Question

I would like to pass parameters instantly to function, without creating additional variable like this:

void DrawSprite( float a[2], float b[2] ) { /* */ }

DrawSprite( (0.0f, 0.0f), (50.0f, 50.0f) );

Is it possible, and if yes, how to do it?

Answer 1

Some of the answers to this important question seem to be misleading. The issue is: passing a fixed size array to a function. First of all something like

  float x[2] = {0.1F, 0.2F};
  float y[2];
  y=x;

is a error: invalid array assignment (on gcc 4.6.1; but we should be speaking on “standard” here, not particular compiler features). Quite rightly so — what should the operation accomplish? Both x and y point to a piece of memory on stack, so copying the float* x to y would make little sense, as the values will not be copied to the place reserved initially, y[0],y[1].

The only sensible cast from float x[2] is to float *y. This is precisely what happens when a function is called. Your function DrawSprite(float a[2], float b[2]) is really treated as DrawSprite(float* a, float* b) (you might print out typeid(a).name() to check). Therefore you can pass x to DrawSprite, and this x will simply be cast to float*.

The second issue is initialization. The code (float[2]){1.0, 2.0} really constructs a fixed-size temporary array of float (again can be checked by typeid((float[2]){1.0, 2.0}).name(), at least on gcc), but there is not much you can do with this array, as assignments to float [2] are forbidden. You may cast it to float* (e.g. assign by z=(float[2]){1.0,2.0};, but to float* z;, so it is really z=(float*)(float[2]){1.0,2.0};), or you may pass it to your function, which does the same, i.e. casts to float*.