C语言中的简单堆栈损坏

Question

Can somebody explain me, why in this program if I call the printf in that way the flag will be win? but without will not? Why this printf allow such things i can't understand thanks. Why without the printf the array can't overwrite the variable flag?

#include <stdio.h>
#include <stdbool.h>

int main() {

    int flag = false;

    int arr[10] = {0};
    int siz = sizeof(arr) / sizeof(* arr);

    printf("%p", &flag);
    arr[10] = 1; // Without the printf call can't get the win. Why?
    puts("");

    if(flag == true)
    {
        printf("win !");
    }
    else
    {
        printf("lose");
    }



    return 0;
}

Answer 1

Your program acesses the array beyond it's bounds. The array indexes start at 0 and end at N - 1 where N is the size of the array.

Doing this invokes undefined behavior, so your prediction of program's behavior will be wrong after this. Adding the printf() can change this behavior and it does, and that is what undefined behavior means, it should not affect the behavior of the program but once you have caused the undefined behavior at

arr[10] = 1;

you cannot know how the program will behave anymore.

Answer 2

This is causing your problem

arr[10] = 1;

You only allocate 10 elements in your array

int arr[10] = { 0 };

arr[10] is actually trying to access the 11th element in the array because array indices start at 0 .