Incorrect result when using defines in an expression

Question

The order of magnitude is not correct:

#include <stdio.h>
#include <math.h>
#define M_H 1.007975/(6.02214076*pow(10,23)*1000)
#define k_B 1.380649*pow(10,-23)
int main()
{
    printf("%G\n",M_H);
    printf("%G\n",k_B);
    printf("%G\n",M_H/k_B);
    return 0;
}

Because it gives:

1.67378E-27
1.38065E-23
1.21232E-50

And i need that constants as defined.

I used gcc 12.2.1-1 with the -lm flags. I was expecting:

1.67378E-27
1.38065E-23
1.21232E-4

operator precedence, reminder that macros are just text substitution — pm100, Jan 26 '23 at 18:25
In `M_H/k_B`, do you really want `1.007975 / (6.02214076 * pow(10,23) * 1000) / 1.380649 * pow(10,-23)` or do you want `(1.007975 / (6.02214076 * pow(10,23) * 1000)) / (1.380649 * pow(10,-23))`? — Ted Lyngmo, Jan 26 '23 at 18:25
GERMÁN ORLANDO CUEVA ESTRADA, Why code `1.380649*pow(10,-23)` and not `1.380649e-23`? — chux - Reinstate Monica, Jan 26 '23 at 18:30
You macros shouldn't be calling `pow` at all. It makes no sense. `pow(10, -23)` is just `1e-23`. Have you never seen scientific notation before? — Tom Karzes, Jan 26 '23 at 18:33
GERMÁN ORLANDO CUEVA ESTRADA, `f*pow(10, p)` risks loss of precision. Note that `1.380649e-17 != (1.380649*pow(10,-17))`. Best to form your constants directly. — chux - Reinstate Monica, Jan 26 '23 at 18:37
Unrelated: The Avogadro constant is 6.02214076E+23. Why do you divide the atomic weight of hydrogen by 6.02214076E+26? — Ted Lyngmo, Jan 26 '23 at 18:57

Yakov Galka · Answer 1 · 2023-01-26T18:37:49.783

M_H/k_B expands to:

1.007975/(6.02214076*pow(10,23)*1000)/1.380649*pow(10,-23)

Note that the 2nd pow is not enclosed in parentheses, and therefore the multiplication is done after the preceding division. You should wrap your defines in parentheses to prevent that:

#define M_H (1.007975/(6.02214076*pow(10,23)*1000))
#define k_B (1.380649*pow(10,-23))

I'd also recommend you to replace the pow function call with the 1e-23 syntax (aka scientific notation); this way the number in k_B will be parsed as a single token, and the parentheses would not be necessary there:

#define M_H (1.007975/6.02214076e26)
#define k_B 1.380649e-23

and i need that constants as defined

If, for some reason, you cannot modify those definitions, you can still add the parentheses at the point of use:

printf("%G\n",(M_H)/(k_B));

score 1 · Answer 2 · answered Jan 26 '23 at 18:26

1

do

#define M_H (1.007975/(6.02214076*pow(10,23)*1000))
#define k_B (1.380649*pow(10,-23))

to ensure correct order

answered Jan 26 '23 at 18:26

pm100

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Incorrect result when using defines in an expression

2 Answers2