CODEWITHSUNDEEP
objective-cpointersmemory-address
Cornstalks
Cornstalks

Reputation: 38228

Why does &object evaluate to two different values in Objective-C?

Consider the following program:

#include <objc/nsobject.h>
#include <stdio.h>

void function(NSObject** object) {
  printf("value of NSObject** in function: %p\n", object);
}

int main() {
  NSObject* object;
  printf("value of NSObject** in main:     %p\n", &object);
  function(&object);
}

Program output (compiled with -fobjc-arc on x86_64):

value of NSObject** in main: 0x7ffee1217668
value of NSObject** in function: 0x7ffee1217660

Why are the values not equal? They're offset by 8 bytes, which I find unexpected. If I change it to a plain C type (e.g. int**) then they both print the same values. But double pointers to Objetive-C objects always give different values.

What's going on here, and why does &object evaluate to two different values?

Upvotes: 1

Views: 49

Answers (1)

Ken Thomases
Ken Thomases

Reputation: 90611

ARC is creating an autoreleasing temporary object and passing the address of that and then, after function returns, assigning that to the strong object variable.

See Ownership Inference: Indirect parameters to see why function's parameter is inferred to be __autoreleasing. See Passing to an out parameter by writeback to see why ARC creates the temporary.

I believe you could declare an explicit __strong qualifier on function's parameter to avoid that.

Upvotes: 2

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