Difference in casting float to int, 32-bit C

Question

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 (bhcompile@porky.devel.redhat.com) (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.

welcome to the world of floating point math - you may want to read https://stackoverflow.com/questions/588004/is-floating-point-math-broken — Support Ukraine, Feb 26 '19 at 08:55
Cannot reproduce, what is your platform/compiler/version etc.? — Jabberwocky, Feb 26 '19 at 08:55
Insert this `printf("%15.15f %15.15f\n", f1, f10);` in the loop and show the output. — Jabberwocky, Feb 26 '19 at 08:59
Cannot reproduce either (windows/intel gcc 7.1.0 or Rasbian linux/ARM gcc 6.3.0) — Euan Smith, Feb 26 '19 at 09:02
MSVC cannot reproduce but I get one warning about loss of data converting `double` to `float`. — Weather Vane, Feb 26 '19 at 09:20
@WeatherVane that warning is here: `f1 = 3 + i * 0.1;`. Putting an `f` after `0.1` makes it go away, but I doubt it will change the behaviour on the OP's platform. — Jabberwocky, Feb 26 '19 at 09:28
@Jabberwocky the difference is that in one instance the product is placed in a `float` variable and in the other instance the product goes directly to an `int` variable. Are `floats` promoted to `double`for the multiplication by an `int`? If the multiplication was by `10.0` (a `double`) instead of `10` that could be, but need not be if the result will be the same. — Weather Vane, Feb 26 '19 at 09:38
Please read the 3rd comment and act accordingly. This might help to find out what's going on. — Jabberwocky, Feb 26 '19 at 09:41
Please read [Change floating point rounding mode](https://stackoverflow.com/questions/6867693/change-floating-point-rounding-mode). — Weather Vane, Feb 26 '19 at 09:48
Floating point results can change depending on your compiler settings. Please include *complete* compilation commands to your question. — user694733, Feb 26 '19 at 09:52
@user694733 complete compilation commands have been added to post — cpaitor, Feb 26 '19 at 10:02
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`. — Weather Vane, Feb 26 '19 at 10:05
@Wheather Vane, it makes no difference if the number is multiplied by 10 or 10.0, the update to 10.0 were merely to check this — cpaitor, Feb 26 '19 at 10:10
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. — Nikos C., Feb 26 '19 at 10:13
Godbolt.org does not have a version of GCC as far back as 3.2.2, so I am unable to explore this, but [the code generated by 4.1.2](https://godbolt.org/z/mB2f70) looks like it would behave the same as shown in the question. Note that the differences between `f1*10.0` and `f10` arise entirely from the conversion to `float` during assignment to `f10`, and this conversion is mandated by the C standard; a compiler may not disregard it by the license to use more precision in expressions. Therefore, if GCC 3.2.2 behaved differently, either a flag deviating from the C standard was used or it is a bug. — Eric Postpischil, Feb 26 '19 at 14:04
a lot of similar/duplicate questions: [floating-point difference between x86 and x64](https://stackoverflow.com/q/22710272/995714), [the same code yield different numeric results on 32 vs 64-bit](https://stackoverflow.com/q/7847274/995714), [different behaviour or sqrt when compiled with 64 or 32 bits](https://stackoverflow.com/q/10808601/995714), [different result in 32 bit and 64 bit application](https://stackoverflow.com/q/33887108/995714), [acos(double) gives different result on x64 and x32 Visual Studio](https://stackoverflow.com/q/45734549/995714). Note that there's no "32-bit C" — phuclv, Feb 26 '19 at 14:40
Possible duplicate of [Difference in floating point arithmetics between x86 and x64](https://stackoverflow.com/questions/22710272/difference-in-floating-point-arithmetics-between-x86-and-x64) — phuclv, Feb 26 '19 at 14:40
@phuclv Not sure how you figure those posts to be duplicates as they consider differences between 32 vs 64 bit. The difference here is for 32 bit — cpaitor, Feb 26 '19 at 17:11
why not? it's basically about an expression that prints different values on 32 and 64-bit builds, and so are the other questions with the same explanation which is the higher evaluated precision in x87. Think horizontally across targets instead and they all boil down to the same answer — phuclv, Feb 27 '19 at 02:10

score 7 · Accepted Answer · answered Feb 26 '19 at 10:10

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

answered Feb 26 '19 at 10:10

Paul Ogilvie

25,048
4
23
41

3

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`. – Eric Postpischil Feb 26 '19 at 13:07
@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). – Martin Bonner supports Monica Feb 26 '19 at 13:15
@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. – Eric Postpischil Feb 26 '19 at 13:46
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". – Paul Ogilvie Feb 26 '19 at 13:50
@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). – Eric Postpischil Feb 26 '19 at 13:57

Eric Postpischil · Answer 2 · 2019-02-26T14:10:19.037

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.

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. — Nikos C., Feb 26 '19 at 13:48
@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. — Eric Postpischil, Feb 26 '19 at 13:53

Support Ukraine · Answer 3 · 2019-02-26T10:35:13.320

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.

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. — Nikos C., Feb 26 '19 at 10:12
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. — Eric Postpischil, Feb 26 '19 at 13:05
@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"" — Support Ukraine, Feb 26 '19 at 14:43

ccxxshow · Answer 4 · 2019-02-27T09:47:17.467

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

Difference in casting float to int, 32-bit C

4 Answers4

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