Q63053: Operands Reversed in C 6.00 Divide Operation
Article: Q63053
Product(s): See article
Version(s): 6.00 | 6.00
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | buglist6.00 | mspl13_c
Last Modified: 25-JUL-1990
Under certain situations, the Microsoft C version 6.00 compiler can
generate code that causes the denominator in a division operation to
be divided by the numerator instead of the other way around.
The following code, compiled under default optimizations with
Microsoft C version 6.00, demonstrates the problem and causes the
following error:
run-time error M6103: Math
- floating-point error: divide by 0
The problem apparently occur only when the denominator contains both a
double value and a function parameter, and the numerator is a global
float. Also, the next statement apparently must contain an expression
utilizing the result of the operation. This prerequisites to the
problem seem to indicate that the problem is directly tied to
optimizations removing common subexpressions.
In the example below, the denominator in the first assignment
statement in set_adc() is divided by the numerator. Therefore, instead
of receiving 26.0/2.56 cast to an int, i gets a value of 0, or
2.56/26.0 cast to an int.
The following are several possible workarounds to the problem:
1. Disable optimizations.
2. Declare the numerator locally or as a double instead of a float.
3. Use a global float variable in the denominator instead of parameter.
4. Don't use a double value in the denominator.
5. Cast the denominator as a float before division.
6. Break up assignment statements with a function call.
7. Use the /qc compiler option.
Code Example
------------
float f1;
float f2=26.0f; // Works if f2 is declared locally or as double.
void set_adc(float width)
{
// Works if width declared as local variable instead of parameter.
int i;
i=(int)(f2/(2.56*width));
// Works if used with float constant 2.56f.
// Works if denominator cast as float.
// Works if broken up with function call
// such as printf("hello");.
f1=f2/i;
}
void main(void)
{
set_adc(1.0f);
}
The section of mixed source and assembly below shows the assembly
instructions generated by default optimizations. The troublesome
instruction [FDIVP ST(1),ST] is at offset 0023.
Following the logic below, width is pushed onto the coprocessor
stack, ST. Then it is multiplied by 2.56. Then _f2 is pushed onto the
coprocessor stack, ST making the above result ST(1). Finally, the
FDIVP instruction takes ST(1), the denominator, and divides it by ST,
_f2 - the numerator.
The rest is to be expected, __ftol is called to convert the float to
an integer. The result, 0, is moved from the AX register into the
local variable i. Then i is pushed onto the coprocessor stack, ST,
and then the FDIVR instruction divides _f2 by this value causing the
divide by 0 error.
8: i=(int)(f2/(2.56*width));
9: f1=f2/i;
0047:0014 9B WAIT
0047:0015 D94604 FLD DWord Ptr [width]
0047:0018 9B WAIT
0047:0019 DC0EB802 FMUL QWord Ptr [__fpinit+e (02B8)]
0047:001D 9B WAIT
0047:001E D9064200 FLD DWord Ptr [_f2 (0042)]
0047:0022 9B WAIT
0047:0023 DEF9 FDIVP ST(1),ST ; wrong
0047:0025 E8001B CALL __ftol (1B28)
0047:0028 8946FE MOV Word Ptr [i],AX
0047:002B 9B WAIT
0047:002C DF46FE FILD Word Ptr [i]
0047:002F 9B WAIT
0047:0030 D83E4200 FDIVR DWord Ptr [_f2 (0042)]
0047:0034 9B WAIT
0047:0035 D91ED004 FSTP DWord Ptr [_f1 (04D0)]
0047:0039 90 NOP
0047:003A 9B WAIT
The following code generated with disabled optimizations shows the
correct method of doing this. Width is pushed onto the coprocessor
stack, ST. Width is then multiplied by 2.56 with the result stored in
ST. The FDIVR instruction then divides _f2 by the above value, and
after conversion, i equals 10 as it is supposed to.
8: i=(int)(f2/(2.56*width));
0047:0016 9B WAIT
0047:0017 D94604 FLD DWord Ptr [width]
0047:001A 9B WAIT
0047:001B DC0EB802 FMUL QWord Ptr [__fpinit+e (02B8)]
0047:001F 9B WAIT
0047:0020 D83E4200 FDIVR DWord Ptr [_f2 (0042)] ; right
0047:0024 E8091B CALL __ftol (1B30)
0047:0027 8946FE MOV Word Ptr [i],AX
9: f1=f2/i;
0047:002A 9B WAIT
0047:002B D9064200 FLD DWord Ptr [_f2 (0042)]
0047:002F 9B WAIT
0047:0030 DE76FE FIDIV Word Ptr [i]
0047:0033 9B WAIT
0047:0034 D91ED004 FSTP DWord Ptr [_f1 (04D0)]
0047:0038 90 NOP
0047:0039 9B WAIT
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