sprintf replacement for embedded systems [duplicate] - c

I'm coding for a microcontroller-based application and I need to convert a float to a character string, but I do not need the heavy overhead associated with sprintf(). Is there any eloquent way to do this? I don't need too much. I only need 2 digits of precision.

Here's a version optimized for embedded systems that doesn't require any stdio or memset, and has low memory footprint. You're responsible for passing a char buffer initialized with zeros (with pointer p) where you want to store your string, and defining CHAR_BUFF_SIZE when you make said buffer (so the returned string will be null terminated).
static char * _float_to_char(float x, char *p) {
char *s = p + CHAR_BUFF_SIZE; // go to end of buffer
uint16_t decimals; // variable to store the decimals
int units; // variable to store the units (part to left of decimal place)
if (x < 0) { // take care of negative numbers
decimals = (int)(x * -100) % 100; // make 1000 for 3 decimals etc.
units = (int)(-1 * x);
} else { // positive numbers
decimals = (int)(x * 100) % 100;
units = (int)x;
}
*--s = (decimals % 10) + '0';
decimals /= 10; // repeat for as many decimal places as you need
*--s = (decimals % 10) + '0';
*--s = '.';
while (units > 0) {
*--s = (units % 10) + '0';
units /= 10;
}
if (x < 0) *--s = '-'; // unary minus sign for negative numbers
return s;
}
Tested on ARM Cortex M0 & M4. Rounds correctly.

Try this. It should be nice and small. I've output the string directly - doing a printf, rather than a sprintf. I'll leave it to you to allocate space for the return string, as well as copying the result into it.
// prints a number with 2 digits following the decimal place
// creates the string backwards, before printing it character-by-character from
// the end to the start
//
// Usage: myPrintf(270.458)
// Output: 270.45
void myPrintf(float fVal)
{
char result[100];
int dVal, dec, i;
fVal += 0.005; // added after a comment from Matt McNabb, see below.
dVal = fVal;
dec = (int)(fVal * 100) % 100;
memset(result, 0, 100);
result[0] = (dec % 10) + '0';
result[1] = (dec / 10) + '0';
result[2] = '.';
i = 3;
while (dVal > 0)
{
result[i] = (dVal % 10) + '0';
dVal /= 10;
i++;
}
for (i=strlen(result)-1; i>=0; i--)
putc(result[i], stdout);
}

// convert float to string one decimal digit at a time
// assumes float is < 65536 and ARRAYSIZE is big enough
// problem: it truncates numbers at size without rounding
// str is a char array to hold the result, float is the number to convert
// size is the number of decimal digits you want
void FloatToStringNew(char *str, float f, char size)
{
char pos; // position in string
char len; // length of decimal part of result
char* curr; // temp holder for next digit
int value; // decimal digit(s) to convert
pos = 0; // initialize pos, just to be sure
value = (int)f; // truncate the floating point number
itoa(value,str); // this is kinda dangerous depending on the length of str
// now str array has the digits before the decimal
if (f < 0 ) // handle negative numbers
{
f *= -1;
value *= -1;
}
len = strlen(str); // find out how big the integer part was
pos = len; // position the pointer to the end of the integer part
str[pos++] = '.'; // add decimal point to string
while(pos < (size + len + 1) ) // process remaining digits
{
f = f - (float)value; // hack off the whole part of the number
f *= 10; // move next digit over
value = (int)f; // get next digit
itoa(value, curr); // convert digit to string
str[pos++] = *curr; // add digit to result string and increment pointer
}
}

While you guys were answering I've come up with my own solution which that works better for my application and I figure I'd share. It doesn't convert the float to a string, but rather 8-bit integers. My range of numbers is very small (0-15) and always non-negative, so this will allow me to send the data over bluetooth to my android app.
//Assumes bytes* is at least 2-bytes long
void floatToBytes(byte_t* bytes, float flt)
{
bytes[1] = (byte_t) flt; //truncate whole numbers
flt = (flt - bytes[1])*100; //remove whole part of flt and shift 2 places over
bytes[0] = (byte_t) flt; //truncate the fractional part from the new "whole" part
}
//Example: 144.2345 -> bytes[1] = 144; -> bytes[0] = 23

I can't comment on enhzflep's response, but to handle negative numbers correctly (which the current version does not), you only need to add
if (fVal < 0) {
putc('-', stdout);
fVal = -fVal;
}
at the beginning of the function.

Its a Liitle large method, but It would work for both int and float, decimalPoint parameter is passed with zero value for Integer, Please let me know if you have smaller function than this.
void floatToStr(uint8_t *out, float x,int decimalPoint)
{
uint16_t absval = fabs(x);
uint16_t absvalcopy = absval;
int decimalcount = 0;
while(absvalcopy != 0)
{
absvalcopy /= 10;
decimalcount ++;
}
uint8_t *absbuffer = malloc(sizeof(uint8_t) * (decimalcount + decimalPoint + 1));
int absbufferindex = 0;
absvalcopy = absval;
uint8_t temp;
int i = 0;
for(i = decimalcount; i > 0; i--)
{
uint16_t frst1 = fabs((absvalcopy / pow(10.0, i-1)));
temp = (frst1 % 10) + 0x30;
*(absbuffer + absbufferindex) = temp;
absbufferindex++;
}
if(decimalPoint > 0)
{
*(absbuffer + absbufferindex) = '.';
absbufferindex ++;
//------------------- Decimal Extractor ---------------------//
for(i = 1; i < decimalPoint + 1; i++)
{
uint32_t valueFloat = (x - (float)absval)*pow(10,i);
*(absbuffer + absbufferindex) = ((valueFloat) % 10) + 0x30;
absbufferindex++;
}
}
for(i=0; i< (decimalcount + decimalPoint + 1); i++)
{
*(out + i) = *(absbuffer + i);
}
i=0;
if(decimalPoint > 0)
i = 1;
*(out + decimalcount + decimalPoint + i) = 0;
}

