I need help trying to fix the second part of my program, converting decimal to binary, this is what I have so far and when i compile it i keep getting 0 so im not sure what i did wrong. any help please?
#include <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
char string[100];
int s;
char a;
char j;
int sum = 0;
int r;
int q;
printf("B = B to D\n");
printf("D = D to B\n");
printf("choose which one to convert to:");
scanf("%c%c", &a, &j);
if (a == 'B')
{
printf("enter binary number to convert to decimal: ");
scanf("%s", string);
for(s = strlen(string)-1; s >= 0; s--)
{
if(string[s] == '1')
{
sum = sum + pow(2, strlen(string) - (s +1));
}
}
printf("the decimal number is: %d\n", sum);
}
if (a == 'D')
{
printf("enter decimal number to convert to binary: ");
scanf("%s", string);
while (r > 0)
{
r = q%2;
q = q%2;
}
printf("the binary number is: %d\n", r);
}
return 0;
}
There are a few problems here. For one thing, the first time that you check r, it is uninitialized. Another problem is that you're setting both r and q to the same value every time you go through the while loop. You probably want q = q/2 instead of q = q%2. Finally, you're overwriting r every pass through the loop, instead of building up a string of bits. Here's some pseudocode for what you want to do:
output_string = ""
while input > 0:
output_string = concat(input%2, output_string)
input /= 2
print output_string
Note that you're also never converting the string you read in to an integer and putting that in q, so you'll need to do that as well.
This C99 code will do the trick if you want a negative number to be printed as a string of binary digits with a sign:
if (a == 'D')
{
int r;
printf("enter decimal number to convert to binary: ");
scanf("%d", &r);
int i = 0;
int p = (r >= 0) ? (r = -r, 1) : 0;
string[i++] = '\0';
do
{
string[i++] = (r % 2) == 0 ? '0' : '1';
r /= 2;
} while (r != 0);
if (!p)
string[i++] = '-';
int k = 0;
while (--i > k)
{
char t = string[i];
string[i] = string[k];
string[k++] = t;
}
printf("the binary number is: %s\n", string);
}
For example, given -1234 (decimal), the output is -10011010010 (binary). It also handles both the extremes: INT_MAX, -INT_MAX and INT_MIN (assuming 32-bit int):
B = B to D
D = D to B
choose which one to convert to: D
enter decimal number to convert to binary: 2147483647
the binary number is: 1111111111111111111111111111111
B = B to D
D = D to B
choose which one to convert to: D
enter decimal number to convert to binary: -2147483647
the binary number is: -1111111111111111111111111111111
B = B to D
D = D to B
choose which one to convert to: D
enter decimal number to convert to binary: -2147483648
the binary number is: -10000000000000000000000000000000
If, on the other hand, you want the bit pattern corresponding to the value, then Joachim Pileborg's answer does that for you.
(It's C99 code because it declares variables at convenient points part way through a block, rather than at the start of a block as C89 requires.)
The simplest thing is probably to convert the string input to a proper integer (using e.g. strtol), and the convert that number to a string containing only ones and zeroes.
Something like:
/* Convert a (possibly signed) decimal number in a string to a long integer */
unsigned long number = (unsigned long) strtol(string, NULL, 10);
char output_string[65]; /* If longs are 64 bits, plus one for terminator */
char *output_ptr = output_string;
/* Start with the highest bit, go down to the lowest */
/* sizeof(long) is either 4 or 8 depending on 32 or 64 bit platforms */
/* Multiply with 8 to get the number of bits */
/* -1 because bits are numbered from 0 to 31 (or 63) */
for (int bit = (sizeof(unsigned long) * 8) - 1; bit >= 0; bit--)
{
/* Using right shift to get the current bit into the lowest position */
/* Doing bitwise AND to see if the lowest bit is a one or a zero */
/* Adding '0' makes a a printable ASCII value of a digit */
*output_ptr++ = ((number >> bit) & 1) + '0';
/* `*output_ptr` gets the value that `output_ptr` points to */
/* Then use the `++` operator to increase the pointer */
/* Now `output_ptr` points to the next character in `output_string` */
}
/* Terminate string */
*output_ptr = '\0';
printf("%ld in binary is %s\n", number, output_string);
Related
We were assigned a task to convert octal numbers to binary and decimal. Smaller numbers works just fine but it then gives a different output at a higher input. Here is the code:
#include <stdio.h>
void main() {
unsigned long n, g, ans = 1, r = 0, dec = 0, place = 1, bin = 0;
printf("Conversion: Octal to decimal and binary.\n");
printf("Enter number: ");
scanf("%lu", &n);
printf("%lu is ", n);
for (g = n; g != 0; ans = ans * 8) {
r = g % 10;
dec = dec + r * ans;
g = g / 10;
}
printf("%lu in Decimal Form. \n", dec);
printf("%lu is ", n);
for (; dec != 0; place = place * 10) {
r = dec % 2;
bin = bin + (r * place);
dec = dec / 2;
}
printf("%lu in Binary Form.\n", bin);
}
We were only required to use limited data types and control structures. No arrays, strings, functions or such.
