Does a object stores lambda function have it's own address?

Question

Based on the result of this topic

I have written a program like

void test1(){}
int main() {
    auto test2 = [](){};
    printf("%p\n%p\n", test1, &test1);
    printf("%p\n%p", test2, &test2);

    return 0;
}

the result are

0x561cac7d91a9

0x7ffe9e565397

(nil)

So the test2 is storing a function pointer to the lambda function?

and my question is that object test2, which stores the lambda data, not have its own address?

I thought this test2 should have its own address.

`test2` stores a callable object. Its type is a class with an `operator()`. A funciton, much similar to an array, can decay to a pointer, thats why you see the same address in the first two outputs — 463035818_is_not_an_ai, Nov 08 '22 at 10:18
@VioletGiraffe I refer to `test1` vs `&test1`. I dont think they can be different — 463035818_is_not_an_ai, Nov 08 '22 at 10:22

score 5 · Accepted Answer · answered Nov 08 '22 at 10:39

Does a object stores lambda function have it's own address?

Yes, like all objects in C++ the variable test2 also has a unique address. You can see this by printing &test uisng cout as shown below:

int main() {
    auto test2 = [](){};
     
    std::cout << &test2;  //prints address of test2
    return 0;
}

The output of the above modified program is:

0x7ffe4d4b5fcf

1 Answers1