Re: Shifting: what's going on?

[Andrew Haley <aph at redhat dot com>]

Facundo Ciccioli writes:
 > Hi. What follows applies to gcc (GCC) 3.4.2 (mingw-special), platform
 > x86-32bits.
 > 
 > This code:
 > 
 > int main() {
 >     unsigned long long a= 1;
 >     unsigned s= 63;
 > 
 >     a= (a << s);
 > 
 >     return 0;
 > }
 > generates this assembler (removing preamble and prologue stuff):
 > 
 >     movl    $1, -8(%ebp)
 >     movl    $0, -4(%ebp)
 >     movl    $63, -12(%ebp)
 >     movl    -12(%ebp), %ecx
 >     movl    -8(%ebp), %eax
 >     movl    -4(%ebp), %edx
 >     shldl   %cl,%eax, %edx
 >     sall    %cl, %eax
 >     testb   $32, %cl
 >     je  L2
 >     movl    %eax, %edx
 >     movl    $0, %eax
 > L2:
 >     movl    %eax, -8(%ebp)
 >     movl    %edx, -4(%ebp)
 > 
 > This is fine, and works as I expect.
 > However, this code:
 > 
 > int main() {
 >     unsigned long long a;
 >     unsigned s= 63;
 > 
 >     a= (1 << s);
 > 
 >     return 0;
 > }
 > generates this assembler:
 > 
 >     movl    $63, -12(%ebp)
 >     movl    -12(%ebp), %ecx
 >     movl    $1, %eax
 >     sall    %cl, %eax
 >     cltd
 >     movl    %eax, -8(%ebp)
 >     movl    %edx, -4(%ebp)
 > which is... wrong. Not only the results aren't what they should, but
 > looking a little harder, the assembler generated doesn't seem to make
 > sense. What's with the CLTD instruction there? If I change s (the
 > shifted amount) to be 63 (31 is equivalent, because only 5 bits of %cl
 > are used) then the result are a bunch of 1's because the sign of %eax
 > goes all over %edx. And that's only one problem. If s is between 62
 > and 32, the 1 appears in %eax instead of being in %edx, where it
 > should.
 > 
 > This is obviusly not urgent, since the first code is perfectly
 > acceptable and applicable to what I am doing, but I just got curious.

This is an arithmetic overflow, and is explicitly undefined.  One
result is as good as any other.  What do you expect the code to do?

Andrew.

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