The input in our teacher's test case is 575360400 which must print an output of 101111101011110000100000000 in binary and 100000000 in decimal. But the output in binary is 14184298036271661312. I used unsigned long already and it just won't work.
I don't know how this is possible with the given restrictions and your comments and answers will be really much of a help.
Smaller numbers works just fine but it then gives a different output at a higher input.
Overflow
Input "575360400" (base 8) converts to a 27-bit value. For place = place * 10 ... bin = bin + (r * place); to work, bin needs to be a 90-bit unsigned long. unsigned long is certainly 64-bit or less.
OP needs a new approach.
I'd start with reading the octal input with scanf("%lo", &n); and printing the decimal with printf("%lu\n", n);.
To print in binary, without arrays, functions, etc., consider a mask, first with the most significant bit, that is shifted right each iteration.
bool significant_digit = false;
// Form 0b1000...0000
// 0b1111... - 0b01111..
unsigned long mask = ULONG_MAX - ULONG_MAX/2;
while (mask) {
bool bit = mask & n;
if (bit || significant_digit || mask == 1) {
significant_digit = true;
printf("%d", bit);
}
mask >>= 1;
}
printf("\n", bit);
}
Better approaches exist. Yet with OP's artificial limitations: "No arrays, strings, functions or such.", I opted for something illustrative and simple.
Or wait until 2023
C2x expected to support printf("%lb\n", n);. Just ask the professor for an extension 😉.
/* hexadecimal to decimal conversion */
#include <stdio.h>
#include <math.h>
#include <string.h>
int main()
{
char hex[17];
long long decimal;
int i , val, len;
decimal = 0;
// Input hexadecimal number from user
printf("Enter any hexadecimal number: ");
gets(hex);
//Find the length of total number of hex digit
len = strlen(hex);
len--;
for(i=0; hex[i]!='\0'; i++)
{
// Find the decimal representation of hex[i]
if(hex[i]>='0' && hex[i]<='9')
{
val = hex[i] - 48;
}
else if(hex[i]>='a' && hex[i]<='f')
{
val = hex[i] - 97 + 10;
}
else if(hex[i]>='A' && hex[i]<='F')
{
val = hex[i] - 65 + 10;
}
decimal += val * pow(16, len);
len--;
}
printf("Hexadecimal number = %s\n", hex);
printf("Decimal number = %lld", decimal);
return 0;
}
In the above program when i have used scanf instead of gets,it doesn't give the result.why? i used scanf("%x",hex); . please explain me decimal += val * pow(16, len); too.thank you so much in advance.
Because if you use scanf(), it does the conversion from string for you, that's sort of its entire point.
unsigned int x;
if(scanf("%x", &x) == 1)
printf("you entered %d (hex 0x%x)\n", x, x);
You can't combine %x a pointer to a character array, it requires a pointer to an unsigned integer. This is of course well documented in the manual page.
