[英]The + operator, difference between class types and built-in types?
I'm new to C++. 我是C ++的新手。 The book I read tells me that if the plus (
+
) operator has been overloaded for some class object, say, the string
class, to make this problem more concrete. 我读过的这本书告诉我,如果加号(
+
)运算符已经为某个类对象(比如string
类)重载,那么这个问题就更具体了。
#include<iostream>
#include<string>
using namespace std;
int main()
{
string s1("abc");
string s2("def");
string s3("def");
cout<<(s1+s2=s3)<<endl;
int x=1;
int y=2
int z=3;
cout<<(x+y=z)<<endl;
return 0;
}
As you may expect, the first cout
statement is correct while the second is wrong. 正如您所料,第一个
cout
语句是正确的,而第二个是错误的。 The compiler complaints x+y
is not a modifiable lvalue. 编译器投诉
x+y
不是可修改的左值。 My question is why the +
operator returns a modifiable lvalue for string
objects but not for int
? 我的问题是为什么
+
运算符为string
对象返回一个可修改的左值但不为int
?
It does not return a modifiable lvalue for string. 它不会为字符串返回可修改的左值。 It returns a temporary object, and
s1+s2
and x+y
are both rvalues. 它返回一个临时对象,
s1+s2
和x+y
都是rvalues。
However, objects of class type may have overloaded operator=
, which string
does. 但是,类类型的对象可能具有重载的
operator=
,即string
。 You are allowed to call member functions on rvalues. 您可以在rvalues上调用成员函数。
The difference between the two cases is in =
(not +
) 这两种情况之间的区别在于
=
(未+
)
For std::string
, s1 + s2 = s3
is in fact: 对于
std::string
, s1 + s2 = s3
实际上是:
(operator+(s1, s2)).operator =(s3)
s1 + s2
returns a rvalue s1 + s2
返回一个右值
Member methods can be applied to temporary also. 成员方法也可以应用于临时。
Since C++11, we have the lvalue/rvalue qualifier for method, 从C ++ 11开始,我们有方法的左值/右值限定符,
so you may forbid o1 + o2 = o3
for your custom type with: 所以你可以禁止
o1 + o2 = o3
为你的自定义类型:
struct Object
{
Object& operator =(const Object& rhs) & ; // Note the & here
};
so Object::operator =
can only be applied to lvalue. 所以
Object::operator =
只能应用于左值。
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