11

I currently working with an old code that needs to run a 32-bit system. During this work I stumbled across an issue that (out of academic interest) I would like to understand the cause of.

It seems that casting from float to int in 32-bit C behaves differently if the cast is done on a variable or on an expression. Consider the program:

#include <stdio.h> int main() { int i,c1,c2; float f1,f10; for (i=0; i< 21; i++) { f1 = 3+i*0.1; f10 = f1*10.0; c1 = (int)f10; c2 = (int)(f1*10.0); printf("%d, %d, %d, %11.9f, %11.9f\n",c1,c2,c1-c2,f10,f1*10.0); } }

Compiled (using gcc) either directly on a 32-bit system or on a 64-bit system using the -m32 modifier the output of the program is:

30, 30, 0, 30.000000000 30.000000000 31, 30, 1, 31.000000000 30.999999046 32, 32, 0, 32.000000000 32.000000477 33, 32, 1, 33.000000000 32.999999523 34, 34, 0, 34.000000000 34.000000954 35, 35, 0, 35.000000000 35.000000000 36, 35, 1, 36.000000000 35.999999046 37, 37, 0, 37.000000000 37.000000477 38, 37, 1, 38.000000000 37.999999523 39, 39, 0, 39.000000000 39.000000954 40, 40, 0, 40.000000000 40.000000000 41, 40, 1, 41.000000000 40.999999046 42, 41, 1, 42.000000000 41.999998093 43, 43, 0, 43.000000000 43.000001907 44, 44, 0, 44.000000000 44.000000954 45, 45, 0, 45.000000000 45.000000000 46, 45, 1, 46.000000000 45.999999046 47, 46, 1, 47.000000000 46.999998093 48, 48, 0, 48.000000000 48.000001907 49, 49, 0, 49.000000000 49.000000954 50, 50, 0, 50.000000000 50.000000000

Hence, it is clear that a difference exists between casting a variable and an expression. Note, that the issue exists also if float is changed to double and/or int is changed to short or long, also the issue do not manifest if program is compiled as 64-bit.

To clarify, the issue that I'm trying to understand here is not about floating-point arithmetic/rounding, but rather differences in memory handling in 32-bit.

The issue were tested on:

  • Linux version 4.15.0-45-generic (buildd@lgw01-amd64-031) (gcc version 7.3.0 (Ubuntu 7.3.0-16ubuntu3)), program compiled using: gcc -m32 Cast32int.c

  • Linux version 2.4.20-8 ([email protected]) (gcc version 3.2.2 20030222 (Red Hat Linux 3.2.2-5)), program compiled using: gcc Cast32int.c

Any pointers to help me understand what is going on here are appreciated.

Improve this question
22
  • 2
    welcome to the world of floating point math - you may want to read stackoverflow.com/questions/588004/… Commented Feb 26, 2019 at 8:55
  • 1
    Cannot reproduce, what is your platform/compiler/version etc.? Commented Feb 26, 2019 at 8:55
  • 2
    Cannot reproduce either (windows/intel gcc 7.1.0 or Rasbian linux/ARM gcc 6.3.0) Commented Feb 26, 2019 at 9:02
  • 4
    The updated code multiplies by 10.0 which is a double. It may be that the float is promoted to double before the multiplication and then back to float before being cast to int. In the other case the double product is converted directly to int. Commented Feb 26, 2019 at 10:05
  • 3
    On some platforms and GCC versions, the floating point co-processor is used for intermediate computations (what you called "expressions") when targeting 32-bit x86. These computations are 80-bit. On 64-bit systems though this is never used, and you get SSE 64-bit computations instead. Commented Feb 26, 2019 at 10:13

4 Answers 4

Reset to default
7

With MS Visual C 2008 I was able to reproduce this.

Inspecting the assembler, the difference between the two is an intermediate store and fetch of a result with intermediate conversions:

 f10 = f1*10.0; // double result f10 converted to float and stored c1 = (int)f10; // float result f10 fetched and converted to double c2 = (int)(f1*10.0); // no store/fetch/convert

The assembler generated pushes values onto the FPU stack that get converted to 64 bits and then are multiplied. For c1 the result is then converted back to float and stored and is then retrieved again and placed on the FPU stack (and converted to double again) for a call to __ftol2_sse, a run-time function to convert a double to int.

