如何在另一个函数内部使用带有参数的函数的返回值 [英] How to use the return value of a function with parameters inside another function

问题描述

所有这些函数都在int main()之外:

All these functions are outside the int main():

int func1(int x) {

    int v1 = 6 * x;
    return v1; // the input argument will be 2, so v1 = 12
}

int func2(){
    int v2 = func1() / 4; // It's suppose to be 12 / 4
    //I get "too few arguments in function call"
 }

我知道第二个函数中的 func1()缺少参数，这就是发出参数过多"警告的原因.我想知道的是如何使用 func2()内的 func1()的返回值作为变量的值.

I know func1() inside 2nd function lacks arguments, that's the reason for the "too few arguments" warning. What I want to know is how can I use the return value of func1() inside func2(), as a value to a variable.

像这样，没有参数的函数可以正常工作:

Like this, with functions without parameters, which work fine:

     int sum() {

        int v1 = 10;
        int v2 = 4;
        int v3 = v1 + v2; // 10 + 4 = 14
        return v3; // v3 = 14
    }

    int sub() {

        int v4 = sum() - 7; // 14 - 7 = 7;
        return v4; // v4 = 7
 }

谢谢.

对不起，当我说我了解警告的原因时，我以为自己很清楚.我的错.

EDIT : Sorry, I thought I had made myself clear when I said I understood the reason for the warning. My mistake.

我需要用户在功能1中的输入是可变的.

I need the input from the user in function 1 to be variable.

因此，在main()中，当提示用户提供数字时，它将被传递给函数1.参数不是固定的，而是可变的.

So, in main(), when the user is prompted to give a number, it will be passed to function 1. The argument is not fixed, it is variable.

这是主体背后的想法:

int main() {
    int x = 0;
    cin >> x;
    func1(x); // send user input to func1, then to func2
    int e = func2(); // return result of above to int e
    cout << e << "\n";


}

推荐答案

您会收到此错误，因为您必须为函数提供所需的所有参数.您没有为函数提供参数，因此该函数不起作用.sum()不接收任何参数，因此必须这样调用

You get this error because you have to give the function all the arguments it needs. You are not giving an argument to a function so it will not work. sum() doesnt receive any arguments so it has to be called like this

sum()

func1()接受一个参数，因此请将其更改为不接受参数的函数

func1() takes one argument so either change it to a function that takes no arguments

int func1() {

    int v1 = 6 * 2;
    return v1; // the input argument will be 2, so v1 = 12
}

但是它表明您没有考虑自己在做什么，这一点都没用

but it shows that you havent thought about what you are doing and its kinda useless

或这样调用它而不更改功能

or call it like this without changing the function

int v2 = func1(2) / 4;

编辑
因此，您可以更改func2以获得一个参数并将其传递给func1

EDIT
So you can change func2 to also get one argument and pass it to func1

int func2(int x){
    int v2 = func1(x) / 4;
 }

主要是

int x;
cin>>x;
func2(x);

这也是您的func2()应该返回int但又没有return语句的小东西

also small thing your func2() should return int but you dont have a return statement

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