功能参数太多了

时间:2014-09-24 01:26:29

标签: c++ header

我从头文件中收到此错误:too many arguments to function void printCandidateReport();。我是C ++的新手,只需要一些正确的指导来解决这个错误。

我的头文件如下所示:

#ifndef CANDIDATE_H_INCLUDED
#define CANDIDATE_H_INCLUDED

// Max # of candidates permitted by this program
const int maxCandidates = 10;

// How many candidates in the national election?
int nCandidates;

// How many candidates in the primary for the state being processed
int nCandidatesInPrimary;

// Names of the candidates participating in this state's primary
extern std::string candidate[maxCandidates];

// Names of all candidates participating in the national election
std::string candidateNames[maxCandidates];

// How many votes wone by each candiate in this state's primary
int votesForCandidate[maxCandidates];

void readCandidates ();
void printCandidateReport ();
int findCandidate();
#endif

和调用此头文件的文件:

#include <iostream>
#include "candidate.h"
/**
* Find the candidate with the indicated name. Returns the array index
* for the candidate if found, nCandidates if it cannot be found.
*/
int findCandidate(std::string name) {
    int result = nCandidates;
    for (int i = 0; i < nCandidates && result == nCandidates; ++i)
        if (candidateNames[i] == name)
            result = i;
    return result;
}

/**
* Print the report line for the indicated candidate
*/
void printCandidateReport(int candidateNum) {
    int requiredToWin = (2 * totalDelegates + 2) / 3; // Note: the +2 rounds up
    if (delegatesWon[candidateNum] >= requiredToWin)
        cout << "* ";
    else
        cout << "  ";
    cout << delegatesWon[candidateNum] << " " << candidateNames[candidateNum]
         << endl;
}

/**
* read the list of candidate names, initializing their delegate counts to 0.
*/
void readCandidates() {
    cin >> nCandidates;
    string line;
    getline(cin, line);

    for (int i = 0; i < nCandidates; ++i) {
        getline(cin, candidateNames[i]);
        delegatesWon[i] = 0;
    }
}

为什么我会收到此错误,我该如何解决?

4 个答案:

答案 0 :(得分:7)

在头文件中声明:

void printCandidateReport ();

但实施是:

void printCandidateReport(int candidateNum){...}

将标题文件更改为

void printCandidateReport(int candidateNum);

答案 1 :(得分:3)

错误消息正好告诉您问题所在。

在头文件中,声明没有参数的函数:

void printCandidateReport ();

在源文件中,您使用int类型的参数定义它:

void printCandidateReport(int candidateNum){

将缺少的参数添加到声明中,或将其从定义中删除。

答案 2 :(得分:2)

错误too many arguments to function可以通过消除多余的参数来解决 函数中的(参数)。

发生此错误是因为您的头文件没有参数值,而在实际源代码中您使用的是int参数。

您有两个选择,您可以在函数声明中添加缺少的int参数,或者从函数中完全删除它。

答案 3 :(得分:2)

头文件声明函数printCandidateReport()没有参数,cpp文件使用int参数定义函数。只需将int参数添加到头文件中的函数声明中即可修复它