helper function that converts a numerical value to the corresponding character, and digit_to_value() converts a character to the corresponding numerical value.
All operations (from parsing to arithmetic operators) really need a "constructor", that creates a new digit string with sufficient number of digits. (The number of digits is known beforehand for each operation, and depends only on the number of significant digits in the terms.) For this, I created a function that constructs a zero of desired size:
/* Clear the specified digit string to zero. */
static inline void digitstring_zero(digitstring *ds, int idigits, int fdigits)
{
int size;
char *data;
if (!ds) {
fprintf(stderr, "digitstring_zero(): No digitstring specified.\n");
exit(EXIT_FAILURE);
}
/* Require at least one integral digit. */
if (idigits < 1)
idigits = 1;
if (fdigits < 0)
fdigits = 0;
/* Total number of chars needed, including decimal point
and string-terminating nul char. */
size = idigits + 1 + fdigits + 1;
/* Check if dynamically allocated buffer needs resizing. */
if (ds->size < size) {
if (ds->data)
data = realloc(ds->data, size);
else
data = malloc(size);
if (!data) {
fprintf(stderr, "digitstring_zero(): Out of memory.\n");
exit(EXIT_FAILURE);
}
ds->data = data;
ds->size = size;
} else {
data = ds->data;
size = ds->size;
}
/* Fill it with zeroes. */
memset(data, value_to_digit(0), idigits + 1 + fdigits);
/* Pad the unused space with nul chars, terminating the string. */
memset(data + idigits + 1 + fdigits, '\0', size - idigits - 1 - fdigits);
/* Assign the decimal point. */
ds->point = data + idigits;
/* If there are no decimals, no need for a decimal point either. */
if (fdigits > 0)
ds->point[0] = decimal_point;
else
ds->point[0] = '\0';
/* After setting the desired digits, use digitstring_trim(). */
ds->idigits = idigits;
ds->fdigits = fdigits;
}
It will ensure the digit string has enough room for the specified number of digits, reallocating its dynamically allocated buffer if necessary, reusing it if already large enough.
The idea is that to implement an operation, you first find out the maximum number of integral and fractional digits the result can have. You use the above to create the result digit string, then digitstring_set() to set each digit to their respective values. You will typically operate in increasing digit significance, which means increasing digit "positions".
If we have additional helper functions int digits(const char *src), which returns the number of consecutive valid digit characters starting at src, and int decimal_points(const char *src), which returns 1 if src points to a decimal point, and 0 otherwise, we can parse input strings into digit strings using
/* Parse a string into a digit string, returning the pointer
to the first unparsed character, or NULL if an error occurs. */
static const char *digitstring_parse(digitstring *ds, const char *src)
{
const int zero = value_to_digit(0);
const char *idigit, *fdigit;
int idigits, fdigits, fextra, n;
/* Fail if nothing to parse. */
if (!src)
return NULL;
/* Skip leading whitespace. */
while (isspace((unsigned char)(*src)))
src++;
/* Fail if nothing to parse. */
if (*src == '\0')
return NULL;
/* Scan integer digits. */
idigit = src;
src += digits(src);
idigits = (int)(src - idigit);
/* Decimal point? */
fextra = 0;
n = decimal_points(src);
if (n > 0) {
src += n;
/* Scan fractional digits. */
fdigit = src;
src += digits(src);
fdigits = (int)(src - fdigit);
if (fdigits < 1)
fextra = 1;
} else {
fdigit = src;
fdigits = 0;
}
/* No digits? */
if (idigit == 0 && fdigit == 0)
return NULL;
/* Trim leading zeroes. */
while (idigits > 1 && *idigit == zero) {
idigits--;
idigit++;
}
/* Trim trailing zeroes. */
while (fdigits > 1 && fdigit[fdigits - 1] == zero)
fdigits--;
/* Create the necessary digit string, */
digitstring_zero(ds, idigits, fdigits + fextra);
/* copy the integer digits, if any, */
if (idigits > 0)
memcpy(ds->point - idigits, idigit, idigits);
/* and the fractional digits, if any. */
if (fdigits > 0)
memcpy(ds->point + 1, fdigit, fdigits);
/* Return a pointer to the first unparsed character. */
return src;
}
After updating its digits, one can call a helper function to remove any extra leading zeroes:
static inline void digitstring_ltrim(digitstring *ds)
{
if (ds && ds->point) {
const int zero = value_to_digit(0);
while (ds->idigits > 1 && ds->point[-ds->idigits] == zero)
ds->idigits--;
}
}
Adding two (unsigned) digit strings, possibly reusing one of the terms, is now quite simple to implement:
static void digitstring_add(digitstring *to, const digitstring *src1, const digitstring *src2)
{
digitstring result = DIGITSTRING_INIT;
unsigned int carry = 0;
int i, idigits, fdigits;
if (!to || !src1 || !src2) {
fprintf(stderr, "digitstring_add(): NULL digitstring specified.\n");
exit(EXIT_FAILURE);
}
/* For addition, the result has as many digits
as the longer source term. */
idigits = (src1->idigits >= src2->idigits) ? src1->idigits : src2->idigits;
fdigits = (src1->fdigits >= src2->fdigits) ? src1->fdigits : src2->fdigits;
/* Result needs possibly one more integer digit,
in case carry overflows. */
digitstring_zero(&result, idigits + 1, fdigits);
/* Addition loop, in order of increasing digit significance. */
for (i = -fdigits; i < idigits; i++) {
const unsigned int sum = digitstring_get(src1, i)
+ digitstring_get(src2, i)
+ carry;
digitstring_set(&result, i, sum % RADIX);
carry = sum / RADIX;
}
digitstring_set(&result, idigits, carry);
/* Trim leading zeroes. */
digitstring_ltrim(&result);
/* At this point, we can discard the target, even if it is actually
one of the sources, and copy the result to it. */
digitstring_free(to);
*to = result;
}
where RADIX is the radix used (the number of unique digits, 2 for binary). Pay extra attention to the digit loop. -fdigits is the least significant position in the result, and idigits-1 the most significant position. We need the accessor functions, because the source terms might not contain those digits at all (they are logically zero then).
