Why doesn't my program seg fault when I dereference a NULL pointer inside of malloc?

Question

I use this malloc style all the time

int *rc = 0;
rc = malloc(sizeof(*rc));

However, it doesn't seg fault even though when I call sizeof(*rc) I assume that rc==0, and I am dereferencing a NULL pointer.

Answer 1

You are not really dereferencing anything. The argument of sizeof is not evaluated, unless it is a VLA. It is explicitly allowed by the language to put whatever "garbage" you want as the argument of sizeof. The language guarantees that it will not evaluate anything, just perform compile-time analysis of the type of the expression. For example, expression sizeof i++ is guaranteed not to change the value of i.

The only exception from that rule is Variable Length Arrays. The result of sizeof for VLAs is a run-time value, which means that the argument is evaluated and must be valid.

Answer 2

The sizeof operator doesn't actually evaluate its operand, it only looks at its type. The type of *rc is int, so it's equivalent to sizeof (int). This all happens at compile time.

(Also, this is not "inside of malloc".)