Re: Why are these two functions compiled differently?

[Ian Lance Taylor <iant at google dot com>]

"Bingfeng Mei" <bmei@broadcom.com> writes:

> #define A  255
>
> int tst1(short a, short b){
>   if(a > (b - A))
>     return 0;
>   else
>     return 1;  
>
> }
>
>
> int tst2(short a, short b){
>   short c = b - A;
>   if(a > c)
>     return 0;
>   else
>     return 1;  
>
> }

These computations are different.  Assume short is 16 bits and int is 32
bits.  Consider the case of b == 0x8000.  The tst1 function will sign
extend that to 0xffff8000, subtract 255, sign extend a to int, and
compare the sign extended a with 0xffff7f01.  The tst2 function will
compute 0x8000 - 255 as a 16 bit value, getting 0x7f01.  It will sign
extend that to 0x7f01, and compare the sign extended a with 0x7f01.

It may seem that there is an undefined signed overflow when using this
value, but there actually isn't.  All computations in C are done in type
int.  So the computations have no overflow.  The truncation from int to
short is not undefined, it is implementation defined.

Ian