PRB: Underflow Caused By exp() Function Fails to Set errno

• The C Run Time (CRT) included with: - Microsoft Visual C++, 32-bit Editions, version 5.0

SYMPTOMS

When using the exp() function and an underflow error occurs, the errno variable is not set to a value.

RESOLUTION

   #include <stdio.h>
   #include <math.h>
   #include <ERRNO.H>
   extern int errno;

   void main( void )
   {
      double x = -1.0e+3,y;

      y = exp( x );

      // The exp function returns the exponential value of the
      // floating-point parameter, x, if successful. On overflow,
      // the function returns INF (infinite) and on underflow, exp
      // returns 0.

      // Check for error conditions.
      // Underflow error or other.
      if(y==0 || (errno!=0))
      {
         if (y==0)
            printf("ERROR: Underflow Error\n");
         else
            perror("ERROR");

         printf("Errno: %i\n",errno);
      }
      // No Errors--print results.
      else
      {
         printf( "exp( %f ) = %f\n", x, y );
      }
   }

STATUS

This behavior is by design.

MORE INFORMATION

The errno variable is a global variable, which is given an integer value upon an error. Math functions, such as exp(), set this value to indicate an error state; library math routines set errno by calling _matherr. However, the errno variable is not set when an underflow occurs. According to the ANSI C specification, whether or not errno acquires the value of the macro ERANGE is implementation-defined.

Steps to Reproduce Behavior

Type in and execute the following program [when using the exp() function and an underflow error occurs, the errno variable is not set to a value]:

   #include <stdio.h>
   #include <math.h>
   #include <ERRNO.H>
   extern int errno;

   void main( void )
   {
      double x = -1.0e+3,y;

      y = exp( x );

      // Check for error conditions.
      // perror() returns the error message associated with errno.
      if(errno!=0)
      {
            perror("ERROR");
      }
      // No Errors--print results. But error did occur because we have an
      // underflow error.
      // The underflow goes undetected because errno wasn't set to any
      // value.
      else
      {
         printf( "exp( %f ) = %f\n", x, y );
      }
   }