C frexp() function

C frexp() function - Extract mantissa and exponent from a double precision number


double frexp(double x, int *exponent)

The frexp() function is used to break down the floating-point value x into a term m for the mantissa and another term n for the exponent.


Name Description Required /Optional
x Floating-point value. Required
exponent Pointer to stored integer exponent. Required

Return value from frexp()

  • Returns the mantissa term m.
  • If x is 0, frexp() returns 0 for both the mantissa and exponent. The mantissa has the same sign as the argument x.

Example: frexp() function

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

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

int main( void )
   double x, y;
   int n;

   x = 16.4;
   y = frexp( x, &n );
   printf( "frexp( %f, &n ) = %f, n = %d\n", x, y, n );
   x = 25.9;
   y = frexp( x, &n );
   printf( "frexp( %f, &n ) = %f, n = %d\n", x, y, n );
   x = 100.2;
   y = frexp( x, &n );
   printf( "frexp( %f, &n ) = %f, n = %d\n", x, y, n );


frexp( 16.400000, &n ) = 0.512500, n = 5
frexp( 25.900000, &n ) = 0.809375, n = 5
frexp( 100.200000, &n ) = 0.782813, n = 7

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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