For c2 the intermediate value is not converted to and from float and passed immediately to the __ftol2_sse function. For this function see also the answer at Convert double to int?.

Assembler:

 f10 = f1*10; fld dword ptr [f1] fmul qword ptr [__real@4024000000000000 (496190h)] fstp dword ptr [f10] c2 = (int)(f1*10); fld dword ptr [f1] fmul qword ptr [__real@4024000000000000 (496190h)] call __ftol2_sse mov dword ptr [c2],eax c1 = (int)f10; fld dword ptr [f10] call __ftol2_sse mov dword ptr [c1],eax
Improve this answer
Sign up to request clarification or add additional context in comments.

5 Comments

The stores and fetches are immaterial; they are merely implementations of the C semantics. The key reason differences are observed between f1*10.0 and f10 is the former is a double expression while the latter is a float. The assignment f10 = f1*10.0; changes the value when it converts the double to float.
@EricPostpischil Are you sure? I am pretty sure that f1*10.0 will have the precision of a long double (80 bits), rather than a double (just 64 bits).
@MartinBonner: It is irrelevant. f10 is a float; its significand is 24 bits on this platform. 10 has three significant bits, so multiplying by it produces a number with at most 27 bits. So its exact value will fit into the 53-bit significand of a double. Extended it to 64 bits will not change anything.
I think the correct answer is indeed @EricPostpischil's observation that in C all floating point calculations are performed in double precision. Hence the original code behaves as designed. My answer is just the identification of what happens "under the hood".
@PaulOgilvie: It is not that all floating-point calculations are performed in double but that using one operand of type double causes other operands to be converted to double, per the usual arithmetic conversions. If all operands are of type float, as with using 10.0f instead of 10.0, the expression type remains float (although an implementation is permitted to use double to evaluate it, but that does not seem to be happening in OP’s code).
5

In the “32-bit system,” the difference is caused by the fact that f1*10.0 uses full double precision, while f10 has only float precision because that is its type. f1*10.0 uses double precision because 10.0 is a double constant. When f1*10.0 is assigned to f10, the value changes because it is implicitly converted to float, which has less precision.

If you use the float constant 10.0f instead, the differences vanish.

Consider the first case, when i is 1. Then:

  • In f1 = 3+i*0.1, 0.1 is a double constant, so the arithmetic is performed in double, and the result is 3.100000000000000088817841970012523233890533447265625. Then, to assign this to f1, it is converted to float, which produces 3.099999904632568359375.
  • In f10 = f1*10.0;, 10.0 is a double constant, so the arithmetic is again performed in double, and the result is 30.99999904632568359375. For assignment to f10, this is converted to float, and the result is 31.
  • Later, when f10 and f1*10.0 are printed, we see the values given above, with nine digits after the decimal point, “31.000000000” for f10, and “30.999999046”.

If you print f1*10.0f, with the float constant 10.0f instead of the double constant 10.0, the result will be “31.000000000” rather than “30.999999046”.

(The above uses IEEE-754 basic 32-bit and 64-bit binary floating-point arithmetic.)

In particular, note this: The difference between f1*10.0 and f10 arises when f1*10.0 is converted to float for assignment to f10. While C permits implementations to use extra precision in evaluating expressions, it requires implementations to discard this precision in assignments and casts. Therefore, in a standard-conforming compiler, the assignment to f10 must use float precision. This means, even when the program is compiled for a “64-bit system,” the differences should occur. If they do not, the compiler does not conforming to the C standard.

Furthermore, if float is changed to double, the conversion to float does not happen, and the value will not be changed. In this case, no differences between f1*10.0 and f10 should manifest.

Given that the question reports the differences do not manifest with a “64-bit” compilation and do manifest with double, it is questionable whether the observations have been reported accurately. To clarify this, the exact code should be shown, and the observations should be reproduced by a third party.

Improve this answer

2 Comments

The question was why is there a difference between an -m32 and an -m64 build of the same source code, not what the difference between 10.0 and 10.0f is.
@NikosC.: Indeed, but the question falsely states “it is clear that a difference exists between casting a variable and an expression.” This answer disproves that and explains the difference. The question fails to present convincing evidence there is a difference between the 32-bit and 64-bit compilations—i suspect user error. I have updated the answer to note the issue.
2

The C standard is not very strict about how floating point math is to be carried out. The standard allows an implementation to do calculation with higher precision than the involved types.