Related

convert a decimal number to string in c

I need to redo printf for a projet, so I actually have a problem with the conversion of float.
I managed to convert almost everything but for the number 1254451555.6
I got an issue: I got 1254451555.59999.
I think it's the calculation to keep the part after the . that doesnt work.
nbr = ((n - nbr) * 100000000);
I tried different things but I haven't managed to fix it yet.
Do you have any idea?
int getlenghtitoa(long long n, int nbase)
{
int i;
i = 0;
while (n >= 0)
{
n /= nbase;
i++;
if (n == 0)
break ;
}
return (i);
}
float ft_nbconv(float n, int i)
{
while (i-- > 0)
n = n *10;
return (n);
}
int ft_power(long long nbr)
{
int i;
i = 1;
while(nbr > 10)
{
i *= 10;
nbr = nbr / 10;
}
return (i);
}
char *ft_conver_f(long double n)
{
char *dest;
int i;
int a;
long long int nbr;
int power;
nbr = (long long) n;
i = getlenghtitoa((long long )n, 10);
if (!(dest = malloc(sizeof(char) * (i + 8))))
return (0);
a = i;
i = 0;
power = ft_power(nbr);
while (a--)
{
dest[i++] = ((nbr / power) % 10) + '0';
if (power != 1)
power /= 10;
}
dest[i++] = '.';
nbr = ((n - nbr) * 100000000);
power = 10000000;
while (a++ < 5)
{
if (a == 5)
if ((((nbr / power)) % 10) >= 5)
{
dest[i++] = ((nbr / power) % 10 + 1) + '0';
break;
}
dest[i++] = ((nbr / power) % 10) + '0';
power /= 10;
}
dest[i] = '\0';
return (dest);
}
Most decimal fractions cannot be represented exactly as binary fractions. A consequence is that, in general, the decimal floating-point numbers you enter are only approximated by the binary floating-point numbers actually stored in the machine.
That's why when implementing a printf, the only way to really be able to convert a floating number to a 2-seperated-by-point integers, is by using the precision factor and rounding manually.
If you are not required to implement the precision, the default is 6.
(Precision is the number of places to print after the dot (and it's rounded)).
And that's what's missing in your implementation.
Let's call the digits before the dot the ipart and the digits after the fpart .
nbr = ((n - nbr) * 100000000);
This should be
nbr = ((n - nbr) * 10000000); // 7 zeros
// nbr is now equal to 5999999
if (nbr % 10 >= 5)
{
nbr = nbr / 10 + 1;
}
else
nbr = nbr / 10;
This way, you get 7 digits after the dot, see if the last one is higher than 5, if it is, you add +1 to nbr (after dividing by 10 to make sure nbr has 6 digits), if it's not, you just divide by 10.
One more note about this rounding method, It will not be able to carry the rounding from the fpart to the ipart .
what if you want to print 3.9999999 ? It should print 4.000000. That means that can't just convert the ipart to a string from the beginning, because sometimes rounding the fpart will add +1 to your ipart
So think about creating a function ltoa for example that takes a long long int and converts it to a string, complete the piece of code about rounding i just gave you to make sure rounding can be carried to the ipart , then convert the whole thing to string using something like
dest = join(ltoa(ipart), ".", ltoa(fpart)).
A couple more notes, your function does not handle negative numbers.
And your int ft_pow can be easily flooded, so consider changing to long long ft_pow

Convert float to string without sprintf()

