返回类型以返回所有字符串

问题描述

我有这个代码，它返回成员函数，但只返回emp_name的最后一个字符。

非常感谢任何帮助。



我尝试过：

I have this code and it returns the member function as written but return only the last character of emp_name.
any assistance is highly appreciated.

What I have tried:

#include <iostream>

using namespace std;

class emp {
public:

	int getInfo(char emp_name, int emp_id_number, int emp_age) {
		cout << "Employee Name: " << emp_name << "\n";
		cout << "Employee Id Number: " << emp_id_number << "\n";
		cout << "Employee Age: " << emp_age;
	}
};
int main() {
	emp emp1;
	emp1.getInfo('Felix', 635932, 23);
}

推荐答案

int getInfo(char emp_name, int emp_id_number, int emp_age) {

您已将 emp_name 参数声明为单个字符，而不是字符串。它应该是：

You have declared the emp_name parameter as a single character, rather than a character string. It should be:

int getInfo(char* emp_name, int emp_id_number, int emp_age) { // char*

// Must be char* here
int getInfo(char *emp_name, int emp_id_number, int emp_age)

// Must pass a string here using double quotes
emp1.getInfo("Felix", 635932, 23);

您已定义 getInfo（）返回 int 但你没有 return 语句。



我建议在编译时使用高警告级别（-Wall with G ++，/ W4在项目设置中使用VS）。那么你的代码至少会有两个警告。

You have defined getInfo() to return an int but you have no return statement.

I suggest to use a high warning level when compiling (-Wall with G++, /W4 in the project settings with VS). Then you would have got at least two warnings for your code.

#include <iostream>

using namespace std;

class emp {
public:

	int getInfo(char* emp_name, int emp_id_number, int emp_age) {
		cout << "Employee Name: " << emp_name << "\n";
		cout << "Employee Id Number: " << emp_id_number << "\n";
		cout << "Employee Age: " << emp_age;
	}
};
int main() {
	emp emp1;
	emp1.getInfo("Felix", 635932, 23);
}

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