These functions have been tested to work on both binary and octal number strings. I like this implementation, because it omits the decimal point if all terms are integers (so you get 12 + 33 = 45), but (due to fextra in digitstring_parse()) if any of the terms have a decimal point, then the result will have at least one fractional digit (so 12. + 33 = 45.0).
After all the beautiful ideas in Animals' answer I felt the strange urge, to present my own solution:
#include <stdbool.h>
#include <stddef.h>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#define MAX(x, y) ((x) > (y) ? (x) : (y))
size_t gpp(char const *s)
{
char *n = strchr(s, '.');
return n ? n - s + 1 : 0;
}
char* bin_add(char const *a, char const *b)
{
char const *inp[] = { a, b };
size_t ll[] = { strlen(a), strlen(b) };
size_t pp[] = { gpp(a), gpp(b) };
size_t OO[2], off[2];
for (size_t i = 0; i < 2; ++i) {
OO[i] = pp[i] ? pp[i] - 1 : ll[i];
pp[i] = pp[i] ? ll[i] - pp[i] : 0;}
for (size_t i = 0; i < 2; ++i)
off[i] = OO[i] < OO[!i] ? OO[!i] - OO[i] : 0;
size_t ML = MAX(OO[0], OO[1]) + MAX(pp[0], pp[1]) + (!!pp[0] || !!pp[1]);
char *Ol = calloc(ML + 2, 1);
if(!Ol) return Ol;
char ops[2];
int xc = 0;
size_t lO = ML;
unsigned cc[2] = { 0 };
for (size_t i = ML; i; --i) {
bool pt = false;
for (size_t l = 0; l < 2; ++l) {
ops[l] = i <= ll[l] + off[l] && i - off[l] - 1
< ll[l] ? inp[l][i - off[l] - 1] : '0';
if (ops[l] == '.') {
if (cc[l]) {
free(Ol);
return NULL;
}
pt = true;
++cc[l];
}
ops[l] -= '0';
}
if (pt) {
Ol[i] = '.';
continue;
}
if ((Ol[i] = ops[0] + ops[1] + xc) > 1) {
Ol[i] = 0;
xc = 1;
}
else xc = 0;
lO = (Ol[i] += '0') == '1' ? i : lO;
}
if((Ol[0] = '0' + xc) == '1') return Ol;
for (size_t i = 0; i <= ML - lO + 1; ++i)
Ol[i] = Ol[lO + i];
return Ol;
}
int main(void)
{
char a[81], b[81];
while (scanf(" %80[0.1] %80[0.1]", a, b) & 1 << 1) {
char *c = bin_add(a, b);
if (!c && errno == ENOMEM) {
fputs("Not enough memory :(\n\n", stderr);
return EXIT_FAILURE;
}
else if (!c) {
fputs("Input error :(\n\n", stderr);
goto clear;
}
char* O[] = { a, b, c };
size_t lO[3], Ol = 0;
for (size_t i = 0; i < 3; ++i) {
lO[i] = gpp(O[i]);
lO[i] = lO[i] ? lO[i] : strlen(i[O]) + 1;
Ol = lO[i] > Ol ? lO[i] : Ol;
}
putchar('\n');
for (size_t i = 0; i < 3; ++i) {
for (size_t l = 0; l < Ol - lO[i]; ++l, putchar(' '));
puts(O[i]);
}
putchar('\n');
free(c);
clear :{ int c; while ((c = getchar()) != '\n' && c != EOF); }
}
}
Sample Output:
11001001 .11001001
11001001
.11001001
11001001.11001001
11001001 1010
11001001
1010
11010011
111111 1
111111
1
1000000
1010101 010111001.0101110101010
1010101
010111001.0101110101010
100001110.0101110101010
1001001.010111010101 10100101100.10010111101
1001001.010111010101
10100101100.10010111101
10101110101.111000001111
. .
.
.
0
.. .
Input error :(
A
Press any key to continue . . .
I contemplated wheter I should ask for a review at codereview. But I think I schould rather not.
There are two answers, depending upon whether you desire fixed- or floating- point arithmetic.
The first issue is reading the number. strtol() is your friend here:
char input[BUFFER_SIZE];
char * tmp;
long integral, fractional;
fgets(input, BUFFER_SIZE-1, stdin);
integral = strtol(input, &tmp, 2); /* Read BINARY integral part */
tmp++; /* Pass over the binary point. You may want to check that it is actually a dot */
fractional = strtol(tmp, null, 2); /* Read BINARY fractional part */
The next issue is figuring out how you will do the arithmetic. fractional must be bit-shifted an amount depending on how many digits past the point were provided and your desired precision. Fixed point arithmetic is simple: fractional <<= FRAC_BITS - strlen(tmp) then add the fractional parts together. Mask by ((1<<FRAC_BITS)-1) for the fractional part of the sum, shift the remaining bits and add them to the integral parts for the integral part of the sum. Floating-point is a little more finicky, but not too much harder.
