Discussion
What will be the output of the following C code?
#include <stdio.h>
int main()
{
int i = 10;
int *const p = &i;
foo(&p);
printf("%d\n", *p);
}
void foo(int **p)
{
int j = 11;
*p = &j;
printf("%d\n", **p);
}
Answer: A
Correct answer: 11 11
In main, int *const p makes the pointer itself const, so &p has type int *const *. Passing that to a parameter of type int ** discards the qualifier, which is a constraint violation - but compilers generally warn rather than refuse, so the program runs.
foo then writes through the pointer, redirecting main's p to j, and both prints show 11. The exercise illustrates that const can be circumvented through a pointer conversion the compiler only warns about.
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