I'm coding for a microcontroller-based application and I need to convert a float to a character string, but I do not need the heavy overhead associated with sprintf(). Is there any eloquent way to do this? I don't need too much. I only need 2 digits of precision.
Here's a version optimized for embedded systems that doesn't require any stdio or memset, and has low memory footprint. You're responsible for passing a char buffer initialized with zeros (with pointer p) where you want to store your string, and defining CHAR_BUFF_SIZE when you make said buffer (so the returned string will be null terminated).
static char * _float_to_char(float x, char *p) {
char *s = p + CHAR_BUFF_SIZE; // go to end of buffer
uint16_t decimals; // variable to store the decimals
int units; // variable to store the units (part to left of decimal place)
if (x < 0) { // take care of negative numbers
decimals = (int)(x * -100) % 100; // make 1000 for 3 decimals etc.
units = (int)(-1 * x);
} else { // positive numbers
decimals = (int)(x * 100) % 100;
units = (int)x;
}
*--s = (decimals % 10) + '0';
decimals /= 10; // repeat for as many decimal places as you need
*--s = (decimals % 10) + '0';
*--s = '.';
while (units > 0) {
*--s = (units % 10) + '0';
units /= 10;
}
if (x < 0) *--s = '-'; // unary minus sign for negative numbers
return s;
}
Tested on ARM Cortex M0 & M4. Rounds correctly.
Try this. It should be nice and small. I've output the string directly - doing a printf, rather than a sprintf. I'll leave it to you to allocate space for the return string, as well as copying the result into it.
// prints a number with 2 digits following the decimal place
// creates the string backwards, before printing it character-by-character from
// the end to the start
//
// Usage: myPrintf(270.458)
// Output: 270.45
void myPrintf(float fVal)
{
char result[100];
int dVal, dec, i;
fVal += 0.005; // added after a comment from Matt McNabb, see below.
dVal = fVal;
dec = (int)(fVal * 100) % 100;
memset(result, 0, 100);
result[0] = (dec % 10) + '0';
result[1] = (dec / 10) + '0';
result[2] = '.';
i = 3;
while (dVal > 0)
{
result[i] = (dVal % 10) + '0';
dVal /= 10;
i++;
}
for (i=strlen(result)-1; i>=0; i--)
putc(result[i], stdout);
}
// convert float to string one decimal digit at a time
// assumes float is < 65536 and ARRAYSIZE is big enough
// problem: it truncates numbers at size without rounding
// str is a char array to hold the result, float is the number to convert
// size is the number of decimal digits you want
void FloatToStringNew(char *str, float f, char size)
{
char pos; // position in string
char len; // length of decimal part of result
char* curr; // temp holder for next digit
int value; // decimal digit(s) to convert
pos = 0; // initialize pos, just to be sure
value = (int)f; // truncate the floating point number
itoa(value,str); // this is kinda dangerous depending on the length of str
// now str array has the digits before the decimal
if (f < 0 ) // handle negative numbers
{
f *= -1;
value *= -1;
}
len = strlen(str); // find out how big the integer part was
pos = len; // position the pointer to the end of the integer part
str[pos++] = '.'; // add decimal point to string
while(pos < (size + len + 1) ) // process remaining digits
{
f = f - (float)value; // hack off the whole part of the number
f *= 10; // move next digit over
value = (int)f; // get next digit
itoa(value, curr); // convert digit to string
str[pos++] = *curr; // add digit to result string and increment pointer
}
}
While you guys were answering I've come up with my own solution which that works better for my application and I figure I'd share. It doesn't convert the float to a string, but rather 8-bit integers. My range of numbers is very small (0-15) and always non-negative, so this will allow me to send the data over bluetooth to my android app.
//Assumes bytes* is at least 2-bytes long
void floatToBytes(byte_t* bytes, float flt)
{
bytes[1] = (byte_t) flt; //truncate whole numbers
flt = (flt - bytes[1])*100; //remove whole part of flt and shift 2 places over
bytes[0] = (byte_t) flt; //truncate the fractional part from the new "whole" part
}
//Example: 144.2345 -> bytes[1] = 144; -> bytes[0] = 23
I can't comment on enhzflep's response, but to handle negative numbers correctly (which the current version does not), you only need to add
if (fVal < 0) {
putc('-', stdout);
fVal = -fVal;
}
at the beginning of the function.
Its a Liitle large method, but It would work for both int and float, decimalPoint parameter is passed with zero value for Integer, Please let me know if you have smaller function than this.
void floatToStr(uint8_t *out, float x,int decimalPoint)
{
uint16_t absval = fabs(x);
uint16_t absvalcopy = absval;
int decimalcount = 0;
while(absvalcopy != 0)
{
absvalcopy /= 10;
decimalcount ++;
}
uint8_t *absbuffer = malloc(sizeof(uint8_t) * (decimalcount + decimalPoint + 1));
int absbufferindex = 0;
absvalcopy = absval;
uint8_t temp;
int i = 0;
for(i = decimalcount; i > 0; i--)
{
uint16_t frst1 = fabs((absvalcopy / pow(10.0, i-1)));
temp = (frst1 % 10) + 0x30;
*(absbuffer + absbufferindex) = temp;
absbufferindex++;
}
if(decimalPoint > 0)
{
*(absbuffer + absbufferindex) = '.';
absbufferindex ++;
//------------------- Decimal Extractor ---------------------//
for(i = 1; i < decimalPoint + 1; i++)
{
uint32_t valueFloat = (x - (float)absval)*pow(10,i);
*(absbuffer + absbufferindex) = ((valueFloat) % 10) + 0x30;
absbufferindex++;
}
}
for(i=0; i< (decimalcount + decimalPoint + 1); i++)
{
*(out + i) = *(absbuffer + i);
}
i=0;
if(decimalPoint > 0)
i = 1;
*(out + decimalcount + decimalPoint + i) = 0;
}

Any decent way of printing out floats and doubles with commas?