For real numbers, convert non-fraction and fraction part to decimal, do the addition and print it as binary. This will require function to convert a number to binary string. Just a note that real numbers are float numbers in C and they are represented in binary with mantessa form like 2e^3 which is 2 multiplied by exponent to the power of 3.
Related
I am trying to code a program that will take a floating point number in base 10 and convert its fractional part in base 2. In the following code, I am intending to call my converting function into a printf, and format the output; the issue I have lies in my fra_binary() where I can't figure out the best way to return an integer made of the result of the conversion at each turn respectively (concatenation). Here is what I have done now (the code is not optimized because I am still working on it) :
#include <stdio.h>
#include <math.h>
int fra_binary(double fract) ;
int main()
{
long double n ;
double fract, deci ;
printf("base 10 :\n") ;
scanf("%Lf", &n) ;
fract = modf(n, &deci) ;
int d = deci ;
printf("base 2: %d.%d\n", d, fra_binary(fract)) ;
return(0) ;
}
int fra_binary(double F)
{
double fl ;
double decimal ;
int array[30] ;
for (int i = 0 ; i < 30 ; i++) {
fl = F * 2 ;
F = modf(fl, &decimal) ;
array[i] = decimal ;
if (F == 0) break ;
}
return array[0] ;
}
Obviously this returns partly the desired output, because I would need the whole array concatenated as one int or char to display the series of 1 and 0s I need. So at each turn, I want to use the decimal part of the number I work on as the binary number to concatenate (1 + 0 = 10 and not 1). How would I go about it?
Hope this makes sense!
return array[0] ; is only the first value of int array[30] set in fra_binary(). Code discards all but the first calculation of the loop for (int i = 0 ; i < 30 ; i++).
convert its fractional part in base 2
OP's loop idea is a good starting point. Yet int array[30] is insufficient to encode the fractional portion of all double into a "binary".
can't figure out the best way to return an integer
Returning an int will be insufficient. Instead consider using a string - or manage an integer array in a likewise fashion.
Use defines from <float.h> to drive the buffer requirements.
#include <stdio.h>
#include <math.h>
#include <float.h>
char *fra_binary(char *dest, double x) {
_Static_assert(FLT_RADIX == 2, "Unexpected FP base");
double deci;
double fract = modf(x, &deci);
fract = fabs(fract);
char *s = dest;
do {
double d;
fract = modf(fract * 2.0, &d);
*s++ = "01"[(int) d];
} while (fract);
*s = '\0';
// For debug
printf("%*.*g --> %.0f and .", DBL_DECIMAL_DIG + 8, DBL_DECIMAL_DIG, x,
deci);
return dest;
}
int main(void) {
// Perhaps 53 - -1021 + 1
char fraction_string[DBL_MANT_DIG - DBL_MIN_EXP + 1];
puts(fra_binary(fraction_string, -0.0));
puts(fra_binary(fraction_string, 1.0));
puts(fra_binary(fraction_string, asin(-1))); // machine pi
puts(fra_binary(fraction_string, -0.1));
puts(fra_binary(fraction_string, DBL_MAX));
puts(fra_binary(fraction_string, DBL_MIN));
puts(fra_binary(fraction_string, DBL_TRUE_MIN));
}
Output
-0 --> -0 and .0
1 --> 1 and .0
3.1415926535897931 --> 3 and .001001000011111101101010100010001000010110100011
-0.10000000000000001 --> -0 and .0001100110011001100110011001100110011001100110011001101
1.7976931348623157e+308 --> 179769313486231570814527423731704356798070600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 and .0
2.2250738585072014e-308 --> 0 and .00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
4.9406564584124654e-324 --> 0 and .000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
Also unclear why input is long double, yet processing is with double. Recommend using just one FP type.
Note that your algorithm finds out the binary representation of the fraction most significant bit first.
One way to convert the fractional part to a binary string, would be to supply the function with a string and a string length, and have the function fill it with up to that many binary digits:
/* This function returns the number of chars needed in dst
to describe the fractional part of value in binary,
not including the trailing NUL ('\0').
Returns zero in case of an error (non-finite value).
*/
size_t fractional_bits(char *dst, size_t len, double value)
{
double fraction, integral;
size_t i = 0;
if (!isfinite(value))
return 0;
if (value > 0.0)
fraction = modf(value, &integral);
else
if (value < 0.0)
fraction = modf(-value, &integral);
else {
/* Zero fraction. */
if (len > 1) {
dst[0] = '0';
dst[1] = '\0';
} else
if (len > 0)
dst[0] = '\0';
/* One binary digit was needed for exact representation. */
return 1;
}
while (fraction > 0.0) {
fraction = fraction * 2.0;
if (fraction >= 1.0) {
fraction = fraction - 1.0;
if (i < len)
dst[i] = '1';
} else
if (i < len)
dst[i] = '0';
i++;
}
if (i < len)
dst[i] = '\0';
else
if (len > 0)
dst[len - 1] = '\0';
return i;
}
The above function works very much like snprintf(), except it takes only the double whose fractional bits are to be stored as a string of binary digits (0 or 1). and returns 0 in case of an error (non-finite double value).
Another option is to use an unsigned integer type to hold the bits. For example, if your code is intended to work on architectures where double is an IEEE-754 Binary64 type or similar, the mantissa has up to 53 bits of precision, and an uint64_t would suffice.
