Why does pow(n,2) return 24 when n=5, with my compiler and OS?

问题

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
    int n,i,ele;
    n=5;
    ele=pow(n,2);
    printf(\"%d\",ele);
    return 0;
}

The output is 24.

I\'m using GNU/GCC in Code::Blocks.

What is happening?

I know the pow function returns a double , but 25 fits an int type so why does this code print a 24 instead of a 25? If n=4; n=6; n=3; n=2; the code works, but with the five it doesn\'t.

回答1:

Here is what may be happening here. You should be able to confirm this by looking at your compiler's implementation of the pow function:

Assuming you have the correct #include's, (all the previous answers and comments about this are correct -- don't take the #include files for granted), the prototype for the standard pow function is this:

double pow(double, double);

and you're calling pow like this:

pow(5,2);

The pow function goes through an algorithm (probably using logarithms), thus uses floating point functions and values to compute the power value.

The pow function does not go through a naive "multiply the value of x a total of n times", since it has to also compute pow using fractional exponents, and you can't compute fractional powers that way.

So more than likely, the computation of pow using the parameters 5 and 2 resulted in a slight rounding error. When you assigned to an int, you truncated the fractional value, thus yielding 24.

If you are using integers, you might as well write your own "intpow" or similar function that simply multiplies the value the requisite number of times. The benefits of this are:

You won't get into the situation where you may get subtle rounding errors using pow.
Your intpow function will more than likely run faster than an equivalent call to pow.

回答2:

You want int result from a function meant for doubles.

You should perhaps use

ele=(int)(0.5 + pow(n,2));
/*    ^    ^              */
/* casting and rounding   */

回答3:

Floating-point arithmetic is not exact.

Although small values can be added and subtracted exactly, the pow() function normally works by multiplying logarithms, so even if the inputs are both exact, the result is not. Assigning to int always truncates, so if the inexactness is negative, you'll get 24 rather than 25.

The moral of this story is to use integer operations on integers, and be suspicious of <math.h> functions when the actual arguments are to be promoted or truncated. It's unfortunate that GCC doesn't warn unless you add -Wfloat-conversion (it's not in -Wall -Wextra, probably because there are many cases where such conversion is anticipated and wanted).

For integer powers, it's always safer and faster to use multiplication (division if negative) rather than pow() - reserve the latter for where it's needed! Do be aware of the risk of overflow, though.

回答4:

If you do not #include <math.h> then the compiler does not know the types of the arguments to pow() which are both double not int -- so you get an undefined result. You get 24, I get 16418.

回答5:

When you use pow with variables, its result is double. Assigning to an int truncates it.

So you can avoid this error by assigning result of pow to double or float variable.

So basically

It translates to exp(log(x) * y) which will produce a result that isn't precisely the same as x^y - just a near approximation as a floating point value,. So for example 5^2 will become 24.9999996 or 25.00002

来源：https://stackoverflow.com/questions/25678481/why-does-pown-2-return-24-when-n-5-with-my-compiler-and-os

标签

math.h

c-standard-library