Thread: Expected expression error

Hey there guys! New to cboard. Got tired of the people over at stack overflow, so I'm hoping you guys can help me out. So I'm pretty new to C, and keep getting an error. Basically I'm trying to convert a ppm image to all red. Here is my code that I can't get to work:

Code:

change(pixel_t *PIXEL, &w, &h,pixel_t *buffer);

And here is the beginning of my convert function:

Code:

void change(pixel_t *pixel, int w, int h, pixel_t *buffer) {

     int average, sum;
     int i;
     pixel_t *pointer;

Everything else works fine. I keep getting an error when I call the convert function. It says "expected expression before pixel_t" and "too few arguments to function "change".

I know that everything else in the main is working. I can post more if needed, but don't think it is necessary. I could be wrong though. Does anyone see my problem?

Sorry if it is super simple and I'm just too dumb to see it.

When you call change:

Code:

change(pixel_t *PIXEL

the part after the ( is technically a declaration, the parser wants a pointer type expression.

You need to allocate a pixel and a pixel buffer, and end up with something like this:

Code:

pixel_t pixel;
pixel_t *buffer = /* allocate buffer */;
change(&pixel, &w, &h, buffer);

If there is documentation available, it should explain what change does and how to call it, specifically.

In the prototype you are essentially passing pointers to w and h instead of the ints you have defined in the actual function.