％d和％c的char的scanf的不同答案

Question

This is a program used for swapping the nibbles of a byte, which is perfect for a byte but I faced a problem.

Code:

#include<stdio.h>  

void main()
{
    unsigned char a = 0;
    scanf("%d", &a);    
    a = ((a << 4) | (a >> 4)); 
    printf("the value of a is %d\n\r", a); 
} 

You can see in the scanf statement that I've received it as %d instead of receiving it as %c which is for a char . The above code works perfectly. But if I replace %d with %c , I am getting a different undesired answer

Why?

Answer 1

the correct specifier for scanf ing an unsigned char is %hhu , not %d .

So scanf("%d", &a); should be scanf("%hhu", &a); .

You should also use int main(void) instead of void main() and remove the \\r in the printf, because \\n is a new line on every system.

Answer 2

In your case

 scanf("%d", &a)

is wrong. %d expects a pointer to int , not a pointer to unsigned char . it will invoke undefined behaviour .

The correct way to do it will be

 scanf("%c", &a);

or

scanf("%hhu", &a);