Q57586: Float Data Type May Cause Problems in TSR Applications

Article: Q57586
Product(s): See article
Version(s): 5.00 5.10
Operating System(s): MS-DOS
Keyword(s): ENDUSER | s_quickc s_quickasm | mspl13_c
Last Modified: 26-FEB-1990

The following TSR (terminate and stay resident) interrupt handler does
not execute properly when invoked. The floating point values produced
are invalid. If you change the data type "float" to "double", the
problem disappears.

Code Example
------------

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

void foo (void);
void interrupt far myint (void);
void main (void);

/* Change the next line of code to "double a = 10.0"
   and the problem will go away. */
float a = (float) 10.0;

void foo (void)
   {
   cprintf("The sqrt of %lf is %lf.\n\r",(double) a,sqrt((double) a));
   }

void interrupt far myint (void)
   {
   foo();
   }

void main (void)
   {
   foo();
   _dos_setvect (0x65, myint);
   _dos_keep (0, 2000);
   }

The above program is a TSR interrupt handler for INT 65h, which prints
out the values 10.0 and sqrt(10.0) to the console when invoked. To see
the problem with the above program, follow these steps:

1. Compile with the large memory model (/AL), and run.

2. Invoke interrupt 65h through the DOS DEBUG.EXE utility, as follows:

   a. debug  Press ENTER.

   b. a      Press ENTER.

   c. int 65 Press ENTER.

   d. ret    Press ENTER.

   e. Press ENTER.

   f. g      Press ENTER.

   g. q      Press ENTER.

The problem disappears if you change the variable "a" from type
"float" to type "double".

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