Using #define in c for equations

Question

Can someone explain to me why the value of y here is 13?

#include <stdio.h>
#define avg_sum(n) n * (n-1)/2
int main(){
    int y;
    int z = 9;
    y = avg_sum(z+1);
    printf("y=%i\n",y);
}

That's why you always use parentheses around macro arguments when they are part of an arithmetic expression, and parentheses around the macro body as a whole, too, so that 934 / avg_sum(12) gives you the expected answer too. — Jonathan Leffler
– Jonathan Leffler, Commented Dec 10, 2013 at 21:00

Glenn Teitelbaum · Accepted Answer · 2013-12-10 21:05:07Z

9

avg_sum(9+1) 9+1 * (9+1-1)/2 = 9 + 9/2 = 9+ 4 = 13

macros expand each time so 9+1 is not the same as 10, it might be better with safeguarding parenthesis as follows:

#define avg_sum(n) ((n) * ((n)-1)/2)

but an equivelant function will do you much better and be more intuitive and will evaluate arguments only once

avg_sum(a++) will be ((a++) * ((a++)-1)/2) and will increment a twice whereas a function will not have these issues since all arguments are evaulated before the function is called

edited Dec 10, 2013 at 21:05

answered Dec 10, 2013 at 20:58

Glenn Teitelbaum

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4 Comments

Jonathan Leffler Over a year ago

static inline int avg_sum(int n) { return (n * (n-1)) / 2; } works nicely, unless you're stuck with an archaic C90 compiler (eg the default on Windows, MSVC).

Glenn Teitelbaum Over a year ago

@JonathanLeffler in the function version, I'm not sure you need all the parens

Jonathan Leffler Over a year ago

I was not sure either, but better safe than sorry. Thinking about how things can go wrong if (n-1)/2 is evaluated before it is multiplied by n: suppose n == 4. Then (n-1)/2 is 1 and n*1 is 4; OTOH, (4 * 3)/2 is 6. I think the parentheses do matter, if the compiler is allowed to reorder operations, and I think it is.

Glenn Teitelbaum Over a year ago

@JonathanLeffler 5 | * / % | Multiplication, division, and remainder | left to right No reordering allowed - no parens needed

Ed Swangren · Accepted Answer · 2013-12-11 18:31:20Z

5

The best way to answer these sorts of questions is to simply expand the macro in question:

y = avg_sum(z+1);

y = z + 1 * (z + 1 - 1) / 2

y = 9 + 1 * (9 + 1 - 1) / 2

y == 13

This is why you add parentheses around your macro arguments.

#define avg_sum(n) ((n) * ((n)-1)/2)

edited Dec 11, 2013 at 18:31

answered Dec 10, 2013 at 20:59

Ed Swangren

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Comments

Deck · Accepted Answer · 2013-12-10 20:58:52Z

2

y = avg_sum(z+1);

expands to z + 1 * (z+1-1)/2 but it's wrong. Change your macro to

#define avg_sum(n) ((n) * ((n)-1)/2)

And always parenthesize as well arguments of functional macros as the macro itself . It's an important rule.

answered Dec 10, 2013 at 20:58

Deck

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