函数不接受 1 个参数 c++

Posted 2023-02-22

【中文标题】函数不接受 1 个参数 c++

【英文标题】：function does not take 1 arguments c++

【发布时间】：2014-03-05 17:40:24

【问题描述】：

我的代码有问题，因为我无法弄清楚为什么会收到错误消息。代码如下：

using namespace std;

void presentValue();
bool stringChar();
bool stringVal();
double futureValConv();

int main()

  cout << "Welcome to the Present Value Interest Calculator!\n\"First, let me collect some data." << endl << endl;
  presentValue();
  return 0;


void presentValue()

  //declare variables
  //Response value intialized as x for debugging
  char response = 'x';

  while (response != 'n' || response != 'N')
  
    //declare variables
    double intRate = 0;
    string futureValString;
    double futureVal;
    double years = 0;

    //Simple present value equation
    double presentVal = futureVal / pow((intRate + 1), years);

    cout << "What's your Interest Rate?  ";
    cin >> intRate;
    cout << "OK, and what's your desired Future Value? [Help  ";
    cin >> futureVal;
    //Run descending help program that won't allow escape without a double value
    **futureVal = futureValConv(futureValString);**
    cout << endl << endl << "And finally, how many years would you like to save your money for?  ";
    cin >> years;
    cout << endl << "You've made it this far!!!";
    cout << endl << endl << presentVal;

  

inline double futureValConv(string somestring)

  //delcare variables
  double newString = 0;

  **if (stringChar(somestring))**
  
    cout << endl << "Future Value is the amount you would like to have in your account in the future.\n\n";
    cout << "How much might that be?  ";
    cin >> somestring;
    futureValConv(somestring);

  
  **else if(stringVal(somestring))**
  
    //Convert the Future Value String to a double for future use
    newString = atoi(somestring.c_str());
    
  else
  
    cout << "Please enter a proper value.  ";
    futureValConv(somestring);
  

  return newString;


bool stringChar(string response)

  //declare variables
  char answer = response[0];
  bool status = false;

  if (answer == 'H' || answer == 'h')
  
    status = true;
    return status;
  


bool stringVal(string response)

  //declare varialbes
  int answer = atoi(response.c_str());
  bool status = false;
  int powZero = (answer, 0);

  if (powZero == 1)
  
    status = true;
    return status;
  

我发布了我的大部分代码，因为我无法理解这里发生了什么。它告诉我 stringChar 和 stringVal 不带 1 个参数，以及 futureValConv。我正在尝试运行一个函数来检查字符串的值，并在做出决定之前确定该值是什么。在三个实例中的两个实例中，函数调用自身再次运行，直到用户输入可接受的双精度数（我通过运行 0 的幂来检查结果是否为 1。）我已将生成错误的三行加粗.有趣的是，即使注释掉函数 futureValConv 并且从不调用其他函数，我仍然得到三个错误中的两个。

【参考方案1】：

您在顶部声明了这一点：

bool stringChar();
bool stringVal();

所以编译器期望 stringChar 和 stringVal 函数没有参数。 将声明更改为：

bool stringChar(string response);
bool stringVal(string response);

【讨论】：

特别说明：这不是C，空括号表示参数个数未知。

显然我应该在实现函数原型之前多了解一点。谢谢