I'm working on a program that regards with currency. Ive been finding a solution to display money values decently like this:
9,999.99 USD
Remember when assigning a certain variable with a value (money), you musn't insert commas.
I.e.:
double money=9999.99;
And when accessing it;
printf("%.2l USD",money);
Which will output:
9999.99 USD
This is not what I want, especially on bigger amounts exceeding the hundredth, thousandth, millionth, or even billionth place value.
Now I can't find any solution than printing out the desired output directly on the printf.
printf("9,999.99");
Which is undesirable with many variables.
Can anyone help me out?
Please take a look and printf manual page taking note of the following bit:
*"For some numeric conversions a radix character ("decimal point") or thousands' grouping character is used. The actual character used depends on the LC_NUMERIC part of the locale. The POSIX locale uses '.' as radix character, and does not have a grouping character. Thus,
printf("%'.2f", 1234567.89);
results in "1234567.89" in the POSIX locale, in "1234567,89" in the nl_NL locale, and in "1.234.567,89" in the da_DK locale."*
This can be changed by the function setlocale
There is a function, strfmon which might be able to help you
First, don't use floating-point types to represent money because normally floating-point types are binary and as such cannot represent all decimal fractions (cents) exactly, further these types are prone to rounding errors. Use integers instead and count cents instead of dollars.
#include <stdio.h>
#include <limits.h>
unsigned long long ConstructMoney(unsigned long long dollars, unsigned cents)
{
return dollars * 100 + cents;
}
void PrintWithCommas(unsigned long long n)
{
char s[sizeof n * CHAR_BIT + 1];
char *p = s + sizeof s;
unsigned count = 0;
*--p = '\0';
do
{
*--p = '0' + n % 10;
n /= 10;
if (++count == 3 && n)
{
*--p = ',';
count = 0;
}
} while (n);
printf("%s", p);
}
void PrintMoney(unsigned long long n)
{
PrintWithCommas(n / 100);
putchar('.');
n %= 100;
putchar('0' + n / 10);
putchar('0' + n % 10);
}
int main(void)
{
PrintMoney(ConstructMoney(0, 0)); puts("");
PrintMoney(ConstructMoney(0, 1)); puts("");
PrintMoney(ConstructMoney(1, 0)); puts("");
PrintMoney(ConstructMoney(1, 23)); puts("");
PrintMoney(ConstructMoney(12, 34)); puts("");
PrintMoney(ConstructMoney(123, 45)); puts("");
PrintMoney(ConstructMoney(1234, 56)); puts("");
PrintMoney(ConstructMoney(12345, 67)); puts("");
PrintMoney(ConstructMoney(123456, 78)); puts("");
PrintMoney(ConstructMoney(1234567, 89)); puts("");
return 0;
}
Output (ideone):
0.00
0.01
1.00
1.23
12.34
123.45
1,234.56
12,345.67
123,456.78
1,234,567.89
If you're using the standard library, there's no way to do this -- you have to write some code that does it by hand.
I would recommend multiplying the value by 100, casting to integer, and printing the digits with separators as needed -- it's much easier to handle individual digits on an integer.
The following code, for instance, will fill a char * buffer with the string representation of the value you have:
void formatString (double number, char * buffer) {
if (number < 0) {
*buffer = '-';
formatString(number, buffer + 1);
return;
}
unsigned long long num = (unsigned long long) (number * 100);
unsigned long long x; // temporary storage for counting the digits
unsigned char digits;
for (x = num / 1000, digits = 1; x; digits ++, x /= 10);
// counts the digits, also ensures that there's at least one digit
unsigned char pos; // digit position
for (pos = 1, x = 100; pos < digits; pos ++, x *= 10);
// reuses x as a value for extracting the digit in the needed position;
char * current = buffer;
for (pos = digits; pos; pos --) {
*(current ++) = 48 + (num / x);
// remember 48 + digit gives the ASCII for the digit
if (((pos % 3) == 1) && (pos > 1)) *(current ++) = ',';
num %= x;
x /= 10;
}
*(current ++) = '.';
*(current ++) = 48 + num / 10;
*(current ++) = 48 + num % 10;
*current = 0;
}