The result in your case is likely to come from the fact that c1 is calculated as "float-to-int" while c2 is calculated as "double-to-int" (or even higher precision).

Here is another example showing the same behavior.

#define DD 0.11111111 int main() { int i = 27; int c1,c2,c3; float f1; double d1; printf("%.60f\n", DD); f1 = i * DD; d1 = i * DD; c1 = (int)f1; c2 = (int)(i * DD); c3 = (int)d1; printf("----------------------\n"); printf("f1: %.60f\n", f1); printf("d1: %.60f\n", d1); printf("m : %.60f\n", i * DD); printf("%d, %d, %d\n",c1,c2,c3); }

My output:

0.111111109999999999042863407794357044622302055358886718750000 ---------------------- f1: 3.000000000000000000000000000000000000000000000000000000000000 d1: 2.999999970000000182324129127664491534233093261718750000000000 m : 2.999999970000000182324129127664491534233093261718750000000000 3, 2, 2

The trick here is the number of ones in 0.11111111. The accurate result is "2.99999997". As you change the number of ones the accurate result is still in the form "2.99...997" (i.e. the number of 9 increases when the number of 1 increases).

At some point (aka some number of ones) you will reach a point where storing the result in a float rounds the result to "3.0" while the double is still able to hold "2.999999.....". Then a conversion to int will give different results.

Increasing the number of ones further will lead to a point where the double will also round to "3.0" and the conversion to int will consequently yield the same result.

Improve this answer

3 Comments

On some platforms and GCC versions, the floating point co-processor is used for intermediate computations (what the OP called "expressions"). These are 80-bit. On 64-bit systems though this is never used, and you get SSE 64-bit computations instead.
The differences do not arise in the conversions to int. The difference arises when f1*10.0, which is a double expression, is assigned to f10, which is a float. That assignment changes the value.
@EricPostpischil That was actually what I tried to write. Here: "... reach a point where storing the result in a float rounds the result to "3.0""
1

The main reason is that the rounding-control (RC) field of the x87 FPU control register values are inconsistent in the follow two lines. eventually the values of c1 and c2 are different.

0x08048457 <+58>: fstps 0x44(%esp) 0x0804848b <+110>: fistpl 0x3c(%esp)

Add gcc compile option -mfpmath=387 -mno-sse, it can be reproduced (Even without -m32, or change the float to a double)
Like this:

gcc -otest test.c -g -mfpmath=387 -mno-sse -m32

Then use gdb to debug, breakpoint at 0x0804845b, and run to i=1

 0x08048457 <+58>: fstps 0x44(%esp) 0x0804845b <+62>: flds 0x44(%esp) (gdb) info float =>R7: Valid 0x4003f7ffff8000000000 +30.99999904632568359 R6: Empty 0x4002a000000000000000 R5: Empty 0x00000000000000000000 R4: Empty 0x00000000000000000000 R3: Empty 0x00000000000000000000 R2: Empty 0x00000000000000000000 R1: Empty 0x00000000000000000000 R0: Empty 0x00000000000000000000 Status Word: 0x3820 PE TOP: 7 Control Word: 0x037f IM DM ZM OM UM PM PC: Extended Precision (64-bits) RC: Round to nearest Tag Word: 0x3fff Instruction Pointer: 0x00:0x08048455 Operand Pointer: 0x00:0x00000000 Opcode: 0x0000 (gdb) x /xw 0x44+$esp 0xffffb594: 0x41f80000 ==> 31.0, s=0, M=1.1111 E=4

observe the execution results of fstps,
at this time, the RC value on the control register on the fpu is Round to nearest.
the value on the fpu register is 30.99999904632568359 (80bit).
the value on 0x44(%esp) (variable "f10") is 31.0. (round to nearest)

