How to set a function's return value= to a variable?

Code:

int get_corners(FILE *fp,double x_array[],double y_array[],int macro)
  {
  int i=0;
  int j=0;
 
  while(fscanf(fp,"%lf%lf",&x_array[i],&y_array[j])!=EOF)
  {
  i++;
  j++;
  }
  fclose(fp);
  int actual_size=j;
  printf("actual_size %d", actual_size);
  return actual_size;

So basically i tested it and the return value is 7, which is correct.
How would i be able to pass that value (7) to the next function?
actual_size=7

Code:

void output_corners(FILE*foutput,double x_array[],double y_array[],int *coordinates)
  {
  int i=0;
  int j=0;
 
  for(i=0;i<actual_size;i++)             // But here it says actual_size is undeclared
  fprintf(foutput,"%1lf",*x_array);
 
  }

"actual_size" is declared in the "get_corners()" function, and only lives within the scope of that function. That's why it is saying that it's undeclared in the "output_corners()" function.

The value returned from the "get_corners()" function simply needs to be passed to the "output_corners()" function.

Code:

int size;

size = get_corners(...);

output_corners(...,size);
// be sure to change the argument list in the function
// declaration and definition to match this additional variable

Originally Posted by Matticus

"actual_size" is declared in the "get_corners()" function, and only lives within the scope of that function. That's why it is saying that it's undeclared in the "output_corners()" function.

The value returned from the "get_corners()" function simply needs to be passed to the "output_corners()" function.

Code:

int size;

size = get_corners(...);

output_corners(...,size);
// be sure to change the argument list in the function
// declaration and definition to match this additional variable

Hey,

Code:

int actual_size=get_corners(fp,x_array,y_array,macro);
  printf("coordinates %d",actual_size);

it prints out the memory addresses. why is that?

Thank you Matt. It works!

No reason I can see in the code you've provided. Can you post more of the program?

[edit] Never mind. You're welcome.