How the bool operator of std::shared_ptr is trrigered in the expression(i.e. `bool is_empty = shared_ptr1 && shared_ptr2;` )?

Question

Gaven that cur_front_res and cur_back_res are both shared_ptr, how the bool operator of std::shared_ptr is trigered in the expression(i.e. bool is_empty = cur_front_res && cur_back_res; )?

Just because && always cause built-in conversion if the operands(i.e. before && and after &&) is not an bool type?

The code snippet below works indeed.

#include <iostream>
#include <memory>

int main() {
    std::shared_ptr<int> cur_front_res; // Empty shared_ptr
    std::shared_ptr<int> cur_back_res(new int(42)); // Shared_ptr pointing to an int

    bool is_empty = cur_front_res && cur_back_res;

    if (is_empty) {
        std::cout << "Both cur_front_res and cur_back_res are not empty" << std::endl;
    } else {
        std::cout << "Either cur_front_res or cur_back_res is empty" << std::endl;
    }

    return 0;
}

Answer 1

Just because && always cause built-in conversion if the operands(i.e. before && and after &&) is not an bool type?

Basically yes.

But the bool conversion is somewhat special: even though operator bools are usually explicit (including the one for shared_ptr), there are a few cases where it can be called implicitly ("contextual bool conversion"). Among other things, those include operands of boolean operators, if/while/for conditions, etc.