Curly braces as argument of function

Question

I've noticed in some source code the line:

if(pthread_create((pthread_t[]){}, 0, start_thread, pthread_args)) { ... 

It works correctly, but how to understand the first argument? It seems, that curly braces converts to pthread_t[] type.

P.s. I googled, but didn't find answer, only some guesses (some form of initialization, or legacy feature of c?)

Answer 1

This is a compound literal, with a constraint violation since initializer braces cannot be empty:

(pthread_t[]){} 

Using gcc -std=c99 -Wall -Wextra -Wpedantic this produces the warning:

compound_literal_pthread.c:6:36: warning: ISO C forbids empty initializer braces [-Wpedantic]      pthread_t *ptr = (pthread_t []){}; 

The result seems to be a pointer to pthread_t, though I don't see this behavior documented in the gcc manual. Note that empty braces are allowed as initializers in C++, where they are equivalent to { 0 }. This behavior seems to be supported for C, but undocumented, by gcc. I suspect that is what is happening here, making the above expression equivalent to:

(pthread_t[]){ 0 } 

On my system, pthread_t is a typedef for unsigned long, so this expression would create an array of pthread_t containing only a 0 element. This array would decay to a pointer to pthread_t in the function call.

Answer 2

It's a compound literal as mentioned by @some-programmer-dude.

In this specific case it is use to create an array to store the thread_id and discharge it later without the need of creating an extra variable. That is necessary because pthread_create does not accept NULL as argument for thread_id.