how to print binary equivalent of a given no. without using any predefined functions in c?

Answer 1

You must know the type of the number to extract the right number of bits - but this is only an appearance thing. By using the same 'type' for a mask you get around it

// for int
void printintbits( int x )
{
int mask = 0x01; // LSB = 1
char bs[64]; // worst case scenario is a 64 bit int
int bp = 0; // bit position

for( int i = 0; i < sizeof(int); i++ )
{
bs[bp] = (mask & x) ? '1' : '0';
bp++;
mask <<= 1;
}
bs[bp] = '\0'; // terminate string
printf( "%d = $s\n", x, bs );
}

This gives the result LSB first. For MSB first, change
bs[bp] = (mask & x) ? '1' : '0';
to
bs[sizeof(int)-bp] = (mask & x) ? '1' : '0';

Answer 2

#include
#include

char *strrev(char *dst, char *src) {
unsigned len = strlen(src);
int i = 0;

for(i = 0; i < len; i++) {
dst[i] = src[len-1-i];
}
dst[i] = '\0';
return dst;
}

int main(void) {
int i = 0;
unsigned num = 0;
char bit[33] = {'0'};
char bitr[33] = {'0'};

scanf("%u", &num);

// bit string created in reverse order
for( i = 0; num; i++ ) {
bit[i] = num%2 + '0';
num /= 2;
}
strrev(bitr, bit);

printf("Binary rep of %u is %s\n", num, bitr);
return 0;
}

Answer 3

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