A point from ISO draft n3290 section 3.4.2 paragraph 1:
When the postfix-expression in a function call is an unqualified-id, other namespaces not considered during the usual unqualified lookup may be searched, and in those namespaces, namespace-scope friend function declarations 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).
Here they said aboout "these modifications to the sea开发者_Python百科rch depend on the types of the arguments / template template arguments / namespace of the template argument " ...Can any one expalin with an example please? I tried with argumetn types..please expalin with template template argument types & namespace of the template argument type
Consider a simple unqualified function call:
foo(x);
ADL means that foo
is looked up not just in the enclosing scope, and the namespace that the call is in, but also the namespace of the type of x
. e.g. if x
is a std::vector<int>
then namespace std
is also searched. Thus:
int main() {
std::vector<int> x,y;
swap(x,y);
}
is OK, and will call std::swap()
.
The lookup also depends on the namespace of any template arguments too, so if x
is std::vector<mynamespace::myclass>
then mynamespace
is also included in the lookup. Thus
namespace mynamespace {
struct myclass {};
void foo(std::vector<mynamespace::myclass> const&){}
}
int main() {
std::vector<mynamespace::myclass> x;
foo(x);
}
will call mynamespace::foo()
.
Finally, the lookup also extends to the namespaces of any templates used as template template parameters. e.g.
namespace mynamespace {
template<typename T>
struct mytemplate
{};
template<typename T>
void bar(T const&) {}
}
template<template<typename> class T>
struct wrapper {};
int main() {
wrapper<mynamespace::mytemplate> x;
bar(x);
}
Even though wrapper
is in the global namespace, mynamespace::bar
will be found, because the template template parameter used for x
is mynamespace::mytemplate
.
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