从函数返回指针

Question

Why this code doesn't work? I have already found a workaround by passing a pointer to the function but I wonder if there is an easier solution?

#include <stdio.h>

int* func() {
        int d[3];
        for (int f = 0; f < 3; f++)
                d[f] = 42;
        return d;
}


int main() {
        int* dptr;
        dptr = func();

        printf("Hi\n");

        for (int f = 0; f < 3; f++)
                 printf("%d\n",dptr[f]);
        return 0;
}

Answer 1

With return d; , you are returning a pointer to a local variable, ie array d 's live time will end once function func has finished. Accessing this array afterwards (through the returned pointer) is undefined behaviour.

A simple solution would be to make d a static variable, such that its lifetime is will last until the program ends:

int* func() {
    static int d[3];

But note that this variable d will then exist only once in the program, such that the result of one call may be altered later. Consider to use std::vector<int> as return type or passing the array to be altered as function parameter void func(int[] d) {

Answer 2

Easiest workaround would be to use std::array (which is the prefered method)

#include <array>

std::array<int, 3> do_something() {
   std::array<int, 3> arr = {1, 2, 3};
   return arr;
}

int main() {
    do_something()[0];
}

Best part about this is that the compiler will in-place construct this, meaning no performance was lost due to the copy.