全局范围内可访问的命名空间std中的函数

Question

Under some situations, it seems like I can access functions that should be in the std namespace without a using or std:: qualifier. So far, I've only seen this occur with functions from the algorithm library.

In the following example, I expect all_of() to be in the std namespace, but this code compiles without error in VS2013 (Microsoft Compiler 18).

#include <iostream>
#include <string>
#include <algorithm>

int main() {
    const std::string text = "hey";
    std::cout << all_of(begin(text),end(text),islower);
    return 0;
}

Changing std::cout to cout without adding a using namespace std or using std::cout generates an "undeclared identifier" error as expected.

What's going on here?

Answer 1

This probably happens due to Argument-Dependent Lookup . The iterator returned by begin(text) and end(text) is probably a class defined in namespace std (or nested in a class in namespace std ), which makes namespace std associated with it. Looking up unqualified names for function calls looks into associated namespaces, and finds all_of there.

By the way, this is exactly the same reason why calling begin(text) works, even though the function template begin() is defined in namespace std . text is a std::basic_string , so std is searched.