試圖找到數字的平方根，但代碼不起作用。 C++

Question

我昨天開始C++，不能爲我的生活找出我做錯了什麼。我正在嘗試製作一個小代碼，它需要用戶輸入三個數字，然後計算每個數字的平方根。

#include "stdafx.h" 
#include <iostream> 
#include <cmath> 
using std::cout; 
using std::cin; 

int squareroot(){ //function to find squareroot. 
int sQ1 = sqrt(number1); //variable for the square of number1 
int sQ2 = sqrt(number2); //variable for the square of number2 
int sQ3 = sqrt(number3); //variable for the square of number3 

cout << sQ1 << "\n";//outputs the square of number 1 
cout << sQ2 << "\n";//outputs the square of number 2 
cout << sQ3 << "\n";//outputs the square of number 3 
} 

int main() { // main function 
int number1 = 0; //first number 
int number2 = 0; //second number 
int number3 = 0; //third number 

cout << "type number1"; //asks user to input first number 
cin >> number1; //stores user input into variable number1 

cout << "type number2"; //asks for second number 
cin >> number2; //stores second number into number2 

cout << "type number3"; // asks for third number 
cin >> number3; //stores third number 

cout << number1 << "\n"; //outputs number1 
cout << number2 << "\n"; //ouputs number2 
cout << number3 << "\n"; //outputs number3 

squareroot(); //runs function squareroot() 
} 

Answer 1

您正在嘗試訪問函數中的number1-3而不傳遞其引用。你必須通過PARAMS通過數字或定義number1-3作爲一個全局變量

Answer 2

你應該申報平方根，如：

int squareroot(int number1, int number2, int number3){ //function to find squareroot. 
int sQ1 = sqrt(number1); //variable for the square of number1 
int sQ2 = sqrt(number2); //variable for the square of number2 
int sQ3 = sqrt(number3); //variable for the square of number3 
cout << sQ1 << "\n";//outputs the square of number 1 
cout << sQ2 << "\n";//outputs the square of number 2 
cout << sQ3 << "\n";//outputs the square of number 3 
} 

，然後，把它想：

squareroot(number1, number2, number3); 

無論如何，開方從math.h接受一個double並返回一個double。 參見http://www.cplusplus.com/reference/cmath/sqrt/

Answer 3

現在，這可能是我所能看到的最接近你想要做的事情。我建議閱讀更多的功能和範圍。你應該閱讀的內容並不多。

squareroot()不知道main()中的數字。你將不得不將它們提供給它。此外，避免使用平方根或其他類似函數的整數，整數將截尾（舍入）到最接近的整數，甚至丟失數據。

#include "stdafx.h" 
#include <iostream> 
#include <cmath> 
using std::cout; 
using std::cin; 

void squareroot(float n1, float n2, float n3){ //function to find squareroot. 
    cout << sqrt(n1) << "\n"; //outputs the square of number 1 
    cout << sqrt(n2) << "\n"; //outputs the square of number 2 
    cout << sqrt(n3) << "\n"; //outputs the square of number 3 
} 

int main() { // main function 
    float number1 = 0; //first number 
    float number2 = 0; //second number 
    float number3 = 0; //third number 

    cout << "type number1 "; //asks user to input first number 
    cin >> number1; //stores user input into variable number1 

    cout << "type number2 "; //asks for second number 
    cin >> number2; //stores second number into number2 

    cout << "type number3 "; // asks for third number 
    cin >> number3; //stores third number 

    cout << "\n" << number1 << "\n"; //outputs number1 
    cout << number2 << "\n"; //ouputs number2 
    cout << number3 << "\n"; //outputs number3 

    squareroot(number1, number2, number3); //runs function squareroot() 
} 

Answer 4

給這個例子一拍：

#include <iostream> 
#include <cmath> 
using std::cout; 
using std::cin; 

double squareroot(double number){ //function to find squareroot. 
    return sqrt(number); //variable for the square of number1 
} 

int main() { // main function 

    int number1 = 0; //first number - integer 
    int number2 = 0; //second number - integer 
    int number3 = 0; //third number - integer 

    cout << "\ntype number1"; //asks user to input first number 
    cin >> number1; //stores user input into variable number1 

    cout << "\ntype number2"; //asks for second number 
    cin >> number2; //stores second number into number2 

    cout << "\ntype number3"; // asks for third number 
    cin >> number3; //stores third number 

    cout << squareroot(number1) << "\n"; //outputs number1 
    cout << squareroot(number2) << "\n"; //ouputs number2 
    cout << squareroot(number3) << "\n"; //outputs number3 

}