"不能用作函数错误"

Question

我正在编写一个使用不同.cpp文件中的函数的简单程序.我的所有原型都包含在头文件中.我将一些函数传递给其他函数,我不确定我是否正确执行.我得到的错误是"'functionname'不能用作函数".它说不能使用的growthRate功能是功能和estimatedPopulation功能.数据通过输入函数(我认为它正在工作)进入.

谢谢!

头文件:

#ifndef header_h
#define header_h

#include <iostream>
#include <iomanip>
#include <cstdlib>


using namespace std;

//prototypes
void extern input(int&, float&, float&, int&);
float extern growthRate (float, float);
int extern estimatedPopulation (int, float);
void extern output (int);
void extern myLabel(const char *, const char *);

#endif

growthRate函数:

 #include "header.h"

float growthRate (float birthRate, float deathRate, float growthrt)     
{    
    growthrt = ((birthRate) - (deathRate))
    return growthrt;   
}

估计人口功能:

    #include "header.h"

int estimatedPopulation (int currentPopulation, float growthrt)
{
    return ((currentPopulation) + (currentPopulation) * (growthrt / 100);
}

主要:

#include "header.h"

int main ()
{
    float birthRate, deathRate, growthRate;
    char response; 
    int currentPopulation, years, estimatedPopulation;

    do //main loop
    {  
        input (currentPopulation, birthRate, deathRate, years);
        growthRate (birthRate, deathRate, growthrt);

        estimatedPopulation (currentPopulation, growthrt);
        output (estimatedPopulation (currentPopulation, growthrt));
        cout << "\n Would you like another population estimation? (y,n) ";
        cin >> response;
    }          
    while (response == 'Y' || response == 'y');

    myLabel ("5-19", "12/09/2010");   

    system ("Pause");

    return 0;
}    

Answer 1

您将growRate用作变量名和函数名.该变量隐藏了该函数,然后您尝试使用该变量,就好像它是函数一样 - 这是无效的.

重命名局部变量.