Convert binary format string to int, in C - c

Question

How do I convert a binary string like "010011101" to an int, and how do I convert an int, like 5, to a string "101" in C?

Answer 1

The strtol function in the standard library takes a "base" parameter, which in this case would be 2.
int fromBinary(const char *s) {
return (int) strtol(s, NULL, 2);
}
(first C code I've written in about 8 years :-)

Answer 2

If it is a homework problem they probably want you to implement strtol, you would have a loop something like this:
char* start = &binaryCharArray[0];
int total = 0;
while (*start)
{
total *= 2;
if (*start++ == '1') total += 1;
}
If you wanted to get fancy you could use these in the loop:
total <<= 1;
if (*start++ == '1') total^=1;

Answer 3

I guess it really depends on some questions about your strings/program. If, for example, you knew your number wouldn't be bigger than 255 (IE you were only using 8 bits or 8 0s/1s), you could create a function where you hand it 8 bits from your string, traverse it and add to a sum that you returned everytime you hit a 1. IE if you hit the bit for 2^7 add 128 and the next bit you hit was 2^4 add 16.
This is my quick and dirty idea. I think more and Google for ya while at school. :D

Answer 4

For the 2nd part of the question, i.e. "how do I convert an int, like 5, to a string "101" in C?", try something like:
void
ltostr( unsigned long x, char * s, size_t n )
{
assert( s );
assert( n > 0 );
memset( s, 0, n );
int pos = n - 2;
while( x && (pos >= 0) )
{
s[ pos-- ] = (x & 0x1) ? '1' : '0'; // Check LSb of x
x >>= 1;
}
}

Answer 5

You can use the following coding
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main (void)
{
int nRC = 0;
int nCurVal = 1;
int sum = 0;
char inputArray[9];
memset(inputArray,0,9);
scanf("%s", inputArray);
// now walk the array:
int nPos = strlen(inputArray)-1;
while(nPos >= 0)
{
if( inputArray[nPos] == '1')
{
sum += nCurVal;
}
--nPos;
nCurVal *= 2;
}
printf( "%s converted to decimal is %d\n", inputArray, sum);
return nRC;
}

Answer 6

Use like this:
char c[20];
int s=23;
itoa(s,c,2);
puts(c);
Output:
10111

Answer 7

To answer the second part of the question.
char* get_binary_string(uint16_t data, unsigned char sixteen_bit)
{
char* ret = NULL;
if(sixteen_bit) ret = (char*)malloc(sizeof(char) * 17);
else ret = (char*)malloc(sizeof(char) * 9);
if(ret == NULL) return NULL;
if(sixteen_bit){
for(int8_t i = 15; i >= 0; i--){
*(ret + i) = (char)((data & 1) + '0');
data >>= 1;
}
*(ret + 16) = '\0';
return ret;
}else{
for(int8_t i = 7; i >= 0; i--){
*(ret + i) = (char)((data & 1) + '0');
data >>= 1;
}
*(ret + 8) = '\0';
return ret;
}
return ret;
}

Answer 8

To answer the first part of your question, here is a neat little function I created to convert Binary char strings to integers.
// Function used to change binary character strings to integers
int binToDec(char binCode[])
{
while (binCode != NULL)
{
int base = strlen(binCode) - 1; // the base of 2 to be multiplied, we start of -1 because we dont account for the last bit here
int sum = 0;
for (int i = 0; i < strlen(binCode) - 1; i++) // we do not account for the last bit of the binary code here....
{
int decimal = 1;
if (binCode[i] == '1')
{
for (int j = 0; j < base; j++) // we want to just multiply the number of true bits (not including the 1)
{
decimal = decimal * 2;
}
base = base - 1; // subtract base by 1 since we are moving down the string by 1
}
else // we encounter a zero
{
base = base - 1; // subtract a base multiple every time we encounter a zero...
continue; // carry on with the code
}
sum += decimal;
// starting from the left (higher power) to the end (lowest power or 1)
}
for (int j = strlen(binCode) - 1; j < strlen(binCode) + 1; j++)
{ // accounting for the endian bit that is always 1
if (binCode[j] == '1')
{
sum += 1; // add 1 to the sum total
}
}
return sum; // return the sum as an int
}
return 0;
}