Here is an example of that:
uint64_t fractional_bits(const double val, size_t bits)
{
double fraction, integral;
uint64_t result = 0;
if (bits < 1 || bits > 64) {
errno = EINVAL;
return 0;
}
if (!isfinite(val)) {
errno = EDOM;
return 0;
}
if (val > 0.0)
fraction = modf(val, &integral);
else
if (val < 0.0)
fraction = modf(-val, &integral);
else {
errno = 0;
return 0;
}
while (bits-->0) {
result = result << 1;
fraction = fraction * 2.0;
if (fraction >= 1.0) {
fraction = fraction - 1.0;
result = result + 1;
}
}
errno = 0;
return result;
}
The return value is the binary representation of the fractional part: [i]fractional_part[/i] ≈ [i]result[/i] / 2[sup][i]bits[/i][/sup], where [i]bits[/i] is between 1 and 64, inclusive.
In order for the caller to detect an error, the function clears errno to zero if no error occurred. If an error does occur, the function returns zero with errno set to EDOM if the value is not finite, or to EINVAL if bits is less than 1 or greater than 64.
You can combine the two approaches, if you implement an arbitrary-size unsigned integer type, or a bitmap type.
I'm trying to write my first kernel module so I'm not able to include libraries for atoi, strtol, etc. How can I convert a string to int without these built-in functions? I tried:
int num;
num = string[0] - '0';
which works for the first character, but if I remove the [0] to try and convert the full string it gives me a warning: assignment makes integer from pointer without a cast. So what do I do?
When creating your own string to int function, make sure you check and protect against overflow. For example:
/* an atoi replacement performing the conversion in a single
pass and incorporating 'err' to indicate a failed conversion.
passing NULL as error causes it to be ignored */
int strtoi (const char *s, unsigned char *err)
{
char *p = (char *)s;
int nmax = (1ULL << 31) - 1; /* INT_MAX */
int nmin = -nmax - 1; /* INT_MIN */
long long sum = 0;
char sign = *p;
if (*p == '-' || *p == '+') p++;
while (*p >= '0' && *p <= '9') {
sum = sum * 10 - (*p - '0');
if (sum < nmin || (sign != '-' && -sum > nmax)) goto error;
p++;
}
if (sign != '-') sum = -sum;
return (int)sum;
error:
fprintf (stderr, "strtoi() error: invalid conversion for type int.\n");
if (err) *err = 1;
return 0;
}
You can't remove the [0]. That means that you are subtracting '0' from the pointer string, which is meaningless. You still need to dereference it:
num = string[i] - '0';
A string is an array of characters, represented by an address (a.k.a pointer).
An pointer has an value that might look something like 0xa1de2bdf. This value tells me where the start of the array is.
You cannot subtract a pointer type with a character type (e.g 0xa1de2bdf - 'b' does not really make sense).
To convert a string to a number, you could try this:
//Find the length of the string
int len = 0;
while (str[len] != '\0') {
len++;
}
//Loop through the string
int num = 0, i = 0, digit;
for (i=0; i<len; i++) {
//Extract the digit
digit = ing[i] - '0';
//Multiply the digit with its correct position (ones, tens, hundreds, etc.)
num += digit * pow(10, (len-1)-i);
}
Of course if you are not allowed to use math.h library, you could write your own pow(a,b) function which gives you the value of a^b.
int mypowfunc(int a, int b) {
int i=0, ans=1;
//multiply the value a for b number of times
for (i=0; i<b; i++) {
ans *= a;
}
return ans;
}
I have written the code above in a way that is simple to understand. It assumes that your string has a null character ('\0') right behind the last useful character (which is good practice).
Also, you might want to check that the string is actually a valid string with only digits (e.g '0', '1', '2', etc.). You could do this by including an if... else.. statement while looping through the string.
In modern kernels you want to use kstrto*:
http://lxr.free-electrons.com/source/include/linux/kernel.h#L274
274 /**
275 * kstrtoul - convert a string to an unsigned long
276 * #s: The start of the string. The string must be null-terminated, and may also
277 * include a single newline before its terminating null. The first character
278 * may also be a plus sign, but not a minus sign.
279 * #base: The number base to use. The maximum supported base is 16. If base is
280 * given as 0, then the base of the string is automatically detected with the
281 * conventional semantics - If it begins with 0x the number will be parsed as a
282 * hexadecimal (case insensitive), if it otherwise begins with 0, it will be
283 * parsed as an octal number. Otherwise it will be parsed as a decimal.
284 * #res: Where to write the result of the conversion on success.
285 *
286 * Returns 0 on success, -ERANGE on overflow and -EINVAL on parsing error.
287 * Used as a replacement for the obsolete simple_strtoull. Return code must
288 * be checked.
289 */
This function skips leading and trailing whitespace, handles one optional + / - sign, and returns 0 on invalid input,
// Convert standard null-terminated string to an integer
// - Skips leading whitespaces.
// - Skips trailing whitespaces.
// - Allows for one, optional +/- sign at the front.
// - Returns zero if any non-+/-, non-numeric, non-space character is encountered.
// - Returns zero if digits are separated by spaces (eg "123 45")
// - Range is checked against Overflow/Underflow (INT_MAX / INT_MIN), and returns 0.
int StrToInt(const char* s)
{
int minInt = 1 << (sizeof(int)*CHAR_BIT-1);
int maxInt = -(minInt+1);
char* w;
do { // Skip any leading whitespace
for(w=" \t\n\v\f\r"; *w && *s != *w; ++w) ;
if (*s == *w) ++s; else break;
} while(*s);
int sign = 1;
if ('-' == *s) sign = -1;
if ('+' == *s || '-' == *s) ++s;
long long i=0;
while('0' <= *s && *s <= '9')
{
i = 10*i + *s++ - '0';
if (sign*i < minInt || maxInt < sign*i)
{
i = 0;
break;
}
}
while (*s) // Skip any trailing whitespace
{
for(w=" \t\n\v\f\r"; *w && *s != *w; ++w) ;
if (*w && *s == *w) ++s; else break;
}
return (int)(!*s*sign*i);
}
" not able to include libraries" --> Unclear if code is allowed access to INT_MAX, INT_MIN. There is no way to determine the minimum/maximum signed integer in a completely portable fashion without using the language provided macros like INT_MAX, INT_MIN.
