来自g ++/gcc编译器的"候选函数不可行".这有什么不对？

Question

我正在尝试编译我的main.cpp,但我现在已经持续两个小时了.这里的问题是将函数作为参数传递,但我认为我做错了.编译器说它无法找到该函数,但我已经在"functions.h"中包含了"newt_rhap(params)".

我做了returnType(*functionName)(paramType),但我可能在这里跳过了一些东西.我朋友的代码不需要最近提到的语法.这有什么不对？

我尝试使用-std = c ++ 11和-std = c ++ 98.gcc/g ++编译器来自我的Xcode命令行工具.

g++ (or gcc) -std=c++98(or 11) main.cpp -o main.out

错误没有区别.

**error: no matching function for call to 'newt_rhap'**

./functions.h:5:8: note: candidate function not viable: no known conversion from 'double' to
      'double (*)(double)' for 1st argument

double newt_rhap(double deriv(double), double eq(double), double guess);

这是代码.

// main.cpp
#include <cmath>
#include <cstdlib>
#include <iostream>
#include "functions.h"

using namespace std;


// function declarations
// =============
// void test(double d);
// =============

int main(int argc, char const *argv[])
{
    //
    // stuff here excluded for brevity
    //

    // =============
    do
    {
        // line with error
        guess = newt_rhap(eq1(guess),d1(guess),guess);

        // more brevity

    } while(nSig <= min_nSig);
    // =============

    cout << "Root found: " << guess << endl;

    return 0;
}

然后分别是functions.h和functions.cpp

// functions.h
#ifndef FUNCTIONS_H_INCLUDED
#define FUNCTIONS_H_INCLUDED

// ===========
double newt_rhap(double deriv(double), double eq(double), double guess);
// ===========


// ===========
double eq1(double x);
double d1(double x);
// ===========


#endif

// functions.cpp
#include <cmath>
#include "functions.h"

using namespace std;

// ===========
double newt_rhap(double (*eq)(double ) , double (*deriv)(double ), double guess)
{
    return guess - (eq(guess)/deriv(guess));
}
// ===========

// ===========
double eq1(double x)
{
    return exp(-x) - x;
}

double d1(double x)
{
    return -exp(-x) - 1;
}

Answer 1

代替:

guess = newt_rhap(eq1(guess),d1(guess),guess);

尝试:

guess = newt_rhap(eq1, d1, guess);

该函数将两个函数和一个猜测作为参数.通过传递eq1(guess)你传递一个double,而不是一个函数(eq1带有参数的评估结果guess)