A function defined in c with #Define

Question

#include <stdio.h>
#include <stdlib.h>
#define PATRAT(x) x*x+x;

int main()
{
    int x,y,z,t;

    x = 104;
    y = PATRAT(x - 1);
    z = PATRAT(y++);
    t = ++PATRAT(x);

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

    return 1;
}

Given this code, the results are y = 105, z = 10506, t = 11130 and i can`t understand why. Can you please explain?

`PARAT(x - 1)` expands to `x - 1*x - 1 + x - 1` since `x = 104` this is `104 - 104 - 1 + 104 - 1` or `102` For `z` it's undefined behavior. — Iharob Al Asimi, Feb 10 '16 at 17:15
Realize that `PATRAT` is *not* a true function. It's a preprocessor macro, and those only do textual substitution. Try writing out what the body of `main()` would look like if you expanded the macro invocations. — Steve Summit, Feb 10 '16 at 17:17
I understood the first part with y why it becomes 102. But the result should be 103 since we have y++. — Cioby Andrei, Feb 10 '16 at 17:17
Where did this code come from? Is it just an exercise? `t = ++PATRAT(x);` is a monstrosity. — lurker, Feb 10 '16 at 17:17
z should be PATRAT(103) which is 10712 and t should be PATRAT(104) + 1 which is 10921 but the result is diffrent — Cioby Andrei, Feb 10 '16 at 17:19
@lurker Yes, it is an exercise and i need to find what it returns. But what i am trying to do is to understand why it returns that — Cioby Andrei, Feb 10 '16 at 17:21
Why do you say *`z` should be `PATRAT(103)`*? You don't know that because of the side effects of `y++` inside of the macro. As others have suggested, to understand the behavior of `PATRAT(y++)` you need to expand the macro, which occurs in line in the code. It's not like a function call at all. You'll find that `y++` occurs multiple times. So it's not the same result as, `w = y++; z = PATRAT(w);`. The expression `PATRAT(y++)` expands to, `y++*y+++y++` which the C compiler then compiles and which, among other things, is a mess. — lurker, Feb 10 '16 at 17:28
Possible duplicate of [Why are these constructs (using ++) undefined behavior?](http://stackoverflow.com/questions/949433/why-are-these-constructs-using-undefined-behavior) — too honest for this site, Feb 10 '16 at 17:32
@lurker I didn`t know how the fuction expands and i assumed it was that way but i understand now. So z should be 103 * 104 + 105 right? — Cioby Andrei, Feb 10 '16 at 17:33
Not to belabor the point, but it's **not** a *function*. It's a *macro*. Big difference. And whether `z` *should be* `103 * 104 + 105` is not well-defined due to dependency when `++` occurs. This is the undefined behavior everyone is trying to explain. — lurker, Feb 10 '16 at 17:36

score 3 · Answer 1 · answered Feb 10 '16 at 17:16

3

Macro expansion.

If using GCC compile with -E to see the output after all includes and macros. Then it will become clear.

answered Feb 10 '16 at 17:16

Zan Lynx

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I am using codeblocks, but i don`t know how to compile with -E. And i don`t need that, I need to understand why given that code, i get those results. I don`t know how that function works – Cioby Andrei Feb 10 '16 at 17:23
@CiobyAndrei What zan is suggesting is a method to expand the Macro and see the C pre processor output after that(not the final results).This would be the best method to see how the function actually works. – c_mnon Feb 10 '16 at 17:38
@c_mnon How do i use that method? – Cioby Andrei Feb 10 '16 at 17:40
@CiobyAndrei: Something like `gcc -E my-c-source-file.c` in a terminal window. – Zan Lynx Feb 10 '16 at 17:41

score 3 · Accepted Answer · edited May 23 '17 at 10:30

3

The preprocessor expands it as follows:

y = x - 1*x - 1+x - 1;;
z = y++*y+++y++;;
t = ++x*x+x;;

After we make that human-readable, it says:

y = x - (1 * x) - 1 + x - 1;
z = (y++ * y++) + y++;
t = (++x * x) + x;

Which is undefined behavior because we're modifying y without sequence points in between, so any output is possible.

edited May 23 '17 at 10:30

Community

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answered Feb 10 '16 at 17:17

Emil Laine

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Using `#define PATRAT(x) ((x)*(x)+(x));` will help with the first, by wrapping the argument and the whole expression in parentheses, but you **cannot** pass a pre- or post-increment parameter in to a macro and get reliable results. – TripeHound Feb 10 '16 at 17:29
Thank you! This is what i needed. But for z = PATRAT(y++) it would mean z = 103 * 104 + 105 and this result it is diffrent from 10506. @zenith – Cioby Andrei Feb 10 '16 at 17:30
I need to use the function the way it is @TripeHound – Cioby Andrei Feb 10 '16 at 17:32
@CiobyAndrei: The undefined behavior of incrementing more than once in an expression means that `y` might be the same value during all the operations, then have 3 added at the end. Or it might be y, y+1, y+2, or it might be y then have only 1 added at the end. It is UNDEFINED. – Zan Lynx Feb 10 '16 at 17:47
Sorry, but you cannot do that. Because it is a MACRO, not a function, it just expands the argument-TEXT as shown in the answer, giving something that is BY DEFINITION undefined. One compiler might give 103*103+103, another might give 105*105+105 _or any combination inbetween_. In fact, the same compiler might do it differently at different times. If you NEED to pass a ++ parameter, you MUST use a real function. – TripeHound Feb 10 '16 at 17:51
@ZanLynx I got it! Thank you! – Cioby Andrei Feb 10 '16 at 17:52
@TripeHound I was not trying to solve a problem with this Macro, this was an exercise and i had to tell what the result was and i didn`t know that that was a macro and how it works. – Cioby Andrei Feb 10 '16 at 17:53

A function defined in c with #Define

2 Answers2