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^y);
    printf("%d",z);

I get this error message:

10:11: error: called object is not a function or function pointer 
z=(x^y)+2(x&y); 
        ^

The language is C. Why did this happen?

What is 2(x&y) supposed to do?

UnholySheep
– UnholySheep

2017-03-02 15:53:53 +00:00
Commented Mar 2, 2017 at 15:53 — UnholySheep
– UnholySheep, Commented Mar 2, 2017 at 15:53
change to 2*(x&y)

Roy Avidan
– Roy Avidan

2017-03-02 15:55:44 +00:00
Commented Mar 2, 2017 at 15:55 — Roy Avidan
– Roy Avidan, Commented Mar 2, 2017 at 15:55
Oh Thx a lot!!!

유재영
– 유재영

2017-03-02 15:56:33 +00:00
Commented Mar 2, 2017 at 15:56 — 유재영
– 유재영, Commented Mar 2, 2017 at 15:56

melpomene · Accepted Answer · 2017-03-02 16:19:11Z

1

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.

answered Mar 2, 2017 at 16:19

melpomene

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Comments

anaotha · Accepted Answer · 2017-03-02 16:06:52Z

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.

answered Mar 2, 2017 at 16:06

anaotha

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