字符串作为 C++ 构造函数中的参数
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原文地址: http://stackoverflow.com/questions/8546869/
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String as a parameter in a constructor in C++
提问by System
class example {
private:
char Name[100];
public:
example(){strcpy(Name, "no_name_yet");}
example(char n[100]){strcpy(Name, n);}
};
int main() {
example ex;
char n[100];
cout<<"Give name ";
cin>>n;
example();
}
I want to use the constructor with the parameter so that when the user gives a name it gets copied to the name variable. How can I use the constructoe with the parameter instead of the default one? I tried
我想将构造函数与参数一起使用，以便在用户提供名称时将其复制到 name 变量中。如何使用带有参数的构造函数而不是默认的构造函数？我试过
example(n)
example(char n)
example(*n)
example(n[100])
but none of them work...
但他们都没有工作......
回答by dasblinkenlight
It is example my_instance_of_example(n).
它是example my_instance_of_example(n)。
I must note, however, that using char arrays for strings is not what you do in C++. You should use std::stringinstead, it gives you a lot more flexibility.
但是，我必须注意，对字符串使用 char 数组并不是您在 C++ 中所做的。您应该改用std::string它，它为您提供了更大的灵活性。
回答by Xeo
Easy:
简单：
#include <string>
#include <iostream>
class example {
private:
std::string name;
public:
example() : name("no name yet"){}
example(std::string const& n) : name(n){}
};
int main() {
example ex;
std::string n;
std::cout << "Give name ";
std::cin >> n;
example ex(n); // you have to give your instance a name, "ex" here
// and actually pass the contructor parameter
}
