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:
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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
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