输入负数印出该负数的二进位

[Leaf1029]

老师给出框架了，求高手解答，我们还没学到阵列，应该是用位元对比和循环做的


Complete the source file below to get the correct result.

运行：
input a integer (0~255) : 100
100's binary : 01100100
input a integer (int) (-2^31~2^31-1) : -100
-100's binary : 11111111111111111111111110011100


框架：
#include <stdio.h>
int c2b(unsigned char);
void printi(int);
int main ( void)
{
unsigned char a;
int x;
do
{
printf("input a integer (0~255) : ");
scanf("%u",&x);
}
while ((x > 255 || x < 0));
a = x;
printf("%d's binary : %08d",a,c2b(a) );
printf("\n");

printf("input a integer (int) (-2^31~2^31-1) : ");
scanf("%d",&x);
printi(x);
printf("\n");

return 0;
}

int c2b( unsigned char c )
{
// your code
}

void printi (int x)
{
// your code
}

[mymtom]

#include <stdio.h>
int c2b(unsigned char);
void printi(int);
int main ( void)
{
    unsigned char a;
    int x;
    do
    {
        printf("input a integer (0~255) : ");
        scanf("%u",&x);
    }
    while ((x > 255 || x < 0));
    a = x;
    printf("%d's binary : %08d",a,c2b(a) );
    printf("\n");

    printf("input a integer (int) (-2^31~2^31-1) : ");
    scanf("%d",&x);
    printi(x);
printf("\n");

    return 0;
}

int c2b( unsigned char c )
{
    int i, b;

    b = 0;
    for (i = 0; i < 8; ++i)
        b = 10 * b + ((c >> (7 - i)) & 1U);
    return b;
}

void printi (int x)
{
    int i;
    printf("%d's binary :", x);
    for (i = 0; i < 32; ++i)
        if ((((unsigned int)x) >> (31 - i)) & 1U)
            printf("1");
        else
            printf("0");
}