Why does gcc compiler output pow(10,2) as 99 not 100?

Question

#include <iostream.h>
#include <math.h>
int main()
{
    int j=2;
    int output;
    output=pow(10,j);
    cout<<output;
    return 0;
}

I wrote above code to gcc 12 compiler and got the output 99 instead 100. I don't get the valid reason while searching on various sites. Is there any compiler problem?

Switch on your compiler warnings. And get a standards compliant compiler. — juanchopanza, Aug 24 '14 at 17:30
Perhaps [**this may help**](http://stackoverflow.com/questions/9704195/why-pow10-5-9-999-in-c) — WhozCraig, Aug 24 '14 at 17:34
@PaulR - probably(hopefully!) for the same reason we see this question - inexperience. :shrug: — enhzflep, Aug 24 '14 at 17:42
@enhzflep: you're probably right - I just keep seeing this kind of sloppy usage and it drives me nuts - why would you want to convert an int to a double, call an expensive library function to square that double, then convert the double result back to an int (with truncation and possible loss of accuracy, etc), when all you needed was a simple integer multiply? — Paul R, Aug 24 '14 at 17:47
I got pow(10,2)=100 but in case i declare j=2 and then upon evaluating pow(10,j) i got 99.Why?? — Alok Kumar Maurya, Aug 24 '14 at 20:22
These were just bugs in gcc native libm, not evwn glibc's libm, all fixed now. — Валерий Заподовников, Mar 11 '23 at 23:10

score 8 · Accepted Answer · edited Aug 24 '14 at 17:35

8

Because of integer truncation. pow() returns a floating point value, and due to floating point arithmetic, it is probably ~ 99.999...; however, due to integer truncation, even 99.999... gets truncated down to 99.

edited Aug 24 '14 at 17:35

Jonathan Leffler

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answered Aug 24 '14 at 17:29

yizzlez

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This is most likely the case, and so in this situation to get the more 'correct' integer you could round the result from pow(). `int result = pow(10^2) + 0.5;` would do it. – rsethc Aug 24 '14 at 17:34
4

@rsethc, That method of rounding has problems. Just use `std::round`. – chris Aug 24 '14 at 17:41
Eh, care to give an example? Has always worked for me. – rsethc Aug 24 '14 at 18:58
2

I got pow(10,2)=100 but in case i declare j=2 and then upon evaluating pow(10,j) i got 99. Why is it so then if it returns only floating point value, it must stick to same answer in both case? – Alok Kumar Maurya Aug 24 '14 at 20:23
That was just a bug in pow() implementation in gcc, not even in libm of glibc. – Валерий Заподовников Mar 11 '23 at 23:02

Why does gcc compiler output pow(10,2) as 99 not 100?

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