如何从c ++中的函数返回多个值？

I know this has been asked before but I still don't know how to do it. I Have to write a function which returns the number of times 2, 5 and 9 appear in an array.

我知道以前曾经问过,但我还是不知道怎么做。我必须编写一个函数,它返回数组中出现的2,5和9的次数。

    include <iostream>

    int twofivenine(int array[], int n)
    {
        int i = 0;
        int num_2 = 0;
        int num_5 = 0;
        int num_9 = 0;

        for ( i = 0;  i < n; i++ ){
            switch(){
                case (array[i] == 2):
                    num_2++;
                case (array[i] == 5):
                    num_5++;
                case (array[i] == 9):
                    num_9++;
            }

        }

       return ;
    }


    int main()
    {
        int array[6] = {2,2,3,5,9,9};

        std::cout << "2: 5: 9:" << twofivenine(array, 6) << std::endl;
    }

I'm just not sure how to return (num_2, num_5, and num_9)

我只是不确定如何返回(num_2,num_5和num_9)

4 个解决方案

#1

Can use std::tuple

可以使用std :: tuple

std::tuple<int, int, int > twofivenine( int array[], int n)
{
  // 
  return make_tuple( num_2, num_5, num_9 );
}

  auto x = twofivenine( array, 6 );
  std::cout << std::get<0>( x ) << '\n'
            << std::get<1>( x ) << '\n'
            << std::get<2>( x ) << '\n' ;

#2

There are a number of ways to approach this problem.

有很多方法可以解决这个问题。

1. Pass the values by reference. You can call a function such as the following:

通过引用传递值。您可以调用以下函数:

Example:

void foo(int &a, int &b, int &c)
{
    // modify a, b, and c here
    a = 3
    b = 38
    c = 18
}

int first = 12;
int second = 3;
int third = 27;
foo(first, second, third);
// after calling the function above, first = 3, second = 38, third = 18

2. Store the values to return in a data type. Use a data type from the standard library such as std::vector, std::set, std::tuple, etc. to hold your values then return that entire data member.

存储值以在数据类型中返回。使用标准库中的数据类型(如std :: vector,std :: set,std :: tuple等)来保存值,然后返回整个数据成员。

Example:

std::vector<int> foo()
{
    std::vector<int> myData;
    myData.pushBack(3);
    myData.pushBack(14);
    myData.pushBack(6);
    return myData;
}
// this function returns a vector that contains 3, 14, and 6

3. Create an object to hold your values. Create an object such as a struct or a class to hold your values and return the object in your function.

创建一个对象来保存您的值。创建一个对象(如结构或类)来保存值并返回函数中的对象。

Example:

struct myStruct
{
    int a;
    int b;
    int c;
};

myStruct foo()
{
    // code here that modifies elements of myStruct
    myStruct.a = 13;
    myStruct.b = 2;
    myStruct.c = 29;
    return myStruct;
}
// this function returns a struct with data members a = 13, b = 2, and c = 29

The method you choose will ultimately depend on the situation.

您选择的方法最终取决于具体情况。

#3

Pass objects in by reference, ie

通过引用传递对象,即

void twofivenine(int array[], int n, int &num_2, int &num_5, int &num_9)
{
    //Don't redeclare num_2...
}

Call like so:

像这样打电话:

int num_2, num_5, num_9;
twofivenine(array, 6, num_2, num_5, num_9);

#4

Return a struct by value which has the counts as the data members:

按值返回结构,其中count为数据成员:

struct Result {
   int num_3;
   int num_5;
   int num_9;
};

Result twofivenine(int array[], int n)
{
.
.
.
    return Result{num_2, num_5, num_9};
}

and in main:

在主要:

Result result(twofivenine(array, 6));
std::cout << "2: " << result.num_2 << "5: " << result.num_5 << "9: " << result.num_9 << std::endl;

Most compilers will do RVO (return-value-optimization) where the twofivenine function will directly write to the result struct avoiding a struct copy.

大多数编译器都会执行RVO(返回值优化),其中twofivenine函数将直接写入结果结构,从而避免使用结构副本。