如果一个函数试图一次返回多个值，会发生什么？

先决条件：如何从 C 或 C++ 中的函数返回多个值？

在处理函数调用时，经常会出现使用 return 语句的情况。通常，根据函数的返回类型，无论是原始数据类型（如整数、字符等）还是非原始数据类型（如数组、字符串、向量等），都只会返回一件事。
但是如果我们尝试通过 return 语句返回多个值会发生什么？

本文重点讨论通过 return 语句返回多个值的场景。

预测输出：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function declaration and
// definition
int AddSub(int c, int d)
{
    int x, y;
    x = c - d;
    y = c + d;
 
    // Returning two integers
    // instead of one
    return (x, y);
}
 
// Driver code
int main()
{
    // Initializing the variables
    int i = 100, j = 200, k;
 
    // Calling AddSub function
    k = AddSub(j, i);
 
    // Printing k
    cout << "The value of k = " << k;
}

C

// C program to implement
// the above approach
#include 
 
// Function declaration and
// definition
int AddSub(int c, int d)
{
    int x, y;
    x = c - d;
    y = c + d;
 
    // Returning two integers
    // instead of one
    return (x, y);
}
 
// Driver code
int main()
{
    // Initializing the variables
    int i = 100, j = 200, k;
 
    // Calling AddSub function
    k = AddSub(j, i);
 
    // Printing k
    printf("The value of k = %d", k);
}

Java

// Java program to implement
// the above approach
 
import java.util.*;
 
class GFG{
 
// Function declaration and
// definition
static int[] AddSub(int c, int d)
{
    int x, y;
    x = c - d;
    y = c + d;
 
    // Returning two integers
    // instead of one
    return new int[]{x, y};
}
 
// Driver code
public static void main(String[] args)
{
    // Initializing the variables
    int i = 100, j = 200;
    int []k;
 
    // Calling AddSub function
    k = AddSub(j, i);
 
    // Printing k
    System.out.print("The value of k = " +  (k[k.length-1]));
}
}
// This code is contributed by 29AjayKumar

Python3

# Python program to implement
# the above approach
 
# Function declaration and
# definition
def AddSub(c, d):
    x, y = 0, 0;
    x = c - d;
    y = c + d;
 
    # Returning two integers
    # instead of one
    return [x, y];
 
# Driver code
if __name__ == '__main__':
   
    # Initializing the variables
    i = 100; j = 200;
    k = 0;
 
    # Calling AddSub function
    k = AddSub(j, i);
 
    # Printing k
    print("The value of k = ", (k[len(k) - 1]));
     
# This code is contributed by 29AjayKumar

C#

// C# program to implement
// the above approach
using System;
 
public class GFG{
 
// Function declaration and
// definition
static int[] AddSub(int c, int d)
{
    int x, y;
    x = c - d;
    y = c + d;
 
    // Returning two integers
    // instead of one
    return new int[]{x, y};
}
 
// Driver code
public static void Main(String[] args)
{
    // Initializing the variables
    int i = 100, j = 200;
    int []k;
 
    // Calling AddSub function
    k = AddSub(j, i);
 
    // Printing k
    Console.Write("The value of k = " +  (k[k.Length-1]));
}
}
  
// This code is contributed by shikhasingrajput

Javascript

// javascript program to implement
// the above approach// Function declaration and
// definition
function AddSub(c , d)
{
    var x, y;
    x = c - d;
    y = c + d;
 
    // Returning two integers
    // instead of one
    return [x, y];
}
 
// Driver code
 
// Initializing the variables
var i = 100, j = 200;
var k;
 
// Calling AddSub function
k = AddSub(j, i);
 
// Printing k
document.write("The value of k = " +  (k[k.length-1]));
 
