Hi Stephen ,
Thanks for catching the float to double promotion
in printf . The man page below confirms this .
#> man 3 printf
<snip>
The conversion specifier
f,F -
The double argument is rounded and converted to
decimal notation in the style [-]ddd.ddd, where the
number of digits after the decimal-point character is
equal to the precision specification. If the
precision is missing, it is taken as 6; if the
precision is explicitly zero, no decimal-point
character appears. If a decimal point
appears, at least one digit appears before it.
<snip>
Just to confirm my understanding, if the code was
changed as below :-
printf(" CASE 3 : d-f > %d - %f \n\n", (int)a1, a2);
only the first 32 bits of the float data a1 would be
put on stack & then the 64 bit float value .
This would be because printf is implemented as a
var args function , so everything is literally copied
to the stack.
Hence, the output would be correct.
Thanks for fixing my basics !
--- Stephen Ray <steve@xxxxxxxxxxxx> wrote:
> Just a thought, but not a particularly educated one.
> Compiling with
> -Wall gives me:
>
> test.c:4: warning: return type defaults to `int'
> test.c: In function `main':
> test.c:15: warning: int format, double arg (arg 3)
> test.c:16: warning: int format, double arg (arg 2)
> test.c:18: warning: int format, double arg (arg 2)
> test.c:18: warning: int format, double arg (arg 3)
> test.c:20: warning: control reaches end of non-void
> function
>
> Looks like maybe the floats are automatically
> promoted to doubles before
> they are passed to printf. I don't know if that's
> standard behaviour,
> but it seems reasonable. So then in the first case,
> two 64-bit values
> are put on the stack, and two 64-bit values are
> taken off the stack.
>
> In the second case, two 64-bit values are put on the
> stack, one 64-bit
> value is taken off the stack and presented
> correctly, and one 64-bit
> value has only the first 32 bits read, and
> misinterpreted as a signed int.
>
> In the third case, two 64-bit values are put on the
> stack, and the first
> 32 bits are interpreted as the first signed int, and
> the second next 64
> are interpreted as a double. So case 3 is different
> from case 2 in that
> the double value in case 3 is made up of two halves
> of two different
> doubles, while in case 2 the double is made from an
> actual double.
>
> Or something like that.
>
> --
> Kernelnewbies: Help each other learn about the Linux
> kernel.
> Archive:
> http://mail.nl.linux.org/kernelnewbies/
> FAQ: http://kernelnewbies.org/faq/
>
>
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