storing more than 2 power 31 on a 32-bit system

I have to write a program that can calculate the powers of 2 power 2010 and to find the sum of the digits. eg:
if `2 power 12 => gives 4096 . So 4+0+9+6 = 19 .
Now i need to find the same for 2 power 2010.
Please help me to understand.
Here's something to get you started:
char buf[2010]; // 2^2010 < 10^2010 by a huge margin, so buffer size is safe
snprintf(buf, sizeof buf, "%.0Lf", 0x1p2010L);
You have to either use a library that supplies unlimited integer length types (see http://en.wikipedia.org/wiki/Bignum ), or implement a solution that does not need them (e.g. use a digit array and implement the power calculation on the array yourself, which in your case can be as simple as addition in a loop). Since this is homework, probably the latter.
Knowing 2^32, how would you calculate 2^33 with pen and paper?
2^32 is 4294967296
4294967296
* 2
----------
8589934592
8589934592 is 2^33; sum of digits is 8+5+8+9+...+9+2 (62)
Just be aware that 2^2011 is a number with more than 600 digits: not that many to do by computer
GMP is perhaps the best, fastest free multi-architecture library for this. It provides a solid foundation for such calculations, including not only addition, but parsing from strings, multiplication, division, scientific operations, etc.
For literature on the algorithms themselves, I highly recommend The Art of Computer Programming, Volume 2: Seminumerical Algorithms by Donald Knuth. This book is considered by many to be the best single reference for the topic. This book explains from the ground up how such arithmetic can take place on a machine that can only do 32-bit arithmetic.
If you want to implement this calculation from scratch without using any tools, the following code block requires requires only the following additional methods to be supplied:
unsigned int divModByTen(unsigned int *num, unsigned int length);
bool isZero(unsigned int *num, unsigned int length);
divModByTen should divide replace num in memory with the value of num / 10, and return the remainder. The implementation will take some effort, unless a library is used. isZero just checks if the number is all zero's in memory. Once we have these, we can use the following code sample:
unsigned int div10;
int decimalDigitSum;
unsigned int hugeNumber[64];
memset(twoPow2010, 0, sizeof(twoPow2010));
twoPow2010[63] = 0x4000000;
// at this point, twoPow2010 is 2^2010 encoded in binary stored in memory
decimalDigitSum = 0;
while (!izZero(hugeNumber, 64)) {
mod10 = divModByTen(&hugeNumber[0], 64);
decimalDigitSum += mod10;
}
printf("Digit Sum:%d", decimalDigitSum);
This takes only a few lines of code in Delphi... :)
So in c must be the same or shorter.
function PowerOf2(exp: integer): string;
var
n : integer;
Digit : integer;
begin
result := '1';
while exp <> 0 do
begin
Digit := 0;
for n := Length(result) downto 1 do
begin
Digit := (ord(result[n]) - ord('0')) * 2 + Digit div 10;
result[n] := char(Digit mod 10 + ord('0'))
end;
if Digit > 9 then
result := '1' + result;
dec(exp);
end;
end;
-----EDIT-----
This is 1-to-1 c# version.
string PowerOf2(int exp)
{
int n, digit;
StringBuilder result = new StringBuilder("1");
while (exp != 0)
{
digit = 0;
for (n = result.Length; n >= 1; n--)
{
digit = (result[n-1] - '0') * 2 + digit / 10;
result[n-1] = Convert.ToChar(digit % 10 + '0');
}
if (digit > 9)
{
result = new StringBuilder("1" + result.ToString());
}
exp--;
}
return result.ToString();
}
int Sum(string s)
{
int sum = 0;
for (int i = 0; i < s.Length; i++)
{
sum += s[i] - '0';
}
return sum;
}
for (int i = 1; i < 20; i++)
{
string s1s = PowerOf2(i);
int sum = Sum(s1s);
Console.WriteLine(s1s + " --> " + sum);
}
Here's how you can calculate and print 22010:
#include <stdio.h>
#include <string.h>
void AddNumbers(char* dst, const char* src)
{
char ddigit;
char carry = 0;
while ((ddigit = *dst) != '\0')
{
char sdigit = '0';
if (*src != '\0')
{
sdigit = *src++;
}
ddigit += sdigit - '0' + carry;
if (ddigit > '9')
{
ddigit -= 10;
carry = 1;
}
else
{
carry = 0;
}
*dst++ = ddigit;
}
}
void ReverseString(char* s)
{
size_t i, n = strlen(s);
for (i = 0; i < n / 2; i++)
{
char t = s[i];
s[i] = s[n - 1 - i];
s[n - 1 - i] = t;
}
}
int main(void)
{
char result[607], tmp[sizeof(result)];
int i;
memset (result, '0', sizeof(result));
result[0] = '1';
result[sizeof(result) - 1] = '\0';
for (i = 0; i < 2010; i++)
{
memcpy(tmp, result, sizeof(result));
AddNumbers(result, tmp);
}
ReverseString(result);
printf("%s\n", result);
return 0;
}
You can now sum up the individual digits.

