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how could i use the square root operation in the TC++

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if there is no function which your compiler supports as sqrt() then try to write that function by yourself such as

for i++ ....

//check variable with i*i
if > then try for adding real value btw 0≈1

hard way..
 

I don't know whether there is a function for squareroot in TC++, but instead I will inform you about a really simple way to write your own squareroot function. There are many algorithms to find the squareroot of a number, the simplest being the Newton's iteration method:

6_1216649149.gif


where n is the number whose squareroot would be found, and x0 = 1. So, let's say we will find the squareroot of 2.

First iteration: x1 = 1.5 (not even close)
Second iteration: x2 = 1.41667 (at least we are going somewhere)
Third iteration: x3 = 1.41422 (hmm, really close)

You can go as many iteration as you want (though I think 5 at most is enough, 3 will do a good job), it depends how accurate you want the result, and what are your system requirements.
 

TC++ help file

<math.h>
sqrt, sqrtf, sqrtl
double sqrt(double x);
float sqrt(float x); [C++ only]
long double sqrt(long double x); [C++ only]
float sqrtf(float x); [required with C99]
long double sqrtl(long double x); [required with C99]
The function returns the real square root of x, x^(1/2). A domain error occurs if x < 0.
 

coz we were asked to make a simple program usin TC++ that shall compute values using Quadratic Eq'n..

i've tried
sqrt();
sqrt((b*b)-(4*a*c);

do you think my compiler does not really support sqrt.. or is it just my prog that i hav 2 work on

thnx ^_^
 

zurgh said:
coz we were asked to make a simple program usin TC++ that shall compute values using Quadratic Eq'n..

i've tried
sqrt();
sqrt((b*b)-(4*a*c);

do you think my compiler does not really support sqrt.. or is it just my prog that i hav 2 work on

thnx ^_^
Click to expand...

if the value of sqrt((b*b)-(4*a*c) is negative then it ll stop.
 

    Z

    zurgh

    Points: 2
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i guess.. in a usual or common quadratic eq'n.. it is the 1st and last term that has the larger value.. is it? in that case a negative value shall stop my sqrt prog.. how can i correct that??
 

zurgh said:
i guess.. in a usual or common quadratic eq'n.. it is the 1st and last term that has the larger value.. is it? in that case a negative value shall stop my sqrt prog.. how can i correct that??
Click to expand...

assign ((b*b)-(4*a*c)) to a variable and check it for negativity (<0) if its negative then prompt user such as 'undefined roots'

good luck
 

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