Question

我似乎无法弄清楚如何使这项工作得到任何帮助将会非常感激，因为我已经尝试了几天:(

#include <iostream>
#include <math.h>

using namespace std;

int input;
int sum;
int number1;
int number2;
int number3;

void isArmstrong (int input, int sum)
{
    if (input == sum)
        cout << input << " is an Armstrong number" << endl;

    if (input != sum)
        cout << input << " is not an Armstrong number" << endl;
}

cubeOfDigits不返回输入和总和为isArmstrong，（返回输入，求和）错误如下：表达式结果未使用[-Wunused-value]

int cubeOfDigits (int input, int sum, int number1, int number2, int number3)
{
    cout << "Enter an integer between 0-999" << endl;
    cin >> input;

    number1 = input / 100;
    number2 = input % 100;
    number3 = number2 % 10;

    sum = pow(number1, 3) + pow(number2, 3) + pow(number3, 3);

    isArmstrong(input, sum);

    return input,sum;
}

主要调用cubeOfDigits

int main(void)
{
    cout << "Welcome" << endl;

    cubeOfDigits(input, sum, number1, number2, number3);

    return 0;
}

Answer 1

将cubeofdigits函数从（int）cubeofdigits更改为（void）cubeofdigits，然后从cubeofdigits函数的末尾删除return语句。然后它必须工作。

Answer 2

我认为你的意图是......

#include <iostream>
#include <math.h>

using namespace std;

int input;
int sum;
int number1;
int number2;
int number3;

void isArmstrong(int input, int sum){

if(input == sum)
cout << input << " is an Armstrong number" << endl;
if(input != sum)
cout << input << " is not an Armstrong number" << endl;

}

int cubeOfDigits (int input)
{



number1 = input/100;
number2 = input % 100;
number3 = number2 % 10;
number2 = number2/10;
sum = pow(number1,3) + pow(number2,3) + pow(number3,3);


return sum;

}

int main(void){
cout << "Welcome" << endl;
cout << "Enter an integer between 0-999" << endl;
cin >> input;
isArmstrong(input,cubeOfDigits(input));
return 0;

}

如何将值返回给另一个函数

2 个答案: