[Solved] Can you please explain the output of this C program? [closed]

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The code has undefined behavior, it may print 2 ! on your system, but it may do something entirely different on a different system, and indeed I don’t want to be flying a plane that runs it on its navigation system.

Reformatting the code makes it a little more explicit:

#include <stdio.h>
int main() {
    int b = 10;
    char ch = 33 ^ b & 1;
    for (;"what"[b++ + 21];)
        printf("%c", ch);
}

Here is what is happening:

  • b is initialized with value 10.
  • ch is initialized with value (33 ^ (b & 1)). Since 10 is even, b & 1 is 0, so ch has value 33, in hex: 0x21, which is the character ! in ASCII.
  • the for loops checks the value of an element from the string literal "what", which is an array of 5 char with values { 'w', 'h', 'a', 't', 0 }. The index is computed as b++ + 21. The first value is 31, and b is incremented to 11. Here you have undefined behavior because you are referencing the 32nd element of a 5 byte array.

Anything can happen. Reading this byte from memory can cause a crash or just return some random value. Further iterations of the loop dig deeper into the unknown. On your computer it takes 2 iterations to find a null byte, so the printf is run twice, but on some other machine or just some other time, anything could happen.

This test is bogus, a more reliable alternative would be:

#include <stdio.h>
int main(){
    int b=10,ch=33^b&1;
    for(;"what"[b+++-8];)
    printf("%c",ch);
}

solved Can you please explain the output of this C program? [closed]