Use INT_MAX, INT_MIN is available. Else we could guess the char width is 8. We could guess there are no padding bits. We could guess that integers are 2's complement. With these reasonable assumptions, minimum and maximum are defined below.
Note: Shifting into the sign bit is undefined behavior (UB), so don't do that.
Let us add another restriction: make a solution that works for any signed integer from signed char to intmax_t. This disallows code from using a wider type, as there may not be a wider type.
typedef int Austin_int;
#define Austin_INT_MAXMID ( ((Austin_int)1) << (sizeof(Austin_int)*8 - 2) )
#define Austin_INT_MAX (Austin_INT_MAXMID - 1 + Austin_INT_MAXMID)
#define Austin_INT_MIN (-Austin_INT_MAX - 1)
int Austin_isspace(int ch) {
const char *ws = " \t\n\r\f\v";
while (*ws) {
if (*ws == ch) return 1;
ws++;
}
return 0;
}
// *endptr points to where parsing stopped
// *errorptr indicates overflow
Austin_int Austin_strtoi(const char *s, char **endptr, int *errorptr) {
int error = 0;
while (Austin_isspace(*s)) {
s++;
}
char sign = *s;
if (*s == '-' || *s == '+') {
s++;
}
Austin_int sum = 0;
while (*s >= '0' && *s <= '9') {
int ch = *s - '0';
if (sum <= Austin_INT_MIN / 10 &&
(sum < Austin_INT_MIN / 10 || -ch < Austin_INT_MIN % 10)) {
sum = Austin_INT_MIN;
error = 1;
} else {
sum = sum * 10 - ch;
}
s++;
}
if (sign != '-') {
if (sum < -Austin_INT_MAX) {
sum = Austin_INT_MAX;
error = 1;
} else {
sum = -sum;
}
}
if (endptr) {
*endptr = (char *) s;
}
if (errorptr) {
*errorptr = error;
}
return sum;
}
The above depends on C99 or later in the Austin_INT_MIN Austin_INT_MIN % 10 part.
This is the cleanest and safest way I could come up with
int str_to_int(const char * str, size_t n, int * int_value) {
int i;
int cvalue;
int value_muliplier = 1;
int res_value = 0;
int neg = 1; // -1 for negative and 1 for whole.
size_t str_len; // String length.
int end_at = 0; // Where loop should end.
if (str == NULL || int_value == NULL || n <= 0)
return -1;
// Get string length
str_len = strnlen(str, n);
if (str_len <= 0)
return -1;
// Is negative.
if (str[0] == '-') {
neg = -1;
end_at = 1; // If negative 0 item in 'str' is skipped.
}
// Do the math.
for (i = str_len - 1; i >= end_at; i--) {
cvalue = char_to_int(str[i]);
// Character not a number.
if (cvalue == -1)
return -1;
// Do the same math that is down below.
res_value += cvalue * value_muliplier;
value_muliplier *= 10;
}
/*
* "436"
* res_value = (6 * 1) + (3 * 10) + (4 * 100)
*/
*int_value = (res_value * neg);
return 0;
}
int char_to_int(char c) {
int cvalue = (int)c;
// Not a number.
// 48 to 57 is 0 to 9 in ascii.
if (cvalue < 48 || cvalue > 57)
return -1;
return cvalue - 48; // 48 is the value of zero in ascii.
}
I have a question regarding how the printf() method prints integers, signed or unsigned. One day, I found myself thinking about how difficult it must be to convert a binary sequence into a sequence of decimal digits that a human can understand, given that a computer has no concept of decimal.