// This code is contributed by shikhasingrajput

输出：

The value of k = 300

解释：
你们中的大多数人一定想知道上面代码的输出会是一个“错误”，因为在上面的代码中，超过允许的值被返回给函数调用。
在这种情况下，如果返回多个值而没有采取特殊的预防措施，如数组、指针、结构、引用、元组、类和对象，那么在这种情况下，将返回最后一个值，而之前的所有值都将被忽略。
返回值的数量可以很多，但只会返回最后一个出现的值，也可以忽略括号或大括号“（）”，并且可以不使用大括号编写，只需用逗号分隔多个值，如下面的代码所示：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function declaration and
// definition
int Operations(int c, int d)
{
    int p, q, r, s, t;
    p = c - d;
    q = c + d;
    r = c * d;
    s = c / d;
    t = c % d;
 
    // Returning multiple integers
    // instead of one
    return p, q, r, s, t;
}
 
// Driver code
int main()
{
    // Initializing the variables
    int i = 100, j = 200, k;
 
    // Calling Operations function
    k = Operations(j, i);
 
    // Printing k
    printf("The value of k = %d", k);
}

C

// C program to implement
// the above approach
#include 
 
// Function declaration and
// definition
int Operations(int c, int d)
{
    int p, q, r, s, t;
    p = c - d;
    q = c + d;
    r = c * d;
    s = c / d;
    t = c % d;
 
    // Returning multiple integers
    // instead of one
    return (p, q, r, s, t);
}
 
// Driver code
int main()
{
    // Initializing the variables
    int i = 100, j = 200, k;
 
    // Calling Operations function
    k = Operations(j, i);
 
    // Printing k
    printf("The value of k = %d", k);
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function declaration and
// definition
static int []Operations(int c, int d)
{
    int p, q, r, s, t;
    p = c - d;
    q = c + d;
    r = c * d;
    s = c / d;
    t = c % d;
     
    // Returning multiple integers
    // instead of one
    return new int[]{ p, q, r, s, t };
}
 
// Driver code
public static void main(String[] args)
{
     
    // Initializing the variables
    int i = 100, j = 200;
    int []k;
 
    // Calling Operations function
    k = Operations(j, i);
 
    // Printing k
    System.out.printf("The value of k = %d",
                      k[k.length - 1]);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python program to implement
# the above approach
 
# Function declaration and
# definition
def Operations(c, d):
    p, q, r, s, t = 0,0,0,0,0;
    p = c - d;
    q = c + d;
    r = c * d;
    s = c / d;
    t = c % d;
 
    # Returning multiple integers
    # instead of one
    return [p, q, r, s, t ];
 
# Driver code
if __name__ == '__main__':
 
    # Initializing the variables
    i = 100; j = 200;
    k = 0;
 
    # Calling Operations function
    k = Operations(j, i);
 
    # Printing k
    print("The value of k = ", k[len(k) - 1]);
 
# This code is contributed by shikhasingrajput

C＃

// C# program to implement
// the above approach
using System;
 
public class GFG
{
 
// Function declaration and
// definition
static int []Operations(int c, int d)
{
    int p, q, r, s, t;
    p = c - d;
    q = c + d;
    r = c * d;
    s = c / d;
    t = c % d;
     
    // Returning multiple integers
    // instead of one
    return new int[]{ p, q, r, s, t };
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Initializing the variables
    int i = 100, j = 200;
    int []k;
 
    // Calling Operations function
    k = Operations(j, i);
 
    // Printing k
    Console.Write("The value of k = {0}",
                      k[k.Length - 1]);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program to implement
// the above approach
 
// Function declaration and
// definition
function Operations(c , d)
{
    var p, q, r, s, t;
    p = c - d;
    q = c + d;
    r = c * d;
    s = c / d;
    t = c % d;
     
    // Returning multiple integers
    // instead of one
    return [ p, q, r, s, t ];
}
 
// Driver code
// Initializing the variables
var i = 100, j = 200;
var k;
 
// Calling Operations function
k = Operations(j, i);
 
// Printing k
document.write("The value of k = ",
               k[k.length - 1]);
 
// This code is contributed by shikhasingrajput

输出：

The value of k = 0

解释：
这里输出为 0，因为最后一个变量 t 等于 c % d = 0。所以当返回多个值时， t的值被分配给变量k。