c ++从函数返回类型

Question

In this example I want return int nums[40]
How can I do it?

<RETURNTYPE> test()
{
    int nums[40];

    for (int i=0; i < 40; i++)
    {
        nums[i] = i + 1;
    }

    return nums;
}

Answer 1

You can't do that. For several reasons (including history, and C compatibility, and the rule that decays arrays to pointers in several important cases) you cannot return a raw array; but you can return a struct (or class ....) containing that array.

What you could do in C++11 is to return a std::array<int,40> ; it behaves like your thing, but it is a proper class (with all the usual operations you imagine, and iterator stuff too). So your code becomes:

std::array<int,40> test(void) {
  std::array<int,40> nums;
  for (int i=0; i < 40; i++) {
    nums[i] = i + 1;
  }
  return nums;
}

An optimizing compiler (eg g++ -Wall -O2 -std=c++11 .... using a recent GCC ) will optimize that exactly like your hypothetical return type (or like FISOCPP's answer , which is likely to get compiled into the same object code).

If the size (40) can vary at runtime, use std::vector<int> as the type and declare std::vector> nums{40}; -then you later could nums.resize(17); and the data would then stay in heap (so there is a tiny performance penalty).

Answer 2

The only way to achieve that is to encapsulate your array into a structure (or union - which is not relevant).

typedef struct { int ra[20]; } rvtyp;

rvtyp test()
{
    int nums[40];

    for (int i=0; i < 40; i++)
    {
        nums[i] = i + 1;
    }

    return *(rvtyp*)&nums;
}

Then you can access your returned value by 'test().ra'.

Answer 3

You need dynamically allocate memory (it is on stack now). Return type will be int * then.