Below, I have a printf() method (from here) with its associated methods. I've tried to understand as much as I can about how printi() works, as you can see in the comments:
#define PAD_RIGHT 1
#define PAD_ZERO 2
#include <stdarg.h>
static void printchar(char **str, int c)
{
extern int putchar(int c);
if (str) {
**str = c;
++(*str);
}
else (void)putchar(c);
}
static int prints(char **out, const char *string, int width, int pad)
{
register int pc = 0, padchar = ' ';
if (width > 0) {
register int len = 0;
register const char *ptr;
for (ptr = string; *ptr; ++ptr) ++len;
if (len >= width) width = 0;
else width -= len;
if (pad & PAD_ZERO) padchar = '0';
}
if (!(pad & PAD_RIGHT)) {
for ( ; width > 0; --width) {
printchar (out, padchar);
++pc;
}
}
for ( ; *string ; ++string) {
printchar (out, *string);
++pc;
}
for ( ; width > 0; --width) {
printchar (out, padchar);
++pc;
}
return pc;
}
/* the following should be enough for 32 bit int */
#define PRINT_BUF_LEN 12
static int printi(char **out, int i, int b, int sg, int width, int pad, int letbase)
{
/*
i is the number we are turning into a string
b is the base, i.e. base 10 for decimal
sg is if the number is signed, i.e. 1 for signed (%d), 0 for unsigned (%u)
By default, width and pad are 0, letbase is 97
*/
char print_buf[PRINT_BUF_LEN];
register char *s;
register int t, neg = 0, pc = 0;
register unsigned int u = i;
if (i == 0)
{
print_buf[0] = '0';
print_buf[1] = '\0';
return prints(out, print_buf, width, pad);
}
if (sg && b == 10 && i < 0)
{
neg = 1;
u = -i;
}
s = print_buf + PRINT_BUF_LEN - 1;
*s = '\0';
while (u)
{
t = u % b;
if (t >= 10)
t += letbase - '0' - 10;
*--s = t + '0';
u /= b;
}
if (neg)
{
if (width && (pad & PAD_ZERO))
{
printchar(out, '-');
++pc;
--width;
}
else
*--s = '-';
}
return pc + prints(out, s, width, pad);
}
static int print(char** out, const char* format, va_list args)
{
register int width, pad;
register int pc = 0;
char scr[2];
for (; *format != 0; ++format)
{
if (*format == '%')
{
++format;
width = pad = 0;
if (*format == '\0')
break;
if (*format == '%')
goto out;
if (*format == '-')
{
++format;
pad = PAD_RIGHT;
}
while (*format == '0')
{
++format;
pad |= PAD_ZERO;
}
for (; *format >= '0' && *format <= '9'; ++format)
{
width *= 10;
width += *format - '0';
}
if (*format == 's')
{
register char* s = (char*) va_arg(args, int);
pc += prints(out, s ? s : "(null)", width, pad);
continue;
}
if (*format == 'd')
{
pc += printi(out, va_arg(args, int), 10, 1, width, pad, 'a');
continue;
}
if (*format == 'x')
{
pc += printi(out, va_arg(args, int), 16, 0, width, pad, 'a');
continue;
}
if (*format == 'X')
{
pc += printi(out, va_arg(args, int), 16, 0, width, pad, 'A');
continue;
}
if (*format == 'u')
{
pc += printi(out, va_arg(args, int), 10, 0, width, pad, 'a');
continue;
}
if (*format == 'c')
{
/* char are converted to int then pushed on the stack */
scr[0] = (char) va_arg(args, int);
scr[1] = '\0';
pc += prints(out, scr, width, pad);
continue;
}
}
else
{
out:
printchar (out, *format);
++pc;
}
}
if (out)
**out = '\0';
va_end(args);
return pc;
}
int printf(const char *format, ...)
{
va_list args;
va_start( args, format );
return print( 0, format, args );
}
If there's one thing I hate about reading library source code, it's that it's hardly ever readable. Variable names with one character and no comment to explain them are a pain.
Can you please explain, in a simple way, what exactly the method is doing to convert an integer into a string of decimal digits?
The code you've pasted is not difficult to read. I suspect you may have given up early.
Ignoring the potential for a negative number for a moment, this printi() routine:
creates a buffer to print the number into, 12 characters wide
sets a character pointer s to point to the end of that buffer
** NULL-terminates it, then moves the pointer one character to the "left"
Then the routine enters a loop, for as long as the number remains > 0
MOD by 10 (that is, divide by 10 and take the remainder)
this becomes the digit that s is pointing to, so the ASCII representation is put there
s is moved to the left again
set the number to itself / 10; this removes the digit that was just printed
repeat the loop as long as there are more digits to print
The only tricky thing here is the dealing with negative numbers, but if you understand how negative numbers are stored, it's not really tricky at all.
Maybe I've been staring at template library headers too long, but that library code looks pretty readable to me!
I will explain the main loop, since the rest (juggling the sign around etc.) should be fairly easy to figure out.
while (u)
{
t = u % b;
if (t >= 10)
t += letbase - '0' - 10;
*--s = t + '0';
u /= b;
}
Basically what we are doing is extracting digits one at a time, from right to left. Suppose b == 10 (i.e. the usual case of %d or %u). The % operator, called the modulo operator, calculates the remainder that is left after integer division. The first time the line t = u % b; runs, it calculates the rightmost digit of the output string -- what is left as a remainder after you divide the number u by 10. (Suppose the number u was 493: the remainder after dividing this by 10 is 3, the rightmost digit.)
After extracting this rightmost digit into t, the if statement decides what to "call" this digit if it is 10 or larger. This fixup amounts to adjusting t so that, when '0' (the ASCII value of the digit '0', which is 48) is added in the next line the result will be a letter starting at 'a' or 'A' (to produce hex digits and other digits for bases larger than 10).
The line after that writes the digit into the buffer. It goes into the rightmost character of the print_buf buffer (notice how s is earlier initialised to point to the end of this buffer, not the start as is usually the case). The pointer s is subsequently moved one character to the left in preparation for the next character.
The following line, u /= b, simply divides u by 10, effectively discarding the rightmost digit. (This works because integer division never produces fractions, and always rounds down.) This then opens up the second-rightmost digit for the next loop iteration to process. Rinse, repeat. The loop finally stops when there is nothing left (the condition while (u) is equivalent to the condition while (u != 0)).
The method to convert a positive integer I to base 10 is basically :
if (i == 0)
printf("0");
else while (i != 0) {
unsigned int j = i / 10;
unsigned int digit = i - 10 * j;
printf("%c", digit + '0');
i = j;
}
Except that this prints out the number backward.
I have an integer like 1191223
and I want to iterate over the digits. I am not sure how to do this in C, is there any easy way to do this?
Thanks.
Forwards, or backwards?