Print large base 256 array in base 10 in c

I have an array of unsigned chars in c I am trying to print in base 10, and I am stuck. I think this will be better explained in code, so, given:
unsigned char n[3];
char[0] = 1;
char[1] = 2;
char[2] = 3;
I would like to print 197121.
This is trivial with small base 256 arrays. One can simply 1 * 256 ^ 0 + 2 * 256 ^ 1 + 3 * 256 ^ 2.
However, if my array was 100 bytes large, then this quickly becomes a problem. There is no integral type in C that is 100 bytes large, which is why I'm storing numbers in unsigned char arrays to begin with.
How am I supposed to efficiently print this number out in base 10?
I am a bit lost.
There's no easy way to do it using only the standard C library. You'll either have to write the function yourself (not recommended), or use an external library such as GMP.
For example, using GMP, you could do:
unsigned char n[100]; // number to print
mpz_t num;
mpz_import(num, 100, -1, 1, 0, 0, n); // convert byte array into GMP format
mpz_out_str(stdout, 10, num); // print num to stdout in base 10
mpz_clear(num); // free memory for num
When I saw this question, I purpose to solve it, but at that moment I was very busy.
This last weekend I've could gain some prize hours of free time so I considered my pending challenge.
First of all, I suggest you to considered above response. I never use GMP library but I'm sure that it's better solution than a handmade code.
Also, you could be interest to analyze code of bc calculator; it can works with big numbers and I used to test my own code.
Ok, if you are still interested in a code do it by yourself (only with support C language and Standard C library) may be I can give you something.
Before all, a little bit theory. In basic numeric theory (modular arithmetic level) theres is an algorithm that inspire me to arrive at one solution; Multiply and Power algorithm to solve a^N module m:
Result := 1;
for i := k until i = 0
if n_i = 1 then Result := (Result * a) mod m;
if i != 0 then Result := (Result * Result) mod m;
end for;
Where k is number of digits less one of N in binary representation, and n_i is i binary digit. For instance (N is exponent):
N = 44 -> 1 0 1 1 0 0
k = 5
n_5 = 1
n_4 = 0
n_3 = 1
n_2 = 1
n_1 = 0
n_0 = 0
When we make a module operation, as an integer division, we can lose part of the number, so we only have to modify algorithm to don't miss relevant data.
Here is my code (take care that it is an adhoc code, strong dependency of may computer arch. Basically I play with data length of C language so, be carefully because my data length could not be the same):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
enum { SHF = 31, BMASK = 0x1 << SHF, MODULE = 1000000000UL, LIMIT = 1024 };
unsigned int scaleBigNum(const unsigned short scale, const unsigned int lim, unsigned int *num);
unsigned int pow2BigNum(const unsigned int lim, unsigned int *nsrc, unsigned int *ndst);
unsigned int addBigNum(const unsigned int lim1, unsigned int *num1, const unsigned int lim2, unsigned int *num2);
unsigned int bigNum(const unsigned short int base, const unsigned int exp, unsigned int **num);
int main(void)
{
unsigned int *num, lim;
unsigned int *np, nplim;
int i, j;
for(i = 1; i < LIMIT; ++i)
{
lim = bigNum(i, i, &num);
printf("%i^%i == ", i, i);
for(j = lim - 1; j > -1; --j)
printf("%09u", num[j]);
printf("\n");
free(num);
}
return 0;
}
/*
bigNum: Compute number base^exp and store it in num array
#base: Base number
#exp: Exponent number
#num: Pointer to array where it stores big number
Return: Array length of result number
*/
unsigned int bigNum(const unsigned short int base, const unsigned int exp, unsigned int **num)
{
unsigned int m, lim, mem;
unsigned int *v, *w, *k;
//Note: mem has the exactly amount memory to allocate (dinamic memory version)
mem = ( (unsigned int) (exp * log10( (float) base ) / 9 ) ) + 3;
v = (unsigned int *) malloc( mem * sizeof(unsigned int) );
w = (unsigned int *) malloc( mem * sizeof(unsigned int) );
for(m = BMASK; ( (m & exp) == 0 ) && m; m >>= 1 ) ;
v[0] = (m) ? 1 : 0;
for(lim = 1; m > 1; m >>= 1)
{
if( exp & m )
lim = scaleBigNum(base, lim, v);
lim = pow2BigNum(lim, v, w);
k = v;
v = w;
w = k;
}
if(exp & 0x1)
lim = scaleBigNum(base, lim, v);
free(w);
*num = v;
return lim;
}
/*
scaleBigNum: Make an (num[] <- scale*num[]) big number operation
#scale: Scalar that multiply big number
#lim: Length of source big number
#num: Source big number (array of unsigned int). Update it with new big number value
Return: Array length of operation result
Warning: This method can write in an incorrect position if we don't previous reallocate num (if it's necessary). bigNum method do it for us
*/
unsigned int scaleBigNum(const unsigned short scale, const unsigned int lim, unsigned int *num)
{
unsigned int i;
unsigned long long int n, t;
for(n = 0, t = 0, i = 0; i < lim; ++i)
{
t = (n / MODULE);
n = ( (unsigned long long int) scale * num[i] );
num[i] = (n % MODULE) + t; // (n % MODULE) + t always will be smaller than MODULE
}
num[i] = (n / MODULE);
return ( (num[i]) ? lim + 1 : lim );
}
/*
pow2BigNum: Make a (dst[] <- src[] * src[]) big number operation
#lim: Length of source big number
#src: Source big number (array of unsigned int)
#dst: Destination big number (array of unsigned int)
Return: Array length of operation result
Warning: This method can write in an incorrect position if we don't previous reallocate num (if it's necessary). bigNum method do it for us
*/
unsigned int pow2BigNum(const unsigned int lim, unsigned int *src, unsigned int *dst)
{
unsigned int i, j;
unsigned long long int n, t;
unsigned int k, c;
for(c = 0, dst[0] = 0, i = 0; i < lim; ++i)
{
for(j = i, n = 0; j < lim; ++j)
{
n = ( (unsigned long long int) src[i] * src[j] );
k = i + j;
if(i != j)
{
t = 2 * (n % MODULE);
n = 2 * (n / MODULE);
