Why (5+10)/2 is 7.0 not 7.5? C programming

Question

c code :

 int main(void) {
 int i = 0, j = 0;
 double avg = 0;
 int *pi, *pj;
 double *pavg;

..

    pi = &i;
    pj = &j;
    pavg = &avg;

..

    *pi = 5;
    *pj = 10; 
    *pavg = (*pi + *pj) / 2;

here where it prints :

    printf("%lf\n\n", avg);

it prints 7.000000000

    return 0;
}

Because `(*pi + *pj) / 2;` is an int expression, which produces 7 which is saved to double. — Tony Tannous, May 20 '20 at 06:06
it's doing an integer division `15 / 2 = 7` and only _then_ casting it to a `double` — Aemyl, May 20 '20 at 06:07
We need a canonical dupe for this. It's such an incredibly common FAQ - everyone learning C probably runs into this bug. We have [this](https://stackoverflow.com/questions/4890480/c-program-to-convert-fahrenheit-to-celsius-always-prints-zero) in the C FAQ but I don't think it's generic enough to use as a dupe target for this. — Lundin, May 20 '20 at 06:47

babon · Accepted Answer · 2020-05-20T06:29:51.873

When you do (*pi + *pj) / 2, you are doing integer arithmetic. The numbers after the decimal have already been discarded before you assign to the double variable.

One way is to do (*pi + *pj) / 2.0. One of the operands of the expression being a double, the other ints will be promoted to doubles before the expression is evaluated.

Another option is to typecast one of the variables to a double ((double)*pi + *pj) / 2. Here, the de-referencing operator being of higher precedence will be evaluated before the typecast.

Why (5+10)/2 is 7.0 not 7.5? C programming

1 Answers1