The curious case of the pow function

Revision en3, by misof, 2015-11-28 17:47:52

So, what's up with pow()?

Here at Codeforces it is quite common to see solutions that use pow() fail. Most recently this was the case in round #333 problem div2A. Whose fault is it?

Level 1 answer is that it is obviously the contestant's fault. The contestant should have been aware that pow operates on floating-point numbers and that there can be precision errors. If you expect that a floating-point variable contains an integer, you cannot just cast it to an int. The small precision errors mean that your nice round 100 can actually be stored as 100.00000001 (in which case the typecast to an int still works), but it can also be stored as 99.99999999 (in which case the typecast will produce a 99).

You cannot even expect any kinds of deterministic behavior. For example, consider the following short program:

#include <bits/stdc++.h>
using namespace std;
int main () {
    printf("%d\n", (int)pow(10,2));
    for (int j=0; j<3; ++j) printf("%d\n", (int)pow(10,j));

This code computes 10^2, 10^0, 10^1, and again 10^2. What is its output when using the current g++ version at Codeforces? 100, 1, 10, 99. Fun fun fun :)

For extra fun, change the initialization in the cycle to int j=2. The new output: 100, 100 :)

So, what should you do? Be scared and avoid pow() completely? Nah. Just be aware that precision errors may occur. Instead of truncating the value inside the variable, round it to the nearest integer. See "man llround", for instance.

That being said, it's time for the...

Level 2 answer. Wait a moment. Why the f*#& should there be a precision error when I'm computing something as simple as 10^2? Ten squared is clearly 100. Shouldn't the value returned by pow() be as precise as possible? In this case, 100 can be represented exactly in a double. Why isn't the return value 100? Isn't the compiler violating any standards if my program computes 99.99999999 instead?

Well, kind of.

The standard that actually matters is the C++ standard. Regardless of which one you look into (be it the old one, C++11, or C++14), you will find that it actually never states anything about the required precision of operations such as exp(), log(), pow(), and many others. Nothing at all. (At least to the best of my knowledge.) So, technically, the compiler is still "correct", we cannot claim that it violates the C++ standard here.

However, there is another standard that should apply here: the international standard ISO/IEC/IEEE 60559:2011, previously known as the standard IEEE 754-2008. This is the standard for the floating point arithmetic. What does it say about the situation at hand?

The function pow() is one of about 50 functions that are recommended to be implemented in programming languages. Doing so is optional -- i.e., it is not required to conform to the standard.

However, if a language decides to implement these functions, the standard requires the following: A conforming function shall return results correctly rounded for the applicable rounding direction for all operands in its domain. The preferred quantum is language-defined.

In this particular case, this is violated. Thus, we can claim that the Codeforces g++ compiler's pow() implementation does not conform to the IEEE floating-point number standard.

Hence, if you failed a problem because of this issue, do blame the compiler a little. But regardless of that, in the future take precautions and don't trust anyone.

Tags floating-point, double, pow, rounding


  Rev. Lang. By When Δ Comment
en3 English misof 2015-11-28 17:47:52 103
en2 English misof 2015-11-28 17:46:31 1208 (published)
en1 English misof 2015-11-28 17:36:56 2232 Initial revision (saved to drafts)