Difference in casting float to int, 32-bit C

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.

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