namespace A {
class F {}; // line 2
class H : public F {};
}
namespace B {
void F(A::H x); // line 7
void G(A::H x) {
F(x); // line 9
}
}
我使用的是gcc 4.3.3,错误是：
s3.cpp: In function ‘void B::G(A::H)’: s3.cpp:2: error: ‘class A::F’ is not a function,s3.cpp:7: error: conflict with ‘void B::F(A::H)’ s3.cpp:9: error: in call to ‘F’
我认为,因为在第9行中没有名称空间前缀,F(x)应该明确地仅表示B :: F(x).编译器尝试将x转换为自己的超类.根据我的理解,它不应该.为什么这样做？
解决方法
据我所知,这是标准行为.
3.4.2 Argument-dependent name lookup
When an unqualified name is used as the postfix-expression in a function call (5.2.2),other namespaces not considered during the usual unqualified lookup (3.4.1) may be searched,and namespace-scope friend function declarations (11.4) not otherwise visible may be found. These modifications to the search depend on the types of the arguments (and for template template arguments,the namespace of the template argument).For each argument type T in the function call,there is a set of zero or more associated namespaces and a set of zero or more associated classes to be considered. The sets of namespaces and classes is determined entirely by the types of the function arguments (and the namespace of any template template argument).
Typedef names and using-declarations used to specify the types do not contribute to this set. The sets of namespaces and classes are determined in the following way…
此规则允许您编写以下代码：
std::vector<int> x; // adding some data to x //... // now sort it sort( x.begin(),x.end() ); // no need to write std::sort
最后：由于Core Issue 218,一些编译器会编译有问题的代码而没有任何错误.
