Discussion
What is the output of this C code?
int main()
{
int i = 97, *p = &i;
foo(&i);
printf("%d ", *p);
}
void foo(int *p)
{
int j = 2;
p = &j;
printf("%d ", *p);
}
Answer: A
Correct answer: 2 97
The parameter p inside foo is a local copy of the pointer, so p = &j redirects only that copy. It prints 2. Back in main the original pointer is untouched and still points at i, so the second print gives 97. C passes everything by value, pointers included - which is exactly why the next question passes the address of the pointer instead.
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