Also, using pow() here seems excessive, just multiply what you have by 16 before adding in each new digit:
unsigned int parsehex(const char *s)
{
unsigned int x = 0;
const char *digits = "0123456789abcdef";
const char *p;
while(*s && (p = strchr(digits, tolower(*s++))) != NULL)
{
x *= 16;
x += (unsigned int) (p - digits);
}
return x;
}
This is a bit "heavier" (uses strchr()) than your code, but shorter and perhaps therefore easier to validate. If it's overly performance-critical, I'd consider looking into it.
scanf("%x",hex);
should be
scanf("%s",hex);
you cannot do hex[i] when you read as integer.
decimal += val * pow(16, len); represents decimal = decimal + (val * pow(16, len));
Hopes this answers your question
scanf("%x"...) performs the conversion to integer for you. Therefore, you want to deposit the result in decimal:
scanf("%x", &decimal);
Each iteration of the for loop is generating a nibble (4 bits) of the number into val. The val * pow(16, len); is (in)effectively shifting the nibble into the correct position. However, this code is using floating point math to accomplish this (pow returns a double) instead of simply left shifting by 4*len. A better approach is to simply shift decimal left by 4 bits on each iteration and add (or OR) the nibble into the least significant bits. In this way, the first nibble will ultimately end up where it is supposed to be.
Also, character literals work as numbers, so instead of subtracting 48, 97, 65 it would read better if you subtracted '0', 'f', 'F' respectively.
Can any one help me sort out one problem, i have to reverse a number without using array(int/char) for storing them.
input1 = 123
output1 = 321
input2 = 2300
output2 = 0032
I am trying to find the solution but 0 got erased while printing so i thought of octal conversion but still no solution, so i went with the decimal places and i made the 23 to 0.0032. Now my problem is how can i extract the 0032 from that part.
Is there any possible way to achieve this without using array(int/char), with that it will be easy.
#include<stdio.h>
#include<math.h>
int main()
{
int number =3200;
int temp;
while (number >0)
{
temp= number%10;
printf("%d",temp);
number = number/10;
}
return 0;
}
you could use recursion to solve this problem, without using any array in fact u could also reverse a string without using any array using recursion. This code works for both numbers and strings and it has no arrays:
char reverse(int a)
{
char c,d;
if(a=='\n')
return 0;
c=getchar();
d=reverse(c);
putchar(a);
return (c);
}
int main()
{
char c;
scanf("%c",&c);
reverse(c);
}
for a start try this.
int n, l;
char nBuf[126];
n = 1230010;
l = sprintf(nBuf, "%d", n );
while( l >= 0 )
printf("%c", nBuf[l--] );
Though if you are taking input from stdin take it as string rathar than as int or long.
Edit - for not using array
int n = 123;
while(n) {
printf("%d", n%10);
n/=10;
}
I am assuming to get a value of this sort "output2 = 0032" it is better of being a string, else formatting complications turns up with input value length and format left space with zeros etc etc.
This becomes fairly easy if you know that you can represent numbers like so:
x = a_0 + a_1 * b^1 + a_2 * b^2 + ...
a_i are the digits
b is the base
To extract the lowest digit, you can use the remainder: x % b
Dividing by the base "removes" the last digit. That way you can get the digits in order lowest to highest.
If you reverse the digits then the lowest becomes the highest. Looking at below transformation it's easy to see how to incrementally build up a number when the digits come in order highest to lowest:
x = a_0 + b * (a_1 + b * (a_2 + ...
You start of with 0, and for each digit you multiply with the base and then add the digit.
In pseudo code:
output = 0
while input != 0
digit = input % base
input = input / base ;; integer division
output = output * base + digit
end
If you want to store leading zeros, then you need to either store the digits in an array, or remember for how many steps of above loop the output remained zero:
output = 0
zeros = 0
while input != 0
digit = input % base
input = input / base ;; integer division
output = output * base + digit
if output == 0
zeros = zeros + 1
end
end
To print that you obviously need to print zeros zeros and then the number.
