嗨,我正在制作一个包含3个类的程序,当我使用成员初始化列表时,我收到一条错误,说“没有重载函数的实例”people :: people“匹配指定的类型:
的main.cpp
#include <iostream>
#include "conio.h"
#include <string>
#include "birthday.h"
#include "people.h"
using namespace std;
void main(){
birthday birthObj (30, 06, 1987);
people me("The King",birthObj);
_getch();
}
BIRTHDAY.h
#pragma once
class birthday
{
public:
birthday(int d, int m, int y);
void printdate();
private:
int month;
int day;
int year;
};
BIRTHDAY.cpp
#include "birthday.h"
#include <iostream>
#include "conio.h"
#include <string>
using namespace std;
birthday::birthday(int d, int m, int y)
{
month = m;
day = d;
year = y;
}
void birthday::printdate()
{
cout << day << "/" << month << "/" << year;
}
PEOPLE.h
#pragma once
#include <iostream>
#include "conio.h"
#include <string>
#include "birthday.h"
using namespace std;
class people
{
public:
people(string x, birthday bo);
void printInfo();
private:
string name;
birthday dateOfBirth;
};
PEOPLE.cpp
#include "people.h"
#include <iostream>
#include "conio.h"
#include <string>
#include "birthday.h"
using namespace std;
people::people()
: name(x), dateOfBirth(bo)
{
}
void people::printInfo()
{
cout << name << " was born on ";
dateOfBirth.printdate();
}
答案 0 :(得分:1)
People.cpp应该是:
people::people(string x, birthday bo) : name(x), dateOfBirth(bo) { }
答案 1 :(得分:0)
PEOPLE.cpp中的构造函数具有错误的签名:
应该是
people :: people(string x,birthday bo)
而不是
人::人()
答案 2 :(得分:0)
people::people()
: name(x), dateOfBirth(bo)
{
}
你忘记了这个构造函数的参数。
答案 3 :(得分:0)
您没有实现people(string x, birthday bo);
构造函数。在你的PEOPLE.cpp中,改变
people::people()
: name(x), dateOfBirth(bo)
到
people::people(string x, birthday bo)
: name(x), dateOfBirth(bo)
答案 4 :(得分:-1)
poeple ctor声明和定义不匹配!