Macro argument with different results

Question

So I have been having this rather 'unyielding' problem with a macro call in my C program. The macro I used was :

#define MACRO(X) X*X

And the problem is when I do this

printf("%d",MACRO(3));

it displays 9 as a result (which is correct). Only when I pass 3 as 2+1 as follows :

printf("%d",MACRO(2+1));

It strangely displays an outcome of 5. Can anyone please have me informed why?

Seems like a dupe of [The need for parentheses in macros in C](http://stackoverflow.com/q/10820340/1281433), which uses the macro `#define SQR(x) (x*x)`, I already VTC'ed as a typo, but this would be a better dupe target. — Joshua Taylor, Jan 15 '16 at 21:51
please remember that a macro results in a direct replace of text. To get the right operation, surround each 'x' element in the body of the macro with parens, just as you would if writing the body of the macro at the location where the macro is invoked. — user3629249, Jan 16 '16 at 13:59

score 5 · Answer 1 · answered Jan 15 '16 at 21:50

5

printf("%d",MACRO(2+1));

After preprocessing it will be become

printf("%d",2+1*2+1);

Since multiplication has high precidenece over addition, it will print 5;

To resolve the issue you have to define your macro as follows

#define MACRO(X) (X)*(X)

answered Jan 15 '16 at 21:50

Steephen

2

I would use ((X)*(X)) but maybe is redundant – terence hill Jan 15 '16 at 21:53

score 4 · Answer 2 · edited Jan 15 '16 at 21:58

4

You need to define your macro like so

#define MACRO(X) (X)*(X)

A macro is just a text expansion, so your following code is being interpreted like this...

printf("%d", 2+1*2+1);

And following order of operations you can see how you get 5 as your result.

edited Jan 15 '16 at 21:58

Weather Vane

answered Jan 15 '16 at 21:50

jacob

2 Answers2