// (i + j)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (t % MODULE);
++k; // (i + j + 1)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + ( (t / MODULE) + (n % MODULE) );
++k; // (i + j + 2)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (n / MODULE);
}
else
{
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (n % MODULE);
++k; // (i + j)
dst[k] = ( (k > c) ? ((c = k), 0) : dst[k] ) + (n / MODULE);
}
for(k = i + j; k < (lim + j); ++k)
{
dst[k + 1] += (dst[k] / MODULE);
dst[k] %= MODULE;
}
}
}
i = lim << 1;
return ((dst[i - 1]) ? i : i - 1);
}
/*
addBigNum: Make a (num2[] <- num1[] + num2[]) big number operation
#lim1: Length of source num1 big number
#num1: First source operand big number (array of unsigned int). Should be smaller than second
#lim2: Length of source num2 big number
#num2: Second source operand big number (array of unsigned int). Should be equal or greater than first
Return: Array length of operation result or 0 if num1[] > num2[] (dosen't do any op)
Warning: This method can write in an incorrect position if we don't previous reallocate num2
*/
unsigned int addBigNum(const unsigned int lim1, unsigned int *num1, const unsigned int lim2, unsigned int *num2)
{
unsigned long long int n;
unsigned int i;
if(lim1 > lim2)
return 0;
for(num2[lim2] = 0, n = 0, i = 0; i < lim1; ++i)
{
n = num2[i] + num1[i] + (n / MODULE);
num2[i] = n % MODULE;
}
for(n /= MODULE; n; ++i)
{
num2[i] += n;
n = (num2[i] / MODULE);
}
return (lim2 > i) ? lim2 : i;
}
To compile:
gcc -o bgn <name>.c -Wall -O3 -lm //Math library if you wants to use log func
To check result, use direct output as and input to bc. Easy shell script:
#!/bin/bash
select S in ` awk -F '==' '{print $1 " == " $2 }' | bc`;
do
0;
done;
echo "Test Finished!";
We have and array of unsigned int (4 bytes) where we store at each int of array a number of 9 digits ( % 1000000000UL ); hence num[0] we will have the first 9 digits, num[1] we will have digit 10 to 18, num[2]...
I use convencional memory to work but an improvement can do it with dinamic memory. Ok, but how length It could be the array? (or how many memory we need to allocate?). Using bc calculator (bc -l with mathlib) we can determine how many digits has a number:
l(a^N) / l(10) // Natural logarith to Logarithm base 10
If we know digits, we know amount integers we needed:
( l(a^N) / (9 * l(10)) ) + 1 // Truncate result
If you work with value such as (2^k)^N you can resolve it logarithm with this expression:
( k*N*l(2)/(9*l(10)) ) + 1 // Truncate result
to determine the exactly length of integer array. Example:
256^800 = 2^(8*800) ---> l(2^(8*800))/(9*l(10)) + 1 = 8*800*l(2)/(9*l(10)) + 1
The value 1000000000UL (10^9) constant is very important. A constant like 10000000000UL (10^10) dosen't work because can produce and indetected overflow (try what's happens with number 16^16 and 10^10 constant) and a constant more little such as 1000000000UL (10^8) are correct but we need to reserve more memory and do more steps. 10^9 is key constant for unsigned int of 32 bits and unsigned long long int of 64 bits.
The code has two parts, Multiply (easy) and Power by 2 (more hard). Multiply is just multiplication and scale and propagate the integer overflow. It take the principle of associative property in math to do exactly the inverse principle, so if k(A + B + C) we want kA + kB + kC where number will be k*A*10^18 + k*B*10^9 + kC. Obiously, kC operation can generate a number bigger than 999 999 999, but never more bigger than 0xFF FF FF FF FF FF FF FF. A number bigger than 64 bits can never occur in a multiplication because C is an unsigned integer of 32 bits and k is a unsigned short of 16 bits. In worts case, we will have this number:
k = 0x FF FF;
C = 0x 3B 9A C9 FF; // 999999999
n = k*C = 0x 3B 9A | 8E 64 36 01;
n % 1000000000 = 0x 3B 99 CA 01;
n / 1000000000 = 0x FF FE;
After Mul kB we need to add 0x FF FE from last multiplication of C ( B = kB + (C / module) ), and so on (we have 18 bits arithmetic offset, enough to guarantee correct values).
Power is more complex but is in essencial, the same problem (multiplication and add), so I give some tricks about code power:
Data types are important, very important
If you try to multiplication an unsigned integer with unsigned integer, you get another unsigned integer. Use explicit cast to get unsigned long long int and don't lose data.
Always use unsigned modifier, dont forget it!
Power by 2 can directly modify 2 index ahead of current index
gdb is your friend
I've developed another method that add big numbers. These last I don't prove so much but I think it works well. Don't be cruels with me if it has a bug.
...and that's all!
PD1: Developed in a
Intel(R) Pentium(R) 4 CPU 1.70GHz
Data length:
unsigned short: 2
unsigned int: 4
unsigned long int: 4
unsigned long long int: 8
Numbers such as 256^1024 it spend:
real 0m0.059s
user 0m0.033s
sys 0m0.000s
A bucle that's compute i^i where i goes to i = 1 ... 1024:
real 0m40.716s
user 0m14.952s
sys 0m0.067s
For numbers such as 65355^65355, spent time is insane.
PD2: My response is so late but I hope my code it will be usefull.
PD3: Sorry, explain me in english is one of my worst handicaps!
Last update: I just have had an idea that with same algorithm but other implementation, improve response and reduce amount memory to use (we can use the completely bits of unsigned int). The secret: n^2 = n * n = n * (n - 1 + 1) = n * (n - 1) + n.
(I will not do this new code, but if someone are interested, may be after exams... )
