Expression evaluation in C language

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Hi,

I'm a Computer Science student in Master of Science degree. I'm teaching c language to undergraduate students.
Consider this program :

XML

#include <stdio.h>
#include <conio.h>

int main() {

    clrscr();

    float f=4;
    float g = f / 4 + f * 2 / ++f;
    printf("%5.2f", g);
    getch();
}

To my knowledge expression evaluation is done from left to right and based on operators precedence.
Here precedence of ++ is higher than + - / * operators. And, precedence of * / is higher than + - operators. So, the expression must be computed as:
g = f / 4 + f * 2 / ++f;
g = (f / 4) + ((f * 2) / ++f);
........1....5......2....4...3....
and the result must be:
g = 2.6

but, the compiler computes the expression as:
g = (f / 4) + ((f * 2) / ++f);
........4....5......1....3...2....
and the result is:
g = 2.85

Why this happens with no paranthesis in expression ?
I tested this issue in Turbo C and Borland C compilers.

Special Thanks

Posted 29-Nov-10 20:28pm

VahidKeyany

Updated 29-Nov-10 23:25pm

v2

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OriginalGriff · Answer 1 · 2010-11-29T21:29:00

It happens because of operator precedence - google will help you with that - so it does make some sense.

I wouldn't do that - it isn't obvious what you actually want to happen, and that makes it difficult for a mere mortal to comprehend at a glance - the way most people read code. Re work it to two lines: It both makes it clearer for us, but also for the compiler:

C#

float g = f / 4 + f * 2 / (f + 1);
f++;

I would also tend to add more brackets to make it absolutely obvious - but then, I've translated code between languages before...

C#

float g = (f / 4) + ((f * 2) / (f + 1));
f++;