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10鱼币
我想能直接输出类,但是,如果返回const char* 可以运行,但是返回string就报错
所以cout 输出字符串的时候是把字符串看做char*插入到流中的吗,string不行吗
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
class Date
{
int day,month,year;
string s;
public:
Date()
{
}
Date(int pday,int pmonth,int pyear)
{
day = pday;
month = pmonth;
year = pyear;
}
Date(Date& d)
{
day = d.day;
month = d.month;
year = d.year;
}
~Date()
{
}
/*operator string()
{
ostringstream p;
p << year << "," << month << "," << day;
s = p.str();
return s;
}*/
operator const char*()
{
ostringstream p;
p << year << "," << month << "," << day;
return p.str().c_str();
}
};
int main()
{
Date d(29,2,2004);
cout << d << endl;
return 0;
}
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
class A {};
ostream &operator<<(ostream &os, const A &a) {
return os;
}
class Date {
int day, month, year;
string s;
public:
Date(int pday, int pmonth, int pyear) {
day = pday;
month = pmonth;
year = pyear;
}
Date(Date &d) {
day = d.day;
month = d.month;
year = d.year;
}
/*
operator const string() {
ostringstream p;
p << year << "," << month << "," << day;
s = p.str();
return s;
}
*/
/*
operator const char* ()
{
ostringstream p;
p << year << "," << month << "," << day;
return p.str().c_str();
}
*/
operator const A() {
return A();
}
};
int main() {
Date d(29, 2, 2004);
cout << d << endl;
return 0;
}
上面这个代码可以通过编译,因为有这两个
operator const A();
ostream &operator<<(ostream &os, const A &a);
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
class A {};
ostream &operator<<(ostream &os, const A &a) {
return os;
}
class Date {
int day, month, year;
string s;
public:
Date(int pday, int pmonth, int pyear) {
day = pday;
month = pmonth;
year = pyear;
}
Date(Date &d) {
day = d.day;
month = d.month;
year = d.year;
}
operator const string() {
ostringstream p;
p << year << "," << month << "," << day;
s = p.str();
return s;
}
/*
operator const char* ()
{
ostringstream p;
p << year << "," << month << "," << day;
return p.str().c_str();
}
*/
/*
operator const A() {
return A();
}
*/
};
ostream &operator<<(ostream &os, const string &s) {
return os;
}
int main() {
Date d(29, 2, 2004);
cout << d << endl;
return 0;
}
上面这个代码也可以通过编译
因为这两个
ostream &operator<<(ostream &os, const string &s);
operator const string();
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
class A {};
ostream &operator<<(ostream &os, const A &a) {
return os;
}
class Date {
int day, month, year;
string s;
public:
Date(int pday, int pmonth, int pyear) {
day = pday;
month = pmonth;
year = pyear;
}
Date(Date &d) {
day = d.day;
month = d.month;
year = d.year;
}
operator const string() {
ostringstream p;
p << year << "," << month << "," << day;
s = p.str();
return s;
}
/*
operator const char* ()
{
ostringstream p;
p << year << "," << month << "," << day;
return p.str().c_str();
}
*/
/*
operator const A() {
return A();
}
*/
};
int main() {
Date d(29, 2, 2004);
cout << d << endl;
return 0;
}
上面这段代码就无法通过编译了,因为没有这个
ostream &operator<<(ostream &os, const string &s);
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最佳答案
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上面这个代码可以通过编译,因为有这两个
operator const A();
ostream &operator
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