[SOLVED] How to pass a constant array literal to a function that takes a pointer without using a variable C/C++?

How to pass a constant array literal to a function that takes a pointer without using a variable C/C++?

If I have a prototype that looks like this:

function(float,float,float,float)

I can pass values like this:

function(1,2,3,4);

So if my prototype is this:

function(float*);

Is there any way I can achieve something like this?

function( {1,2,3,4} );

Just looking for a lazy way to do this without creating a temporary variable, but I can't seem to nail the syntax.

Solution

You can do it in C99 (but not ANSI C (C90) or any current variant of C++) with compound literals. See section 6.5.2.5 of the C99 standard for the gory details. Here's an example:

// f is a static array of at least 4 floats
void foo(float f[static 4])
{
   ...
}

int main(void)
{
    foo((float[4]){1.0f, 2.0f, 3.0f, 4.0f});  // OK
    foo((float[5]){1.0f, 2.0f, 3.0f, 4.0f, 5.0f});  // also OK, fifth element is ignored
    foo((float[3]){1.0f, 2.0f, 3.0f});  // error, although the GCC doesn't complain
    return 0;
}

GCC also provides this as an extension to C90. If you compile with -std=gnu90 (the default), -std=c99, or -std=gnu99, it will compile; if you compile with -std=c90, it will not.