Question

I was told this is a buggy code, but I don’t know why, some could explain it to me.
why there would not be an array out of bound exception when I compile?

int a[10];
int j;

void main () {
    int i, j = 42;
    for (i = 0; i <=10; i++) a[i] = i;
    printf("%d", j);
}

Answer 1

Here’s one bug:

void main () {

should be

int main (int argc, char** argv) {

Another bug is in your loop. You write past the end of array a, and if your compiler placed j in memory immediately following a (which based on your question I assume it did), then the out-of-bounds array access will actually end up assigning a value to j. Hence, when you write 10 into a[10] (which doesn’t exist), you are writing it into the memory where j lives (causing this to act like j = 10). However, this behavior is dependent on how your compiler lays out the variables in memory, so you may very well see different behavior if you compiled the same program on a different platform.