How could it be that calling a free function from GLM (like dot) can compile without using its namespace?

Question

Like in the following example:

#include <glm/glm.hpp>

int main()
{
    glm::vec3 a;
    dot(a, a);
}

Without using the glm namespace anywhere, calling its dot function compile and works (altough it does not appears in the intellisense).

How can it be possible?

Do a Google search for argument dependent lookup – Richard Hodges Dec 03 '15 at 20:32 — Richard Hodges, Dec 03 '15 at 20:32
@RichardHodges Cool! Didn't know that was a thing. Thanks! – Fernando Toigo Dec 03 '15 at 21:07 — Fernando Toigo, Dec 03 '15 at 21:07

score 0 · Answer 1 · answered Dec 03 '15 at 20:55

During overload resolution associated namespaces are searched. This argument-dependent look-up (ADL). The namespaces associated with a in the call of dot(a, a) depend on how glm::vec3 is defined. Assuming it is defined in glm (rather than namespace glm merely including a using-declaration or a typedef for a type defined elsewhere), names from glm are found. Additional namespaces may be searched if vec3 inherits from class in other namespaces or is actually a class template instantiated with types defined in other namespaces.

How could it be that calling a free function from GLM (like dot) can compile without using its namespace?

1 Answers1