pow() 返回 0 (C++)

Question

有人可以解释为什么下面代码中的pow()在程序运行时返回0，而不是实际的计算吗？我是编程新手，我完全被难住了。

感谢您的任何帮助。

#include <iostream>
#include <math.h>
#include <windows.h>
using namespace std;
//Prototypes:

double phiExpo;

double phiNegExpo;    

double opt1f(double phi, double userInput){
      return userInput * phi;}    

double opt2f(double phi, double userInput){
      return userInput / phi;}  

double opt3f(){
      return phiExpo;}   

double opt4f(){
      return phiNegExpo;}   

double phiExpof(double phi, double userInput){
      pow(phi, userInput);}

double phiNegExpof(double phi, double userInput){
      pow(phi,-userInput);}

//Execute program:

int main()
{
    double userInput;
    int userChoice;
    double phi = 1.61803399;
    bool quit = false; 
    int userChoice2;
    cout << "I want to (press corresponding number, then enter):" << endl;
    cout << endl;
    startchoices: 
    cout << "1. Multiply by phi:" << endl;
    cout << "2. Divide by phi:" << endl;
    cout << "3. Exponentiate phi:" << endl;
    cout << "4. Negatively exponentiate phi:" << endl;
    cout << "5. Quit." << endl;    
    cout << endl;
    cin >> userChoice;
    cout << endl;  
    do {    
       switch (userChoice){

              case 1:
              cout << "Enter number for multiplication: ";
              cin >> userInput;
              cout << endl;
              cout << "Phi multiplied by " << userInput << ": ";   
              cout << opt1f(phi, userInput) << endl;
              cout << endl;
              Sleep(2000);
              cout << "1. Continue." << endl;
              cout << "2. Return to menu." << endl;
              cout << endl;
              cin >> userChoice2;
              cout << endl;
              if(userChoice2 > 1){
                  goto startchoices;}            
              break;

              case 2:
              cout << "Enter number for division: ";
              cin >> userInput;
              cout << endl;
              cout << "Phi divided by " << userInput << ": ";
              cout << opt2f(phi, userInput); 
              cout << endl;
              Sleep(2000);
              cout << "1. Continue." << endl;
              cout << "2. Return to menu." << endl;
              cout << endl;
              cin >> userChoice2;
              cout << endl;
              if(userChoice2 > 1){goto startchoices;}               
              break; 

              case 3:
              cout << "Enter number to exponentiate phi by: ";
              cin >> userInput;
              cout << endl;
              cout << "Phi to the power of " << userInput << ": ";
              cout << opt3f();
              cout << endl;
              Sleep(2000);
              cout<<endl;
              cout << "1. Continue." << endl;
              cout << "2. Return to menu." << endl;
              cout << endl;
              cin >> userChoice2;
              cout << endl;
              if(userChoice2 > 1){goto startchoices;}               
              break; 
        }
    }
}

Answer 1

你从来没有真正调用过pow。选择 3 时，您仅调用 opt3f，它仅返回全局变量 phiExpo，该变量初始化为 0因为它是全球性的。然后，您还需要从 phiExpof 函数返回，就像其他人已经指出的那样。

Answer 2

它可能不会返回 0。相反，您不会返回 pow 的结果：

double phiExpof(double phi, double userInput){
      return pow(phi, userInput);
}

当您没有显式返回值时，您将得到未定义的行为，在本例中为 0。

注意：
我没有注意到其他代码......这是一个问题。另一个是你实际上并没有调用 phiExpof。相反，您将返回 phiExpo 这是一个全局变量。

Answer 3

你怎么知道它返回0？您的代码甚至不检查返回值。