Question

#include<stdio.h>
main()
{
int i;
char c;
for (i=0;i<5;i++){
        scanf("%d",&c);
        printf("%d",i);
}
printf("\n");
}

I thought it will print 0 1 2 3 4 but it didn’t.
What’s the reason of the strange output?

Answer 1

This program exhibits undefined behavior: The type of &c (char *) does not correspond to the type of the scanf arg (%d wants a signed int *).

What is probably happening is that the scanf is writing 4 bytes to the memory location starting at the address of c. Which is only 1 byte long, so the other 3 bytes overwrite the first 3 bytes of i‘s value. On a little-endian system, that would effectively set i to whatever integer value you enter shifted right by 8 bits.

But, of course, the behavior is undefined. Next time you compile this code, it could do something completely different. A different compiler, or the same compiler with different options, could keep i in a register (where scanf cannot overwrite it) (but it might instead smash the return address on the stack, causing a crash when the program ends), or it could put the values on the stack in the opposite order (same deal), or it could leave 4 bytes on the stack for c (causing no unexpected behavior), or it could detect the situation and abort with an error, or it could even make demons fly out of your nose.