如题所述
第1个回答 推荐于2016-03-16
C++构造函数上唯一的语法限制是,它不能指定返回类型,甚至连void也不行(参考C++ primer14.2类的构造函数)
例子如下:
#include <iostream>
class Account {
public:
// 缺省构造函数
Account();
// 带参数的构造函数
Account(std::string name, unsigned int acct_number, double balance);
void print();
private:
std::string _name;
unsigned int _acct_number;
double _balance;
};
Account::Account()
{
this->_name = "name1";
this->_acct_number = 1;
this->_balance = 1L;
}
Account::Account(std::string name, unsigned int acct_number, double balance)
{
this->_name = name;
this->_acct_number = acct_number;
this->_balance = balance;
}
void Account::print()
{
std::cout<<"name="<<_name
<<"; acct_number="<<_acct_number
<<"; balance="<<_balance<<std::endl;
}
int main()
{
Account a1;
a1.print();
Account a2("name2", 2, 2L);
a2.print();
}
运行结果:
name=name1; acct_number=1; balance=1
name=name2; acct_number=2; balance=2本回答被网友采纳
例子如下:
#include <iostream>
class Account {
public:
// 缺省构造函数
Account();
// 带参数的构造函数
Account(std::string name, unsigned int acct_number, double balance);
void print();
private:
std::string _name;
unsigned int _acct_number;
double _balance;
};
Account::Account()
{
this->_name = "name1";
this->_acct_number = 1;
this->_balance = 1L;
}
Account::Account(std::string name, unsigned int acct_number, double balance)
{
this->_name = name;
this->_acct_number = acct_number;
this->_balance = balance;
}
void Account::print()
{
std::cout<<"name="<<_name
<<"; acct_number="<<_acct_number
<<"; balance="<<_balance<<std::endl;
}
int main()
{
Account a1;
a1.print();
Account a2("name2", 2, 2L);
a2.print();
}
运行结果:
name=name1; acct_number=1; balance=1
name=name2; acct_number=2; balance=2本回答被网友采纳
第2个回答 2011-12-18
int funDisp(int a ,int b)
{
return (a+b);
}
/函数功能,输入两个参数求他们的和。
{
return (a+b);
}
/函数功能,输入两个参数求他们的和。
第3个回答 2011-12-18
你们课本上没有么