The #defines dont matter as cc never sees them so line 11 is seen as
func(0)
0 will be an int by default. When this call is made a 32 bit int with a
value of 0 will be pushed onto the stack. When func is executed this 32
bit value is cast to a 16 bit short and this causes a warning to be
emitted.
Even if line 11 was func((short)0) then the 16 bit 0 would be promoted to
a 32 bit 0 as its pushed on the stack and the same problem would occur.
I think anyway...as I said earlier I haven't actually loooked at the gcc
source.
Andrew
On Wed, 8 Nov 2000, Giorgos Keramidas wrote:
> Agreed. But, obviously, this is not what happens when I compile my test
> program posted earlier in this thread:
>
> 1 #include <stdio.h>
> 2
> 3 #define VALUE 0
> 4
> 5 int func (short x);
> 6
> 7 int main (void)
> 8 {
> 9 int var = 10;
> 10
> 11 func(VALUE);
> 12 return 0;
> 13 }
> 14
> 15 int func (short var)
> 16 {
> 17 printf("var: %d\n", var);
> 18 return 0;
> 19 }
>
> Even though I have a cast at line 9 that one would expect to inhibit
There is?
> such a warning, the warning remains. This is what happens when one
> tries to compile programs that use umask(2) with -Wconversion. Macros
umask is taking a 16 bit number off a 32 bit stack...
> My original question was if this is not behavior that should be
> considered buggy, since the size of VALUE has not been determined to be
> equal to sizeof(int) when the #define occurs, or has it?
No it has. It is compiled in as a 32bit int unless you qualify it. None
the less I dont think a warning should be emitted in these cases, that is
when it leaves one function as 16bit and arrives at another as 16bit.
Andrew
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