Does C++ call the destructor when an object goes out of scope?

Question

I thought yes, but

#include <iostream>

struct S {
    int t;
};

class C {
private:
    S s;
public:
    C() {s.t = 7;}
    ~C(){std::cout << "bye C" << std::endl;}
};

class D {
private:
    S s;
public:
    D(int t) {s.t = t;}
    ~D() {std::cout << "bye D(" << s.t << ")" << std::endl;}
};

int main() {
    C c0();
    C* c1 = new C();
    D d0();
    D d1(42);
    std::cout << __LINE__ << std::endl;
    delete c1;
    std::cout << __LINE__ << std::endl;
}

just prints (https://ideone.com/95DK9E)

28
bye C
30
bye D(42)

So why are c0 and d0 not properly destructed by a call to their destructor?

`c0` and `d0` are not objects, but functions returning objects. — Bo Persson, Apr 05 '18 at 12:40
Change the function declarations to `C c0{};` and `D d0{};`, or to `C c0;` and `D d0;` so they become local variables. — Eljay, Apr 05 '18 at 12:42

score 1 · Accepted Answer · answered Apr 05 '18 at 12:40

1

c0 and d0 are not objects. You've written two function declarations.

answered Apr 05 '18 at 12:40

Jesper Juhl

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Never forget https://en.wikipedia.org/wiki/Most_vexing_parse – not-a-user Apr 05 '18 at 12:53

Does C++ call the destructor when an object goes out of scope?

1 Answers1