Assuming a positive integer:
unsigned int n = 1191223;
while (n != 0) {
doSomething (n % 10);
n /= 10;
}
…will work smallest to largest, or…
EDIT I'd forgotten all about this non-working solution I had here. Note that Very Smart People™ seem to use the smallest-to-largest iteration consistently (both Linux kernel and GLibC's printf, for example, just iterate backwards) but here's a lousy way to do it if you really don't want to use snprintf for some reason…
int left_to_right (unsigned int n) {
unsigned int digit = 0;
if (0 == n) {
doSomething (0);
} else {
digit = pow(10, 1.0+ floor(log10(n)));
while (digit /= 10) {
doSomething ( (n / digit) % 10 );
}
}
}
I assume that it's very silly to assume that you have log10 and pow but not snprintf, so an alternate plan would be
int left_to_right_fixed_max (unsigned int n) {
unsigned int digit = 1000000000; /* make this very big */
unsigned int n10 = 10 * n;
if (0 == n) {
doSomething (0);
} else {
while (digit > n10) { digit /= 10; }
while (digit /= 10) {
doSomething ( (n / digit) % 10 );
}
}
}
… or, if you really don't have hardware multiply/divide, you can resort to using a table of powers of ten.
int left_to_right (unsigned int n) {
static const unsigned int digit [] =
{ 1,
10,
100,
1000,
10000,
100000,
1000000,
10000000,
100000000,
1000000000 /* make this very big */
};
static const unsigned char max_place = 10;
/* length of the above array */
unsigned char decimal;
unsigned char place;
unsigned char significant = 0; /* boolean */
if (0 == n) {
doSomething (0);
} else {
place = max_place;
while (place--) {
decimal = 0;
while (n >= digit[place]) {
decimal++;
n -= digit[place];
}
if (decimal | significant) {
doSomething (decimal);
significant |= decimal;
}
}
}
}
…which I have adapted from http://www.piclist.com/techref/language/ccpp/convertbase.htm into a somewhat more general-purpose version.
In the following I assume you mean decimal digits (base 10). Probably you are able to adapt the solutions to other numeral systems by substituting the 10s.
Note, that the modulo operation is a tricky thing concerning negative operands. Therefore I have chosen the data type to be an unsigned integer.
If you want to process the least significant digit first, you could try the following untested approach:
uint32_t n = 1191223;
do {
uint32_t digit = n%10;
// do something with digit
}
while (n/=10);
If you prefer to walk through the digits starting from the most significant digit, you could try to adapt the following untested code:
uint32_t n = 1191223;
#define MAX_DIGITS 10 // log10((double)UINT32_MAX)+1
uint32_t div = pow(10, MAX_DIGITS);
// skip the leading zero digits
while ( div && !(n/div) ) div/=10;
if ( !div ) div = 10; // allow n being zero
do {
uint32_t digit = (n/div)%10;
// do something with digit
}
while (div/=10);
You want to iterate over base-10 digits, but an integer has no concept of arabic notation and digits. Convert it to a string first:
int i = 1191223;
char buffer[16];
char *j;
snprintf(buffer, 16, "%i", i);
for ( j = buffer; *j; ++j ) { /* digit is in *j - '0' */ }
You can use sprintf() to convert it into a char array, and then iterate through that, like so (untested, just to get you started):
int a = 1191223;
char arr[16];
int rc = sprintf(arr, "%d", a);
if (rc < 0) {
// error
}
for (int i = 0; i < rc; i++) {
printf("digit %d = %d\n", i, arr[i]);
}
void access_digits(int n)
{
int digit;
if (n < 0) n = -n;
do {
digit = n % 10;
/* Here you can do whatever you
want to do with the digit */
} while ((n/=10) > 0);
}
Something like this:
char data[128];
int digits = 1191223;
sprintf(data, "%d", digits);
int length = strlen(data);
for(int i = 0; i < length; i++) {
// iterate through each character representing a digit
}
Notice that if you use an octal number like 0100 you also need to change the sprintf(data, "%d", digits); to sprintf(data, "%o", digits);.
For my purposes the following short code did the trick.
Having a an integer variable the_integer, and an integer variable sum_of_digits initialized. (line 1) You could do the following:
1) Convert the integer variable to a variable of type string with use of the std::to_string(int) function.
2) Iterate of the characters of the resulting string. for(char& c: str::to_string(the_integer))
3) To convert the characters back to integers use c -'0' . For this solution take a look at the discussion in (Convert char to int in C and C++).
4) .. and adding them the digits together: sum_of_digits += c-'0'
*) you can then print your variables: lines 3 and 4.
int the_integer = 123456789; int sum_of_digits;
for (char& c: std::to_string(the_integer)) {sum_of_digits += c-'0';}
std::cout << "Integer: " << the_integer << std::endl;
std::cout << "Sum of Digits << sum_of_digits << std::endl;
Note that std::to_string() has some notes, please consult the c++ references to see if the code is still relevant for your purposes.
A hackish way is to convert this to string (see strtol) and then reconvert this to a number.
you could use something like character you want - '0'
Off the top of my head: "i % 100000", "i % 100000", ...
A recursive solution would let you start from "i%10".
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Is there a printf converter to print in binary format?
Still learning C and I was wondering:
Given a number, is it possible to do something like the following?
char a = 5;
printf("binary representation of a = %b",a);
> 101
Or would i have to write my own method to do the transformation to binary?
There is no direct way (i.e. using printf or another standard library function) to print it. You will have to write your own function.
/* This code has an obvious bug and another non-obvious one :) */
void printbits(unsigned char v) {
for (; v; v >>= 1) putchar('0' + (v & 1));
}
If you're using terminal, you can use control codes to print out bytes in natural order:
void printbits(unsigned char v) {
printf("%*s", (int)ceil(log2(v)) + 1, "");
for (; v; v >>= 1) printf("\x1b[2D%c",'0' + (v & 1));
}
Based on dirkgently's answer, but fixing his two bugs, and always printing a fixed number of digits:
void printbits(unsigned char v) {
int i; // for C89 compatability
for(i = 7; i >= 0; i--) putchar('0' + ((v >> i) & 1));
}
Yes (write your own), something like the following complete function.
