临时对象，函数参数和隐式转换 [英] temporary object, function parameters and implicit cast

问题描述

在以下情形中：

  struct Foo 
 {
 // ... 
 operator Bar（）{...} //隐式转换为bar 
} 
 
 Foo GetFoo（）{...} 
 void CallMeBar（Bar x）{ ...} 
 
 // ... 
 CallMeBar（GetFoo（））; 
  

修改cast操作符，
$ b

GetFoo 返回类型为Foo的临时对象。此对象是否存活直到CallMe返回？标准说了什么？

我明白如果 CallMe 会接受Foo，临时对象不会被销毁直到 CallMe 返回。我不知道，然而，隐式转换改变这，只有临时 Bar 被保证存活。

---

典型的情况是Foo = CString，Bar = char *由Foo保存（和释放）的数据。

解决方案

无论是什么类型的转换，临时对象都会调用 CallMe （）函数因为C ++标准：

12.2.3 [...]临时对象作为最后一步（词法）包含它们被创建的点的fullexpression（1.9）。 [...]

1.9.12一个表达式不是另一个表达式的子表达式。

In the following scenario:

struct Foo
{
   // ...
   operator Bar() {... }  // implicit cast to Bar
}

Foo GetFoo() { ... }
void CallMeBar(Bar x) { ... }

// ...
CallMeBar( GetFoo() );

[edit] fixed the cast operator, d'oh[/edit]

GetFoo returns a temporary object of Type Foo. Does this object survive until after CallMe returns? What does the standard say?

I understand that if CallMe would accept Foo, the temporary object would not be destroyed until after CallMe returns. I am not sure, however, fi the implicit cast changes this, and only the temporary Bar is guaranteed to survive.

---

A typical case would be Foo = CString, Bar = char *, i.e. Bar referencing data held by (and freed by) Foo.

解决方案

Regardless of the cast, the temporary object(s) will "survive" the call to CallMe() function because of the C++ standard:

12.2.3 [...] Temporary objects are destroyed as the last step in evaluating the fullexpression (1.9) that (lexically) contains the point where they were created. [...]

1.9.12 A fullexpression is an expression that is not a subexpression of another expression.

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