Thread: error: lvalue required, pretty sure it's an lvalue

Originally Posted by laserlight

That is not true, because my rewrite does not compile, which is precisely the reason why I wrote it: to demonstrate what the expression would be like if parentheses were used to explicitly group, and hence explain why brandones encountered the "lvalue required" error.

Ahh, that makes sense. Thanks for all the clarification.

Got the code working. There were quite a few problems, including some basic mathematical ones. Thank you for the help!

Code:

	printf("%d", x<0);
	x<0 ? (x=-x) : x;
	for(y=14; y>-1; y--){
		int z = x-(pow(2, y));
		z<0 ? printf("0") : printf("1");
		x = (z<0 ? x : z);

Originally Posted by brandones

Got the code working. There were quite a few problems, including some basic mathematical ones. Thank you for the help!

Code:

	printf("%d", x<0);
	x<0 ? (x=-x) : x;
	for(y=14; y>-1; y--){
		int z = x-(pow(2, y));
		z<0 ? printf("0") : printf("1");
		x = (z<0 ? x : z);

Congrats on getting it working... Here's a complete solution in 20 lines...

Code:

#include <stdio.h>

int main (void)
  { char binary[33] = {0};
    unsigned int number;
    int x = 0;

    printf("Enter a number from 0 to 4294967295 : ");
    scanf("%u", &number);

    while(x < 32)
      binary[x++] = '0';
    do
      binary[--x] = (number & 1) + '0';
    while ( (number /= 2) > 0 );

    printf("32bit binary :  %s\n\n", binary);

    return 0; }