#include <stdio.h> /* only needed for the printf() in main(). */
#include <string.h>
/* Create a string of binary digits based on the input value.
Input:
val: value to convert.
buff: buffer to write to must be >= sz+1 chars.
sz: size of buffer.
Returns address of string or NULL if not enough space provided.
*/
static char *binrep (unsigned int val, char *buff, int sz) {
char *pbuff = buff;
/* Must be able to store one character at least. */
if (sz < 1) return NULL;
/* Special case for zero to ensure some output. */
if (val == 0) {
*pbuff++ = '0';
*pbuff = '\0';
return buff;
}
/* Work from the end of the buffer back. */
pbuff += sz;
*pbuff-- = '\0';
/* For each bit (going backwards) store character. */
while (val != 0) {
if (sz-- == 0) return NULL;
*pbuff-- = ((val & 1) == 1) ? '1' : '0';
/* Get next bit. */
val >>= 1;
}
return pbuff+1;
}
Add this main to the end of it to see it in operation:
#define SZ 32
int main(int argc, char *argv[]) {
int i;
int n;
char buff[SZ+1];
/* Process all arguments, outputting their binary. */
for (i = 1; i < argc; i++) {
n = atoi (argv[i]);
printf("[%3d] %9d -> %s (from '%s')\n", i, n,
binrep(n,buff,SZ), argv[i]);
}
return 0;
}
Run it with "progname 0 7 12 52 123" to get:
[ 1] 0 -> 0 (from '0')
[ 2] 7 -> 111 (from '7')
[ 3] 12 -> 1100 (from '12')
[ 4] 52 -> 110100 (from '52')
[ 5] 123 -> 1111011 (from '123')
#include<iostream>
#include<conio.h>
#include<stdlib.h>
using namespace std;
void displayBinary(int n)
{
char bistr[1000];
itoa(n,bistr,2); //2 means binary u can convert n upto base 36
printf("%s",bistr);
}
int main()
{
int n;
cin>>n;
displayBinary(n);
getch();
return 0;
}
Use a lookup table, like:
char *table[16] = {"0000", "0001", .... "1111"};
then print each nibble like this
printf("%s%s", table[a / 0x10], table[a % 0x10]);
Surely you can use just one table, but it will be marginally faster and too big.
There is no direct format specifier for this in the C language. Although I wrote this quick python snippet to help you understand the process step by step to roll your own.
#!/usr/bin/python
dec = input("Enter a decimal number to convert: ")
base = 2
solution = ""
while dec >= base:
solution = str(dec%base) + solution
dec = dec/base
if dec > 0:
solution = str(dec) + solution
print solution
Explained:
dec = input("Enter a decimal number to convert: ") - prompt the user for numerical input (there are multiple ways to do this in C via scanf for example)
base = 2 - specify our base is 2 (binary)
solution = "" - create an empty string in which we will concatenate our solution
while dec >= base: - while our number is bigger than the base entered
solution = str(dec%base) + solution - get the modulus of the number to the base, and add it to the beginning of our string (we must add numbers right to left using division and remainder method). the str() function converts the result of the operation to a string. You cannot concatenate integers with strings in python without a type conversion.
dec = dec/base - divide the decimal number by the base in preperation to take the next modulo
if dec > 0:
solution = str(dec) + solution - if anything is left over, add it to the beginning (this will be 1, if anything)
print solution - print the final number
This code should handle your needs up to 64 bits.
char* pBinFill(long int x,char *so, char fillChar); // version with fill
char* pBin(long int x, char *so); // version without fill
#define width 64
char* pBin(long int x,char *so)
{
char s[width+1];
int i=width;
s[i--]=0x00; // terminate string
do
{ // fill in array from right to left
s[i--]=(x & 1) ? '1':'0'; // determine bit
x>>=1; // shift right 1 bit
} while( x > 0);
i++; // point to last valid character
sprintf(so,"%s",s+i); // stick it in the temp string string
return so;
}
char* pBinFill(long int x,char *so, char fillChar)
{ // fill in array from right to left
char s[width+1];
int i=width;
s[i--]=0x00; // terminate string
do
{
s[i--]=(x & 1) ? '1':'0';
x>>=1; // shift right 1 bit
} while( x > 0);
while(i>=0) s[i--]=fillChar; // fill with fillChar
sprintf(so,"%s",s);
return so;
}
void test()
{
char so[width+1]; // working buffer for pBin
long int val=1;
do
{
printf("%ld =\t\t%#lx =\t\t0b%s\n",val,val,pBinFill(val,so,0));
val*=11; // generate test data
} while (val < 100000000);
}
Output:
00000001 = 0x000001 = 0b00000000000000000000000000000001
00000011 = 0x00000b = 0b00000000000000000000000000001011
00000121 = 0x000079 = 0b00000000000000000000000001111001
00001331 = 0x000533 = 0b00000000000000000000010100110011
00014641 = 0x003931 = 0b00000000000000000011100100110001
00161051 = 0x02751b = 0b00000000000000100111010100011011
01771561 = 0x1b0829 = 0b00000000000110110000100000101001
19487171 = 0x12959c3 = 0b00000001001010010101100111000011
You have to write your own transformation. Only decimal, hex and octal numbers are supported with format specifiers.