A Qustion about iostream

[want.to.be.professer]

Yesterday, my friend ask me a question about iostream.I know the
result ,but can't explain.The program is below:

---------------------------------------------------------------------------------------------------------------------------------------------------
#include <iostream>

using namespace std;

int fun( int x )
{
cout << "____in fun()____" << endl;
return x;
}

int main()
{
int x = 9;
cout << "b = " << x << fun2( 2 ) << "ads" << endl;
return 0;
}

-------------------------------------------------------------------------------------------------------------------------------------------------------
The program run as below:

1.function "fun()"
2.operator << ( basic_iostream<char, _Traits>& _Ostr, const char*
_Val )
( _Val== "b = " )
3. a member function : operator<<(int _Val)
( _Val == 9 )
4. a member function : operator << ( int _Val )
( _Val == 2 )
5. operator << ( basic_iostream<char, _Traits>& _Ostr, const char*
_Val )
( _Val== "ads" )

But why it run like this? I can't explain.So I put in here to wait a
answer.

 

[want.to.be.professer]

Because the C++ standard does not specify the evaluation order for
individual parts of an expression. A C++ compiler is free to choose any
evaluation order. Consider a simpler example:

a= b() + c();

The C++ standard does not specify whether the b() function should be called
before c(), or if c() should be called first, then b(), then the results be
added. A compiler is free to generate code that calls the two functions in
any order, before adding the results, and storing the result of the addition
in a. The C++ standard only defines "sequence points", where all evaluations
must be complete. In the above example, the semicolon is the sequence point.
This is somewhat of an oversimplification, but it gives you the general
idea.

In your example:

cout << "b = " << x << fun2( 2 ) << "ads" << endl;

The function call to fun2() may be executed at any time before the code that
corresponds to " << "ads" ", as long as the return value from the function
call is sent to the stream in the right "spot".

application_pgp-signature_part
< 1KViewDownload

Thanks.Now I know "sequence points", and I saw a paragraph from other
web site,
which says that C++ group the "sequence points" into 5 situation:

At the end of a full expression
After the evaluation of all function arguments in a function call and
before execution of any expressions in the function body
After copying of a returned value and before execution of any
expressions outside the function
After evaluation of the first expression in a&&b, a||b, a?b:c, or
a,b
After the initialization of each base and member in the constructor
initialization list

Who has the Stardard C++ Doc about "sequence points" ?