Programming Forums > Application Development > C

Fall Back Son · Oct 21st, 2006, 3:56 PM

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, 3:59 PM

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

Sane · Oct 21st, 2006, 4:05 PM

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

(Toggle Plain Text)

#include <math.h>

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

(Toggle Plain Text)

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, 4:43 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, 4:49 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, 4:51 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

(Toggle Plain Text)

#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, 5:00 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:

(Toggle Plain Text)

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:

(Toggle Plain Text)

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, 6:58 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, 7:09 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, 7:14 PM

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

Oct 21st, 2006, 3:56 PM	#1
Fall Back Son Professional Programmer Join Date: Oct 2006 Posts: 311 Rep Power: 3	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!

Oct 21st, 2006, 4:05 PM	#3
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 2,099 Rep Power: 6	What niteice said. But you must also include the math.h library. (Toggle Plain Text) #include <math.h> #include <math.h> Then, for sake of listing alternatives, you can also do... (Toggle Plain Text) x_sqrt = pow(x, 0.5) x_sqrt = pow(x, 0.5) Although I'm not sure if "pow" accepts a float as a second argument.

Oct 21st, 2006, 4:51 PM	#6
Fall Back Son Professional Programmer Join Date: Oct 2006 Posts: 311 Rep Power: 3	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 (Toggle Plain Text) #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 ; } #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 ; }

Oct 21st, 2006, 5:00 PM	#7
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 2,099 Rep Power: 6	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: (Toggle Plain Text) C = ( A * A ) + ( B * B ) ; C = sqrtf (C) ; 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: (Toggle Plain Text) C = sqrtf ( ( A * A ) + ( B * B ) ); 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.

Oct 21st, 2006, 7:14 PM	#10
stevengs Professional Programmer Join Date: May 2005 Location: Bad Nauheim, Germany Posts: 436 Rep Power: 4	you declared sqrt to be a float.. then you try to call it as a function??? __________________ -Steven "Is this a piece of your brain?" - Basil Fawlty

Oct 21st, 2006, 3:59 PM	#2
niteice Programmer Join Date: Aug 2005 Posts: 98 Rep Power: 4	sqrt((aa)+(bb));

Oct 21st, 2006, 4:43 PM	#4
Fall Back Son Professional Programmer Join Date: Oct 2006 Posts: 311 Rep Power: 3	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"

Oct 21st, 2006, 4:49 PM	#5
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 2,099 Rep Power: 6	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.

Oct 21st, 2006, 6:58 PM	#8
Fall Back Son Professional Programmer Join Date: Oct 2006 Posts: 311 Rep Power: 3	Thanks. and for the record, I understand mathematics, I am mostly just a noob at programming & chemistry. Lol. Thanks for all the help though.

Oct 21st, 2006, 7:09 PM	#9
Fall Back Son Professional Programmer Join Date: Oct 2006 Posts: 311 Rep Power: 3	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

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Programming Forums > Application Development > C

How do I take a square root?