将一个十进制数转换为二进制输出

#include <stdio.h>
int main (int argc, char *argv[])
{
	int n, i;
	scanf("%d",&n);
	char bits[33] = {0};
	for(i = 31; i >= 0; i--){
	bits[i] = (n & 1) + ‘0‘;
	n >>= 1;
}
	bits[32] = ‘\0‘;
	puts(bits);
	return 0;
}

#include<stdio.h>
void main()
{
	int a = -1;
	int b[32] = {};
	int i=0;
	for(i=0;i<32;i++)
	b[i] = (a&(0x01<<i))?1:0;




	for(i=31;i>=0;i--)
	printf("%d",b[i]);




printf("\n");
}

#include<stdio.h>
void main()
{
	int temp = -1;
	unsigned int a;
	int b[32] = {};
	int i=0;
	//if(temp<0)
	a=temp; 
	while(a!=0){ 
	b[i] = (a%2)?1:0;
	a = a/2;
	i++; 
	}




	for(i=31;i>=0;i--)
	printf("%d",b[i]);




	printf("\n");
}

将一个十进制数转换为二进制输出

原文：http://blog.csdn.net/a987860319/article/details/18563377