Discussion
What is the output of this C code?
int main()
{
int i = 97, *p = &i;
foo(&p);
printf("%d ", *p);
return 0;
}
void foo(int **p)
{
int j = 2;
*p = &j;
printf("%d ", **p);
}
Answer: A
Correct answer: 2 2
Here the address of the pointer is passed, so *p = &j modifies the caller’s pointer itself and both prints give 2. That is how a function changes a pointer in its caller. Note that it is also a bug in the making: after foo returns, p in main points at j, whose lifetime has ended, so the second dereference is strictly undefined even though it prints 2 in practice.
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