Live example here, relevant code:
unsigned reverse(
unsigned input,
unsigned const base,
unsigned * const zeros) {
unsigned output = 0;
unsigned still_zero = 0;
for (; input != 0; input/=base) {
output *= base;
output += input % base;
if (output == 0) {
++still_zero;
}
}
if (zeros != NULL) {
*zeros = still_zero;
}
return output;
}
void print_zeros(unsigned zeros) {
for (; zeros != 0; --zeros) {
printf("0");
}
}
Recursion allows for a simple solution. A small variation on #vishu rathore
void rev_dec(void) {
int ch = getchar();
if (isdigit(ch)) {
rev_dec();
}
if (ch >= 0) putchar(ch);
}
int main(void) {
rev_dec();
return 0;
}
input
0123456789012345678901234567890123456789
output
9876543210987654321098765432109876543210
I'm doing a homework assignment for my course in C (first programming course).
Part of the assignment is to write code so that a user inputs a number up to 9 digits long, and the program needs to determine whether this number is "increasing"/"truly increasing"/"decreasing"/"truly decreasing"/"increasing and decreasing"/"truly decreasing and truly increasing"/"not decreasing and not increasing". (7 options in total)
Since this is our first assignment we're not allowed to use anything besides what was taught in class:
do-while, for, while loops, else-if, if,
break,continue
scanf, printf ,modulo, and the basic operators
(We can't use any library besides for stdio.h)
That's it. I can't use arrays or getchar or any of that stuff. The only function I can use to receive input from the user is scanf.
So far I've already written the algorithm with a flowchart and everything, but I need to separate the user's input into it's distinct digits.
For example, if the user inputs "1234..." i want to save 1 in a, 2 in b, and so on, and then make comparisons between all the digits to determine for example whether they are all equal (increasing and decreasing) or whether a > b >c ... (decreasing) and so on.
I know how to separate each digit by using the % and / operator, but I can't figure out how to "save" these values in a variable that I can later use for the comparisons.
This is what I have so far:
printf("Enter a positive number : ");
do {
scanf ("%ld", &number);
if (number < 0) {
printf ("invalid input...enter a positive integer: ");
continue;
}
else break;
} while (1);
while (number < 0) {
a = number % 10;
number = number - a;
number = number / 10;
b = a;
}
Why not scan them as characters (string)? Then you can access them via an array offset, by subtracting the offset of 48 from the ASCII character code. You can verify that the character is a digit using isdigit from ctype.h.
EDIT
Because of the incredibly absent-minded limitations that your professor put in place:
#include <stdio.h>
int main()
{
int number;
printf("Enter a positive number: ");
do
{
scanf ("%ld", &number);
if (number < 0)
{
printf ("invalid input...enter a positive integer: ");
continue;
}
else break;
} while (1);
int a = -1;
int b = -1;
int c = -1;
int d = -1;
int e = -1;
int f = -1;
int g = -1;
int h = -1;
int i = -1;
while (number > 0)
{
if (a < 0) a = number % 10;
else if (b < 0) b = number % 10;
else if (c < 0) c = number % 10;
else if (d < 0) d = number % 10;
else if (e < 0) e = number % 10;
else if (f < 0) f = number % 10;
else if (g < 0) g = number % 10;
else if (h < 0) h = number % 10;
else if (i < 0) i = number % 10;
number /= 10;
}
/* Printing for verification. */
printf("%i", a);
printf("%i", b);
printf("%i", c);
printf("%i", d);
printf("%i", e);
printf("%i", f);
printf("%i", g);
printf("%i", h);
printf("%i", i);
return 0;
}
The valid numbers at the end will be positive, so those are the ones you validate to meet your different conditions.
Since you only need to compare consecutive digits, there is an elegant way to do this without arrays:
int decreasing = 2;
int increasing = 2;
while(number > 9)
{
int a = number % 10;
int b = (number / 10) % 10;
if(a == b)
{
decreasing = min(1, decreasing);
increasing = min(1, increasing);
}
else if(a > b)
decreasing = 0;
else if(a < b)
increasing = 0;
number /= 10;
}
Here, we walk through the number (by dividing by 10) until only one digit remains. We store info about the number up to this point in decreasing and increasing - a 2 means truly increasing/decreasing, a 1 means increasing/decreasing, and a 0 means not increasing/decreasing.
At each step, a is the ones digit and b is the tens. Then, we change increasing and decreasing based on a comparison between a and b.