I don't know if you still need a solution, but I wrote an article about this problem. It shows a very simple algorithm which can be used to convert an arbitrary long number with base X to a corresponding number of base Y. The algorithm is written in Python, but it is really only a few lines long and doesn't use any Python magic. I needed such an algorithm for a C implementation, too, but decided to describe it using Python for two reasons. First, Python is very readable by anyone who understands algorithms written in a pseudo programming language and, second, I am not allowed to post the C version, because it I did it for my company. Just have a look and you will see how easy this problem can be solved in general. An implementation in C should be straight forward...
Here is a function that does what you want:
#include <math.h>
#include <stddef.h> // for size_t
double getval(unsigned char *arr, size_t len)
{
double ret = 0;
size_t cur;
for(cur = 0; cur < len; cur++)
ret += arr[cur] * pow(256, cur);
return ret;
}
That looks perfectly readable to me. Just pass the unsigned char * array you want to convert and the size. Note that it won't be perfect - for arbitrary precision, I suggest looking into the GNU MP BigNum library, as has been suggested already.
As a bonus, I don't like your storing your numbers in little-endian order, so here's a version if you want to store base-256 numbers in big-endian order:
#include <stddef.h> // for size_t
double getval_big_endian(unsigned char *arr, size_t len)
{
double ret = 0;
size_t cur;
for(cur = 0; cur < len; cur++)
{
ret *= 256;
ret += arr[cur];
}
return ret;
}
Just things to consider.
It may be too late or too irrelevant to make this suggestion, but could you store each byte as two base 10 digits (or one base 100) instead of one base 256? If you haven't implemented division yet, then that implies all you have is addition, subtraction, and maybe multiplication; those shouldn't be too hard to convert. Once you've done that, printing it would be trivial.
As I was not satisfied with the other answers provided, I decided to write an alternative solution myself:
#include <stdlib.h>
#define BASE_256 256
char *largenum2str(unsigned char *num, unsigned int len_num)
{
int temp;
char *str, *b_256 = NULL, *cur_num = NULL, *prod = NULL, *prod_term = NULL;
unsigned int i, j, carry = 0, len_str = 1, len_b_256, len_cur_num, len_prod, len_prod_term;
//Get 256 as an array of base-10 chars we'll use later as our second operand of the product
for ((len_b_256 = 0, temp = BASE_256); temp > 0; len_b_256++)
{
b_256 = realloc(b_256, sizeof(char) * (len_b_256 + 1));
b_256[len_b_256] = temp % 10;
temp = temp / 10;
}
//Our first operand (prod) is the last element of our num array, which we'll convert to a base-10 array
for ((len_prod = 0, temp = num[len_num - 1]); temp > 0; len_prod++)
{
prod = realloc(prod, sizeof(*prod) * (len_prod + 1));
prod[len_prod] = temp % 10;
temp = temp / 10;
}
while (len_num > 1) //We'll stay in this loop as long as we still have elements in num to read
{
len_num--; //Decrease the length of num to keep track of the current element
//Convert this element to a base-10 unsigned char array
for ((len_cur_num = 0, temp = num[len_num - 1]); temp > 0; len_cur_num++)
{
cur_num = (char *)realloc(cur_num, sizeof(char) * (len_cur_num + 1));
cur_num[len_cur_num] = temp % 10;
temp = temp / 10;
}
//Multiply prod by 256 and save that as prod_term
len_prod_term = 0;
prod_term = NULL;
for (i = 0; i < len_b_256; i++)
{ //Repeat this loop 3 times, one for each element in {6,5,2} (256 as a reversed base-10 unsigned char array)
carry = 0; //Set the carry to 0
prod_term = realloc(prod_term, sizeof(*prod_term) * (len_prod + i)); //Allocate memory to save prod_term
for (j = i; j < (len_prod_term); j++) //If we have digits from the last partial product of the multiplication, add it here
{
prod_term[j] = prod_term[j] + prod[j - i] * b_256[i] + carry;
if (prod_term[j] > 9)
{
carry = prod_term[j] / 10;
prod_term[j] = prod_term[j] % 10;
}
else
{
carry = 0;
}
}
while (j < (len_prod + i)) //No remaining elements of the former prod_term, so take only into account the results of multiplying mult * b_256
{
prod_term[j] = prod[j - i] * b_256[i] + carry;
if (prod_term[j] > 9)
{
carry = prod_term[j] / 10;
prod_term[j] = prod_term[j] % 10;
}
else
{
carry = 0;
}
j++;
}
if (carry) //A carry may be present in the last term. If so, allocate memory to save it and increase the length of prod_term
{
len_prod_term = j + 1;
prod_term = realloc(prod_term, sizeof(*prod_term) * (len_prod_term));
prod_term[j] = carry;
}
else
{
len_prod_term = j;
}
}
free(prod); //We don't need prod anymore, prod will now be prod_term
prod = prod_term;
len_prod = len_prod_term;
//Add prod (formerly prod_term) to our current number of the num array, expressed in a b-10 array
carry = 0;
for (i = 0; i < len_cur_num; i++)
{
prod[i] = prod[i] + cur_num[i] + carry;
if (prod[i] > 9)
{
carry = prod[i] / 10;
prod[i] -= 10;
}
else
{
carry = 0;
}
}
while (carry && (i < len_prod))
{
prod[i] = prod[i] + carry;
if (prod[i] > 9)
{
carry = prod[i] / 10;
prod[i] -= 10;
}
else
{
carry = 0;
}
i++;
}
if (carry)
{
len_prod++;
prod = realloc(prod, sizeof(*prod) * len_prod);
prod[len_prod - 1] = carry;
carry = 0;
}
}
str = malloc(sizeof(char) * (len_prod + 1)); //Allocate memory for the return string
for (i = 0; i < len_prod; i++) //Convert the numeric result to its representation as characters
{
str[len_prod - 1 - i] = prod[i] + '0';
}
str[i] = '\0'; //Terminate our string
free(b_256); //Free memory
free(prod);
free(cur_num);
return str;
}
The idea behind it all derives from simple math. For any base-256 number, its base-10 representation can be calculated as:
num[i]*256^i + num[i-1]*256^(i-1) + (···) + num[2]*256^2 + num[1]*256^1 + num[0]*256^0
which expands to:
(((((num[i])*256 + num[i-1])*256 + (···))*256 + num[2])*256 + num[1])*256 + num[0]
So all we have to do is to multiply, step-by step, each element of the number array by 256 and add to it the next element, and so on... That way we can get the base-10 number.

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