Then use gdb to debug, breakpoint at 0x0804848b, and run to i=1

 0x0804848b <+110>: fistpl 0x3c(%esp) (gdb) info float =>R7: Valid 0x4003f7ffff8000000000 +30.99999904632568359 R6: Empty 0x4002a000000000000000 R5: Empty 0x00000000000000000000 R4: Empty 0x00000000000000000000 R3: Empty 0x00000000000000000000 R2: Empty 0x00000000000000000000 R1: Empty 0x00000000000000000000 R0: Empty 0x00000000000000000000 Status Word: 0x3820 PE TOP: 7 Control Word: 0x0c7f IM DM ZM OM UM PM PC: Single Precision (24-bits) RC: Round toward zero Tag Word: 0x3fff Instruction Pointer: 0x00:0x08048485 Operand Pointer: 0x00:0x00000000 Opcode: 0x0000

at this time, the RC value on the control register on the fpu is Round toward zero.
the value on the fpu register is 30.99999904632568359 (80bit). value is the same as above
obviously when the integer is converted, the decimal point is truncated, and the value is 30.

Below is the main decompiled code

 (gdb) disas main Dump of assembler code for function main: 0x0804841d <+0>: push %ebp 0x0804841e <+1>: mov %esp,%ebp 0x08048420 <+3>: and $0xfffffff0,%esp 0x08048423 <+6>: sub $0x50,%esp 0x08048426 <+9>: movl $0x0,0x4c(%esp) 0x0804842e <+17>: jmp 0x80484de <main+193> 0x08048433 <+22>: fildl 0x4c(%esp) 0x08048437 <+26>: fldl 0x80485a8 0x0804843d <+32>: fmulp %st,%st(1) 0x0804843f <+34>: fldl 0x80485b0 0x08048445 <+40>: faddp %st,%st(1) 0x08048447 <+42>: fstps 0x48(%esp) 0x0804844b <+46>: flds 0x48(%esp) 0x0804844f <+50>: flds 0x80485b8 0x08048455 <+56>: fmulp %st,%st(1) 0x08048457 <+58>: fstps 0x44(%esp) // store to f10 0x0804845b <+62>: flds 0x44(%esp) 0x0804845f <+66>: fnstcw 0x2a(%esp) 0x08048463 <+70>: movzwl 0x2a(%esp),%eax 0x08048468 <+75>: mov $0xc,%ah 0x0804846a <+77>: mov %ax,0x28(%esp) 0x0804846f <+82>: fldcw 0x28(%esp) 0x08048473 <+86>: fistpl 0x40(%esp) 0x08048477 <+90>: fldcw 0x2a(%esp) 0x0804847b <+94>: flds 0x48(%esp) 0x0804847f <+98>: fldl 0x80485c0 0x08048485 <+104>: fmulp %st,%st(1) 0x08048487 <+106>: fldcw 0x28(%esp) 0x0804848b <+110>: fistpl 0x3c(%esp) // f1 * 10 convert int 0x0804848f <+114>: fldcw 0x2a(%esp) 0x08048493 <+118>: flds 0x48(%esp) 0x08048497 <+122>: fldl 0x80485c0 0x0804849d <+128>: fmulp %st,%st(1) 0x0804849f <+130>: flds 0x44(%esp) 0x080484a3 <+134>: fxch %st(1) 0x080484a5 <+136>: mov 0x3c(%esp),%eax 0x080484a9 <+140>: mov 0x40(%esp),%edx 0x080484ad <+144>: sub %eax,%edx 0x080484af <+146>: mov %edx,%eax 0x080484b1 <+148>: fstpl 0x18(%esp) 0x080484b5 <+152>: fstpl 0x10(%esp) 0x080484b9 <+156>: mov %eax,0xc(%esp) 0x080484bd <+160>: mov 0x3c(%esp),%eax 0x080484c1 <+164>: mov %eax,0x8(%esp) 0x080484c5 <+168>: mov 0x40(%esp),%eax 0x080484c9 <+172>: mov %eax,0x4(%esp) 0x080484cd <+176>: movl $0x8048588,(%esp) 0x080484d4 <+183>: call 0x80482f0 <printf@plt> 0x080484d9 <+188>: addl $0x1,0x4c(%esp) 0x080484de <+193>: cmpl $0x14,0x4c(%esp) 0x080484e3 <+198>: jle 0x8048433 <main+22> 0x080484e9 <+204>: leave 0x080484ea <+205>: ret
Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.