what if I ignore the return value of a function with shared_ptr return type

Question

#include <iostream>
#include <memory>

using namespace std;

shared_ptr<string> func()
{
    shared_ptr<string> ptr = make_shared<string>("smart poiter");
    return ptr;
}

int main(int argc, char const *argv[])
{
    func();
    cout << "pause" << endl;
    return 0;
}

like code above, will the memory of string "smart poiter" be released?

score 3 · Answer 1 · answered Sep 26 '19 at 01:52

3

Yes. The internal counter will reach 0 and the memory will be released safely.

answered Sep 26 '19 at 01:52

Ayxan Haqverdili

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score 2 · Accepted Answer · answered Sep 26 '19 at 01:52

Yes. shared_ptrs aren't special here; any instance's destructor will be invoked promptly (when the statement finishes evaluation) if the instance is returned without being assigned; not doing so would break RAII in a critical way. shared_ptr's destructor decrements the reference count, no other instances own a reference, so the destructor will release the associated memory.

what if I ignore the return value of a function with shared_ptr return type

2 Answers2