namelookup与非限定名称:C ++ 0x draft n3290 [英] namelookup with Unqualified name : C++0x draft n3290
问题描述
来自ISO C ++ Draft n3290的点:3.4.0第二点
A point from the ISO C++ Draft n3290 : 3.4.0 2nd point
在表达式的上下文中查找的名称在找到表达式的范围中被查找为不合格的名称。
A name "looked up in the context of an expression" is looked up as an unqualified name in the scope where the expression is found.
有人可以用示例解释这个声明吗?
Would someone please explain this statement with an example?
推荐答案
它表示将搜索包含表达式的作用域的名称。即
It says that the scope which contains the expression will be searched for the name. i.e.
namespace foo {
struct bar {
void foobar() {
do_something();
}
};
}
如果您有此代码名称 do_something <将在
foobar
, bar
, foo
和全局作用域(而不是其他命名空间,结构或函数作用域)
if you have this code the name do_something
will be searched in the scope of foobar
, bar
, foo
and in the global scope (and not in other namespaces, structs or function scopes)
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