Incorrect rounding results occur when you use some functions in Visual C++ (184234)

Microsoft Visual C++ 2.0

Microsoft Visual C++ 2.1

Microsoft Visual C++ 2.2

Microsoft Visual C++ 4.0

Microsoft Visual C++ 4.1

Microsoft Visual C++, 32-bit Enterprise Edition 4.2

Microsoft Visual C++, 32-bit Enterprise Edition 5.0

Microsoft Visual C++, 32-bit Enterprise Edition 6.0

Microsoft Visual C++, 32-bit Professional Edition 4.2

Microsoft Visual C++, 32-bit Professional Edition 5.0

Microsoft Visual C++, 32-bit Professional Edition 6.0

Microsoft Visual C++, 32-bit Learning Edition 6.0

Microsoft Visual C++ 2005 Express Edition

Microsoft Visual C++ .NET (2003)

Microsoft Visual C++ .NET (2002)

This article was previously published under Q184234

SYMPTOMS

You might get incorrect rounding results when you use the following functions:

_fcvt()
printf()
fprintf()
sprintf()
vprintf()
vfprintf()
vsprintf()

CAUSE

In the 16-bit compiler, the floating-point representation for a double data type is in 80 bits. The 32-bit compiler uses the Institute of Electrical and Electronics Engineers, Inc. (IEEE) floating-point specification of 64 bits. Because you cannot always get an exact representation of decimal floating-point numbers in binary form, the reduction in the number of bits affects the rounding result for some numbers.

RESOLUTION

The following code demonstrates this behavior. The results are shown for both Visual C++ 5.0, Visual C++ 6.0, Visual C++ 2005 (32-bit compiler), and Visual C++ .NET (32-bit compiler) and Visual C++ 1.52 (16-bit compiler).

Sample Code

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

void main( void )

{
double Value;
   int    Decimal;
   int    Sign;

Value = 6.6975;
(void) printf( "1)  %.7f  -->  %.3f  -->  %s\n", Value, Value,
_fcvt( Value, 3, &Decimal, &Sign ) );
Value = 6.06975;
(void) printf( "2)  %.7f  -->  %.4f  -->  %s\n", Value, Value,
_fcvt( Value, 4, &Decimal, &Sign ) );
Value = 6.006975;
(void) printf( "3)  %.7f  -->  %.5f  -->  %s\n", Value, Value,
_fcvt( Value, 5, &Decimal, &Sign ) );
Value = 1.2345;
(void) printf( "4)  %.7f  -->  %.3f  -->  %s\n", Value, Value,
_fcvt( Value, 3, &Decimal, &Sign ) );
Value = 1.02345;
(void) printf( "5)  %.7f  -->  %.4f  -->  %s\n", Value, Value,
_fcvt( Value, 4, &Decimal, &Sign ) );
Value = 1.002345;
(void) printf( "6)  %.7f  -->  %.5f  -->  %s\n", Value, Value,
_fcvt( Value, 5, &Decimal, &Sign ) );
}

VC++ 1.52c (16-bit compiler) results:

   1)    6.6975000  -->  6.698  -->  6698
   2)    6.0697500  -->  6.0698  -->  60698
   3)    6.0069750  -->  6.00698  -->  600698
   4)    1.2345000  -->  1.235  -->  1235
   5)    1.0234500  -->  1.0235  -->  10235
   6)    1.0023450  -->  1.00235  -->  100235

VC++ 5.0 (32-bit compiler) results:

   1)    6.6975000  -->  6.697  -->  6697
   2)    6.0697500  -->  6.0698  -->  60698
   3)    6.0069750  -->  6.00697  -->  600697
   4)    1.2345000  -->  1.234  -->  1234
   5)    1.0234500  -->  1.0235  -->  10235
   6)    1.0023450  -->  1.00235  -->  100235

With Visual C++ 5.0, test cases 2, 5, and 6 are correct, while 1, 3, and 4 do not round as expected.

To work around this behavior, add a very small number to the variable used. In the example above, add 1e-10 to Value. Modify each assignment, as shown in the following example:

   Value = 6.06975+1e-10;

STATUS

This behavior is by design.

MORE INFORMATION

By adding the small number, you offset the rounding error that is caused by inexact representation of some decimal floating-point numbers in binary. You can make this number even smaller, such as equal to or greater than 1e- 15.

REFERENCES

For additional information about using floating-point numbers, click the following article number to view the article in the Microsoft Knowledge Base:

145889 INFO: Why Floating Point Numbers May Lose Precision

Modification Type:

Major

Last Reviewed:

1/9/2006

Keywords:

kbfunctions kbtshoot kbcode kbCompiler kbprb KB184234 kbAudDeveloper