error: called object is not a function or function pointer

Question

I have the following code:

  z=x-~y-1;
    printf(\"%d\",z);
  z=(x^y)+2(x&y);
    printf(\"%d\",z);
  z=(x|y)+(x&y);
    printf(\"%d\",z);
  z=2(x|y)-(x         


        

Answer 1

Change

z=(x^y)+2(x&y);

to

z=(x^y)+2*(x&y);

and

z=2(x|y)-(x^y);

to

z=2*(x|y)-(x^y);

You need the multiplication operator if multiplication is what you intended.

Answer 2

As for what the error means: 2(x&y) tells the compiler to call the function 2, passing x&y as an argument (just like printf("hi") means "call printf and pass "hi" as an argument").

But 2 isn't a function, so you get a type error. Syntactically speaking, whenever you have a value followed by (, that's a function call.