pow() function

[littlewontons]

Hi Everybody,

I am new to C and I just encountered a problem. When I use the pow() function, I get some problems. The following is my code:

code:

#include <stdio.h> #include <math.h> int main( void ) {     int i = 0;     for( i = 1; i < 10; i++ )     {         printf( "%d\n", (int)pow( 10, i ) );     }     getch();     return 0; }

My output is the following:

code:

10 99 1000 9999 100000 1000000 9999999 99999999 999999999

Can someone help me find out what my problem is? And could you tell me what I did wrong so I can prevent this in the future? Thanks =)

[OneOffDriveByPoster]

Your compiler/libraries/processor/environment is not up to spec. AFAIK, pow should work fine for those values on modern systems. Write your own integer power function to avoid this.

[CodeMonkey2000]

Wow that is really weird. Oh well, as OneOffDriveByPoster said, make your own pow function that returns i to the j-th power. i to the j-th power is the same as 1*i*i...j times.

[Dragan]

@OneOffDriveByPoster, try this code on any modern system.

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

Quote:

int main( void )
{
int i = 0;
int j = 0;
for( i = 1; i < 100000; i++ )
{
j = j + 0.1;
printf( "%d\n", j ); // or %f, i forgot how to print float number
}
getch();
return 0;
}

you expect:
0.1
0.2
0.3
0.4
0.5
0.6
0.7
....
but, you will see something interesting

[OneOffDriveByPoster]

Dragan @ Sat Apr 12, 2008 4:51 am wrote:

@OneOffDriveByPoster, try this code on any modern system.

You chose 0.1, which we all know is 0.0001100110011... in binary. The OP had nice whole numbers, the operands and the correct result can all be represented as an IEEE double. I don't know what underlying algorithm they are using in the OP's case, but I would not call their implementation good.

Oh, and you are assigning to an int every time. I replaced with double (and it really does not look weird for the number of digits that shows up and the numbers that I bothered to look at).

[Dragan]

OK. j need to be float (that was my mistake), I try that code in Microsoft visual c++ 6.0 on many systems and i got results

0.1
0.2
0.3
0.4
0.5
0.599999999
0.699999999
0.799999999
...

How look your results?

[OneOffDriveByPoster]

0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
0.800000
0.900000
1.000000
...

[CodeMonkey2000]

Isn't Microsoft Visual C++ 6.0 really out dated? Switch to minGW, or update it.

[Dragan]

@OneOffDriveByPoster I don't know why but I get wrong results, maybe because of old Microsoft Visual, I realy need to upate Microsoft Visual c++.

[Saad]

These errors are floating point inaccuracies. No floating point number can accurately represented. Your results can depend on how your CPU handles floating point numbers


As for the original question, it's a casting issue. Try the following

c:

#include <stdio.h> #include <math.h> int main( void ) {     int i;     for( i = 1; i < 10; i++ )     {         //~~ printf( "%d\n", (int)pow( 10, i ) );         printf( "%f\n",pow( 10, i ) );     }     getch();     return 0; }

[OneOffDriveByPoster]

Saad @ Sat Apr 12, 2008 1:20 pm wrote:

No floating point number can accurately represented.

*hack* *cough*

[littlewontons]

Thanks guys, but does that mean my C library is out of date? I am not too sure why it does this exactly. I have thought about alternate ways to fix it, but for the future, can i fix it some how? I am using devc++ 4.9.9.2.

[littlewontons]

Oh yea, Sadd, I have tried that already, it outputs correct, it just does not work with %d and (int)

[Dragan]

@littlewontons, you need to do that on alternate way, I think that is not software problem, that is hardware problem (in processor, in float point unit)

[littlewontons]

But I tried it at school, and I get the same error. They are different machines. Does this mean that they both have problems?