g++中，一對(duì)象可以調(diào)用non-const成員，卻無法被non-const引用，這種行為符合C++標(biāo)準(zhǔn)嗎

bruceteen

在g++中，一個(gè)對(duì)象可以調(diào)用其non-const成員，卻無法被non-const方式所引用，這種行為符合C++標(biāo)準(zhǔn)嗎？

1. struct A
   
2. {
   
3.     void memfun()
   
4.     {
   
5.     }
   
6. 
   
7.     int val;
   
8. };
   
9. 
   
10. A foo( A& a )
    
11. {
    
12.     return a;
    
13. }
    
14. 
    
15. int main()
    
16. {
    
17.     //foo( A() ); // gcc4.7.2報(bào)：error: ……from an rvalue of type 'A'
    
18.     A a;
    
19.     foo(a).memfun(); // 既然是rvalue，為什么卻可以調(diào)用non-const成員memfun？
    
20.    
    
21.     return 0;
    
22. }

復(fù)制代碼

myworkstation

這是符合標(biāo)準(zhǔn)的，C++ 標(biāo)準(zhǔn)并沒有說Rvalue一定是const的，也就是說rvalue有cv的修飾，2003標(biāo)準(zhǔn)中有如下規(guī)定:

"Some rvalue expressions—those of class or cv-qualified class type—also refer to objects"
"Class rvalues can have cv-qualified types;"

而對(duì)于rvalue對(duì)象調(diào)用成員函數(shù)有如下說明：
“Expressions such as invocations of constructors and of functions that return a class type refer to objects, and the implementation
can invoke a member function upon such objects, but the expressions are not lvalues.”
上面這句話說可以在作為rvalue(非lvalue既rvalue)表達(dá)式的（作為返回對(duì)象的函數(shù)結(jié)果值）情況下調(diào)用成員函數(shù)。且沒有標(biāo)準(zhǔn)要求是const成員函數(shù)。

根據(jù)上述rvalue與cv的結(jié)合性及成員函數(shù)的調(diào)用說明上來講g++的行為是合乎標(biāo)準(zhǔn)規(guī)定的。

而在VC下 foo( A() )編譯沒有問題是因?yàn)槟�認(rèn)啟用了Microsoft Extensions to C and C++。其有中這么一項(xiàng)：
Passing a Non-Const Pointer Parameter to a Function that Expects a Reference to a Const Pointer Parameter
因?yàn)橐�用是lvalue所以g++下編譯不過是正常的。

可以參考：http://msdn.microsoft.com/en-us/library/34h23df8.aspx


網(wǎng)絡(luò)上其它的相關(guān)說法：
Well, for an non-const rvalue, you can even call the
non-const member function for that object ... That means,
you can even *change* the rvalue if you want. Therefore,
obtaining the "this" pointer is not so strange compared
with these behaviors in my mind...



For non-class types, the const keyword is ignored for rvalues;
an rvalue of a non-class type is never "cv-qualified". (For
class types, it is relevant, since you can call member functions
on an rvalue; if the rvalue is const, you can only call const
member functions on it.) The return value of a function is an
rvalue, so there's really no point in declaring it const. (Note
that except for calling a member function on it, there is no
possible way to modify an rvalue.)

hellioncu

A foo( const A& a ) 這樣應(yīng)該都可以了

bruceteen

回復(fù) 2# hellioncu
我想知道C++標(biāo)準(zhǔn)是怎么規(guī)定的，因?yàn)閂C9.0中 foo( A() ); 是可以編譯通過的
   

群雄逐鹿中原

foo( A() ); ，參數(shù)非const方式時(shí)，不能通過編譯，的確很不爽。

C++一定藏著一個(gè)羞于啟齒的秘密，這個(gè)秘密違背了其標(biāo)準(zhǔn)的和諧。
這個(gè)秘密是什么呢？？

bruceteen

回復(fù) 2# myworkstation
謝謝，我記得當(dāng)年有人回答過我（我忘了），當(dāng)然，沒你說得這么詳細(xì)。再次感謝！