# C sqrt() function

## C sqrt() function - Calculate Square Root

Syntax:

`double sqrt(double x)`

The sqrt() function is used to calculate the nonnegative value of the square root of x.

Parameters:

Name Description Required /Optional
x Non-negative floating-point value Required

Return value from sqrt()

• Returns the square root result.
• Returns 0 if x is negative.

Example: sqrt() function

The following example shows the usage of sqrt() function.

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

int main( void )
{
double val = 66.34, result;
result = sqrt( val );
if( val < 0 )
printf( "Error: sqrt returns %f\n", result );
else
printf( "The square root of %.2f is %.2f\n", val, result );

val = -100;
result = sqrt( val );
if( val < 0 )
printf( "\nError: sqrt returns %f\n", result );
else
printf( "\nThe square root of %.2f is %.2f\n", val, result );
}
``````

Output:

```The square root of 66.34 is 8.14

Error: sqrt returns -1.#IND00
```

C Programming Code Editor:

Previous C Programming: C pow()
Next C Programming: C ceil()

﻿

## C Programming: Tips of the Day

What's the point of const pointers?

const is a tool which you should use in pursuit of a very important C++ concept:

Find bugs at compile-time, rather than run-time, by getting the compiler to enforce what you mean.

Even though it does not change the functionality, adding const generates a compiler error when you're doing things you didn't mean to do. Imagine the following typo:

```void foo(int* ptr)
{
ptr = 0;// oops, I meant *ptr = 0
}
```

If you use int* const, this would generate a compiler error because you're changing the value to ptr. Adding restrictions via syntax is a good thing in general. Just don't take it too far -- the example you gave is a case where most people don't bother using const.

Ref : https://bit.ly/33Cdn3Q