[英]Problem with chaining operator<< and operator++
I'm learning C++ and I have this problem: 我正在学习C ++,但遇到了以下问题:
#include <iostream>
using namespace std;
class test
{
public:
test(){};
test(int i):var{i}{};
test& operator++(){++var; return this;}
test operator++(int dummy){test tmp =*this;var++;return tmp;}
friend ostream& operator<<(ostream&, const test&);
private:
int var;
};
ostream& operator<<(ostream& o, const test& obj)
{
o<<obj.var;
return o;
}
int main()
{
test obj{2};
cout << obj << endl;
obj++;
cout << obj << endl;
cout << obj <<' '<< ++obj<<endl;
return 0;
}
the output i expected was: 2 3 3 4 我期望的输出是:2 3 3 4
instead i have: 2 3 4 4 相反,我有:2 3 4 4
if i replace the last increment ++obj with obj++ the situation is even more weird: 2 3 4 3 如果我用obj ++替换最后一个增量++ obj,情况将更加奇怪:2 3 4 3
it's like the stream is read in the opposite way, can you help me? 就像以相反的方式读取信息流,您能帮我吗?
Let's examine how the line 让我们检查一下线
cout << obj << ' ' << ++obj << endl;
is translated. 已翻译。
Step 1. 第1步。
cout << obj
becomes 变成
// A non-member function.
operator<<(cout, obj)
Step 2. 第2步。
operator<<(cout, obj) << ' '
becomes 变成
// Also a non-member function.
operator<<(operator<<(cout, obj), ' ')
Step 3. 第三步
operator<<(operator<<(cout, obj), ' ') << ++obj
becomes 变成
// Also a non-member function.
operator<<(operator<<(operator<<(cout, obj), ' '), ++obj)
Step 4. 第四步。
operator<<(operator<<(operator<<(cout, obj), ' '), ++obj) << endl;
becomes 变成
// A member function.
operator<<(operator<<(operator<<(cout, obj), ' '), ++obj).operator<<(endl);
That's the entire line. 这就是整行。
In such an expression there is no guarantee that operator<<(cout, obj)
will be executed before ++obj
. 在这样的表达式中,不能保证operator<<(cout, obj)
将在++obj
之前执行。 It appears that in your platform, ++obj
is executed before operator<<(cout, obj)
is executed. 看来在您的平台上, ++obj
是在执行operator<<(cout, obj)
执行的。 That explains the behavior. 这解释了行为。
Please note that the standard has changed. 请注意,标准已更改。 If you are able to use C++17, you will get the expected behavior. 如果您能够使用C ++ 17,则将获得预期的行为。
For starters the data member i can be uninitialized if the default constructor will be used. 对于初学者,如果使用默认构造函数,则数据成员i可以未初始化。
Either declare the data member like 要么像这样声明数据成员
int var = 0; int var = 0;
Or redefine the default constructor for example by using the delegating constructor. 或重新定义默认构造函数,例如通过使用委托构造函数。
class test
{
public:
test() : test( 0 ){};
test(int i):var{i}{};
// ...
The pre-increment operator should look like 预增量运算符应该看起来像
test& operator++(){++var; return *this;}
^^^^^
In the post-increment operator the identifier dummy
is not used. 在后递增运算符中,不使用标识符dummy
。 So remove it 所以删除它
test operator++( int ){test tmp =*this;var++;return tmp;}
This statement 这个说法
cout << obj <<' '<< ++obj<<endl;
has undefined behavior because reading writing the object obj
are not sequenced. 具有未定义的行为,因为读写对象obj
时未排序。
You have to split this statement into two statements 您必须将此语句分为两个语句
cout << obj <<' ';
cout << ++obj<<endl;
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