Discussion
What is the output of this C code?
int main()
{
int i = 10;
int *p = &i;
foo(&p);
printf("%d ", *p);
printf("%d ", *p);
}
void foo(int **const p)
{
int j = 11;
*p = &j;
printf("%d ", **p);
}
Answer: B
Correct answer: 11 11 Undefined-value
int **const p makes the parameter itself constant, but *p is not, so *p = &j is legal and the caller’s pointer is redirected to j. The first two prints give 11, one inside foo and one immediately after it returns. By the third, j’s lifetime has ended and the value is indeterminate - compare with the previous question, where the const sits in the other position and the code will not compile at all.
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