At the end, it should be easy to turn the values of increasing and decreasing into the final answer you want.
Note: The function min returns the smaller of its 2 arguments. You should be able to write your own, or replace those lines with if statements or conditionals.
It's stupid to ask you to do loops without arrays --- but that's your teacher's fault, not yours.
That being said, I would do something like this:
char c;
while (1) {
scanf("%c", &c);
if (c == '\n') /* encountered newline (end of input) */
break;
if (c < '0' || c > '9')
break; /* do something to handle bad characters? */
c -= '0';
/*
* At this point you've got 0 <= c < 9. This is
* where you do your homework :)
*/
}
The trick here is that when you type numbers into a program, you send the buffer all at once, not one character at a time. That means the first scanf will block until the entire string (i.e. "123823" or whatever) arrives all at once, along with the newline character ( '\n' ). Then this loop parses that string at its leisure.
Edit For testing the increasing/decreasing-ness of the digits, you may think you need to store the entire string, but that's not true. Just define some additional variables to remember the important information, such as:
int largest_digit_ive_seen, smallest_digit_ive_seen, strict_increasing_thus_far;
etc. etc.
Let us suppose you have this number 23654
23654 % 10000 = 2 and 3654
3654 % 1000 = 3 and 654
654 % 100 = 6 and 54
54 % 10 = 5 and 4
4
This way you can get all the digits. Of course, you have to know if the number is greater than 10000, 1000, 100 or 10, in order to know the first divisor.
Play with sizeof to get the size of the integer, in order to avoid a huge if...else statement
EDIT:
Let us see
if (number>0) {
// Well, whe have the first and only digit
} else if (number>10) {
int first_digit = number/10;
int second_digit = number % 10;
} else if (number>100) {
int first_digit = number/100;
int second_digit = (number % 100)/10;
int third_digit = (number % 100) % 10;
} ...
and so on, I suppose
// u_i is the user input, My homework asked me to extract a long long, however, this should also be effective for a long.
int digits = 0;
long long d_base = 1;
int d_arr[20];
while (u_i / d_base > 0)
{
d_arr[digits] = (u_i - u_i / (d_base * 10) * (d_base * 10)) / d_base;
u_i -= d_arr[digits] * d_base;
d_base *= 10;
digits++;
}
EDIT: the extracted individual digit now lives in the int array d_arr. I'm not good at C, so I think the array declaration can be optimized.
Here's a working example in plain C :
#include <stdio.h>
unsigned long alePow (unsigned long int x, unsigned long int y);
int main( int argc, const char* argv[] )
{
int enter_num, temp_num, sum = 0;
int divisor, digit, count = 0;
printf("Please enter number\n");
scanf("%d", &enter_num);
temp_num = enter_num;
// Counting the number of digits in the entered integer
while (temp_num != 0)
{
temp_num = temp_num/10;
count++;
}
temp_num = enter_num;
// Extracting the digits
printf("Individual digits in the entered number are ");
do
{
divisor = (int)(alePow(10.0, --count));
digit = temp_num / divisor;
temp_num = temp_num % divisor;
printf(" %d",digit);
sum = sum + digit;
}
while(count != 0);
printf("\nSum of the digits is = %d\n",sum);
return 0;
}
unsigned long alePow(unsigned long int x, unsigned long int y) {
if (x==0) { return 0; }
if (y==0||x==1) { return 1; }
if (y==1) { return x; }
return alePow(x*x, y/2) * ((y%2==0) ? 1 : x);
}
I would suggest loop-unrolling.
int a=-1, b=-1, c=-1, d=-1, e=1, f=-1, g=-1, h=-1, i=-1; // for holding 9 digits
int count = 0; //for number of digits in the given number
if(number>0) {
i=number%10;
number/=10;
count++;
}
if(number>0) {
h=number%10;
number/=10;
count++;
}
if(number>0) {
g=number%10;
number/=10;
count++;
}
....
....
/* All the way down to the storing variable a */
Now, you know the number of digits (variable count) and they are stored in which of the variables. Now you have all digits and you can check their "decreasing", "increasing" etc with lots of if's !
I can't really think of a better soltion given all your conditions.
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.