如何在执行期间终止（或返回）具有结构类型的自动功能？

Question

How do I terminate an "auto" type function with a structure return type during execution?

I want to escape the auto function 'Func' (not the entire program) when some condition is satisfied as follows:

#include "stdafx.h"
#include <vector>
#include <tchar.h>
#include <iostream>

using namespace std;

auto Func(int B, vector<int> A) {
    for (int a = 0; a < B; a++)
    {
        A.push_back(a);
        if (a == 2)
        {
            cout << " Terminate Func! " << endl;
            // return; I want to terminate 'Func' at this point  
        }
    }

    struct result { vector <int> A; int B; };
    return result{ A, B };
}

int _tmain(int argc, _TCHAR* argv[])
{

    vector<int> A;
    int B = 5;

    auto result = Func(B, A);
    A = result.A;
    B = result.B;

    for (int a = 0; a < A.size(); a++)
    {
        cout << A[a] << endl;
    }
}

I don't want to use "exit()" because I just want to terminate a function not the program.

Answer 1

You can return a define the return type result at the top of the function and then return an "empty" instance like this:

auto Func(int B, vector<int> A) {
    struct result { vector <int> A; int B; };

    for (int a = 0; a < B; a++)
    {
        A.push_back(a);
        if (a == 2)
        {
            return result{{}, 0};
        }
    }

    return result{ A, B };
}

If you don't want a valid object to be returned when the early-return condition is met, consider returning a std::optional<result> from the function, and specifically std::nullopt in the early return branch. But this requires C++17.