pTAL Conversion Guide

Hardware Indicators
pTAL Conversion Guide—527302-002
20-13
Overflow Traps Considerations
•
Your program might trap at slightly different locations for TNS processes than for
native processes, even if the program’s trapping state—enabled or disabled—is
the same for both processes.
For a native process, overflow is detected in a pTAL program only when the
expression on the right side of an assignment statement is assigned to the
destination on the left side. A 16-bit overflow that occurs in the middle of
expression evaluation in a TNS process might not be detected in a native process
until the assignment is made. If the expression includes a call, for example, to a
routine that updates a disk file, the result of the two programs could be very
different if overflow is detected before the routine call in a TNS process but after
the routine call in a native process.
For a native process, a 32-bit overflow always causes a trap at the instruction
which causes the overflow.
•
If overflow traps are disabled when a trap occurs, the result of the operation on
which the overflow occurred might be different for TNS processes than for native
processes.
•
Overflow can also be detected when an actual parameter is assigned to its
corresponding formal parameter during a procedure call. In this case, your
program always traps to the current trap handler, if traps are enabled; the result is
undefined if traps are disabled. Your program cannot detect this overflow condition
and the results of all subsequent operations are undefined.
•
An overflow that a TNS process detects during expression evaluation might not be
detected at all by a native process because:
°
For a native process, pTAL performs 16-bit arithmetic in 32-bit registers.
°
The code generated by the native compiler does not generally check for 16-bit
overflow until the value on the right side of an assignment statement is
assigned to the destination on the left side.
In the Example 20-5 on page 20-13, overflow traps are disabled and the operands are
all 16-bit integers. For a TNS process, adding 1 to a in the assignment statement
causes overflow and the program traps immediately.
If you compile Example 20-5 on page 20-13 by using a native compiler and run the
example as a native process, overflow is not detected because the result of adding a
and 1 is 32768, which is maintained in a 32-bit register. The result of subtracting b
(1000) from 32768 is 31768, which is stored into a with no overflow detected.
Example 20-5. Overflow Traps (TAL Only)
PROC p;
BEGIN
INT a := 32767;
INT b := 1000;
a := a + 1 - b; ! Overflow is detected by a TNS process,
END; ! but not by a native process