sqrt() problem

[kimhoong79]

Recently I found out some problem with sqrt(). Here's the code:

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

main()
{
int Result, Data = 4;

printf("%d\n", sqrt(Data));

Result = sqrt(Data);
printf("%d\n", Result);
}

The output of this code is

0
2

which for me, is not consistent. Can anyone explain this to me? Thanks in advance

 

[olded]

Hi:

Not near a development box so I can't check it out, but the sqrt function returns a double not an int. Your complier is automatically casting the return to an int which is why the second example works.

You might try the %f option with the first example.

Regards,


Ed

 

[jstreich]

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

main()
{
   int Result, Data = 4;

   printf("%d\n", (int)sqrt(Data)); // should work
   printf("%f\n", sqrt(Data)); // should work

   Result = sqrt(Data);
   printf("%d\n", Result);
}

Pragmatic programmer tip related to this issue:
Select is not broken.

http://www.pragmaticprogrammer.com/

In most cases, the libaries are written correctly, it is almost always a case of your code being broken. If the library is broken a quick goole search would yeild a lot of hits about people complaining. The error is generally in your use of the library, or errors arround the use of a call to a library function.

[plug=shameless]

http://game-master.us/phpx-3.4.0/

[/plug]

 

[ArkM]

No any casting in printf() args after the 1st because of this function prototype:

int printf(const char*,...);

It's unsafe output: a compiler can't verify arg types. So printf("%d",(double)...) is wrong in all cases...

Moreover: sqrt(4) may be not equal to 2 exactly, so (int)sqrt(4)) is equal 1 or 2 - both results are correct (it's not a language problem: it's floating point arithmetic...

 

[jstreich]

Thanks ArkM -- I am C++ guy, I don't use printf very often...

Right, but it shouldn't be 0... like he got.

[plug=shameless]

http://game-master.us/phpx-3.4.0/

[/plug]

 

[kimhoong79]

thanks everyone!