Programming Forums - How do I take a square root?

Page 1 of 3

Show 40 post(s) from this thread on one page

Programming Forums (http://www.programmingforums.org/forumindex.php)

- C (http://www.programmingforums.org/forum60.html)

- - How do I take a square root? (http://www.programmingforums.org/showthread.php?t=11656)

Fall Back Son

Oct 21st, 2006 2:56 PM

How do I take a square root?

I was making a program just to make sure I can, and I decided to make a pythagorean theorem program. what happened is I defined

"C" as C = ( A * A ) + ( B * B )

which gives the wrong answer, because the formula is supposed to be C square = A squared + B squared. Anyway, my question is in the title: is there any way I can just take the square root of ( A * A ) + ( B * B ) ? If so, thanks. If not... and anybody has any good ways that I could accomplish the same thing, please let me know. thanks!

niteice

Oct 21st, 2006 2:59 PM

sqrt((a*a)+(b*b));

Sane	Oct 21st, 2006 3:05 PM

What niteice said. But you must also include the math.h library.

#include <math.h>

Then, for sake of listing alternatives, you can also do...

x_sqrt = pow(x, 0.5)

Although I'm not sure if "pow" accepts a float as a second argument.

Fall Back Son

Oct 21st, 2006 3:43 PM

ok, thanks. out of curiosity... what if I wanted to use the variable C.

So I would have

C = ( A * A ) + (B * B )

then I would have sqrt (C). But that doesn't work, due to me not defining the variable or something like that. I'd appreciate it if somebody can tell me the proper way to make that work... and I'd appreciate it even more if somebody can explain WHY it works that way. thanks. you guys are amazing. btw, the reason linux gave me that it would work is because "called object is not a function"

Sane	Oct 21st, 2006 3:49 PM

If C is not defined, then define it! Wherever you define "A" and "B", just add in "C" to define it as well.

If that's not the problem, you can post your code (wrapped in [code][/code] tags), and we could give you further assistance. :)

Fall Back Son

Oct 21st, 2006 3:51 PM

C was defined, as far as I know. It said "float A, B, C ; " which I think, at least, defines C. Anyway, code will be edited in here

#include <stdio.h>

#include <math.h>

int main ()



{

float A, B, C, sqrt ; 



/* get A from user */ ;



printf (" enter variable A : ") ;

scanf ("%f", &A) ; 



/* get B from user */ ;



printf (" enter variable B : ") ; 

scanf ("%f", &B) ; 



/* calculate C */ ; 



C = ( A * A ) + ( B * B ) ; 

C = sqrt (C) ; 



printf (" %.2f C \n", C) ; 



return 0 ; 



}

Sane	Oct 21st, 2006 4:00 PM

The function "sqrt" accepts an integer value. You are giving it a float. This can be fixed by using the function "sqrtf". The "f" on the end is there because it will accept a float as its parameter, just as "sqrtl" accepts a long integer.

New Code:

C = ( A * A ) + ( B * B ) ; 

C = sqrtf (C) ;

And I don't want to confuse you, but it's important to note that it can also be rewritten as:

C = sqrtf ( ( A * A ) + ( B * B ) );

This has to do with the "transitive" property of mathematics, as it applies to computer programming. Essentially, anything that's equivilent, can be restated wherever it applies. :)

Fall Back Son

Oct 21st, 2006 5:58 PM

Thanks. and for the record, I understand mathematics, I am mostly just a noob at programming & chemistry. Lol. Thanks for all the help though.

Fall Back Son

Oct 21st, 2006 6:09 PM

It still does not work though. If i remove "sqrtf" from the variable declaration (where it says float A, B, C, sqrtf ; ) the compiler tells me that there is a "implicit differentiation of function sqrtf" and that there is an "undefined reference to sqrtf."

If I leave sqrtf in the variable declaration, it says that on line 21 (the line where it says C = sqrtf (C) )

there is an "error: called object is not a function."

wth??!!! haha

stevengs

Oct 21st, 2006 6:14 PM

you declared sqrt to be a float.. then you try to call it as a function???

All times are GMT -5. The time now is 10:36 AM.

Page 1 of 3

Show 40 post(s) from this thread on one page