Is it possible to pass a structure variable as a function argument without previously defining it?
Asked Answered
M

1

8

I have two structs defined as so (in color.h):

typedef struct rgb {
  uint8_t r, g, b;
} rgb;

typedef struct hsv {
  float h, s, v;
} hsv;

hsv rgb2hsv(rgb color);
rgb hsv2rgb(hsv color);

I then have the following in main.c which works:

hsv hsvCol = {i/255.0, 1, 1};
rgb col = hsv2rgb(hsvCol);

I want to be able to just create the variable hsvCol inside the parameters for hsv2rgb without having to create the variable and passing it as a parameter.

I've tried the each of the following (in place of the two lines above), sadly none of which compile :(

rgb col = hsv2rgb({i/255.0, 1, 1});
rgb col = hsv2rgb(hsv {i/255.0, 1, 1});
rgb col = hsv2rgb(hsv hsvCol {i/255.0, 1, 1})
rgb col = hsv2rgb(struct hsv {i/255.0, 1, 1});

My question is:

  1. Can I do what I was trying to do at all (but obviously in a different way)?

  2. If 1, how do I go about doing so?

Mendelevium answered 20/2, 2017 at 16:55 Comment(2)
rgb col = hsv2rgb((hsv){i/255.0, 1, 1});Wittenburg
Aw maaan, that was so simple! Thanks!Mendelevium
I
9

You can make use of a compound literal.

Quoting C11, chapter §6.5.2.5, paragraph 3,

A postfix expression that consists of a parenthesized type name followed by a brace enclosed list of initializers is a compound literal. It provides an unnamed object whose value is given by the initializer list.

and, paragraph 5,

The value of the compound literal is that of an unnamed object initialized by the initializer list. [...]

So, in your case, the code like

hsv hsvCol = {i/255.0, 1, 1};
rgb col = hsv2rgb(hsvCol);

can be re-written as

rgb col = hsv2rgb( ( hsv ) {i/255.0, 1, 1} );
                    ^^^^    ^^^^^^^^^^^^^
                    |             |
                    |              -- brace enclosed list of initializers
                    -- parenthesized type name
Irremeable answered 20/2, 2017 at 17:2 Comment(0)

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