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floor, floorf 

Visual Studio 2005

Calculates the floor of a value.

Copy 

double floor(
   double x
);
float floor(
   float x 
); // C++ only
long double floor(
   long double x
); // C++ only
float floorf(
   float x
);

Parameters

x

Floating-point value.

The floor function returns a floating-point value representing the largest integer that is less than or equal to x. There is no error return.

Input SEH Exception Matherr Exception

± QNAN,IND

none

_DOMAIN

floor has an implementation that uses Streaming SIMD Extensions 2 (SSE2). See _set_SSE2_enable for information and restrictions on using the SSE2 implementation.

C++ allows overloading, so you can call overloads of floor. In a C program, floor always takes and returns a double.

Function Required header Compatibility

floor, floorf

<math.h>

ANSI, Windows 95, Windows 98, Windows 98 Second Edition, Windows Millennium Edition, Windows NT 4.0, Windows 2000, Windows XP Home Edition, Windows XP Professional, Windows Server 2003

For additional compatibility information, see Compatibility in the Introduction.

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// crt_floor.c
// This example displays the largest integers
// less than or equal to the floating-point values 2.8
// and -2.8. It then shows the smallest integers greater
// than or equal to 2.8 and -2.8.


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

int main( void )
{
   double y;

   y = floor( 2.8 );
   printf( "The floor of 2.8 is %f\n", y );
   y = floor( -2.8 );
   printf( "The floor of -2.8 is %f\n", y );

   y = ceil( 2.8 );
   printf( "The ceil of 2.8 is %f\n", y );
   y = ceil( -2.8 );
   printf( "The ceil of -2.8 is %f\n", y );
}

Output

The floor of 2.8 is 2.000000
The floor of -2.8 is -3.000000
The ceil of 2.8 is 3.000000
The ceil of -2.8 is -2.000000
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Annotations FAQ
Round-off floating point number to its closest integer without using floor() or ceil()

In case you want want to round-off floating point numbers to closest (upper/lower) integers like this:

1.0 -> round-off to 1
1.2 -> round-off to 1
1.5 -> round-off to 2
1.8 -> round-off to 2
2.4 -> round-off to 2
2.5 -> round-off to 3

Use something like this:

int round ( float f_val )
{
    return ( f_val < 0.0 ) ? (int) ( f_val - 0.5 ) : (int) ( f_val + 0.5 );
}
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