Re: Integer overflow in operator new

["J.C. Pizarro" <jcpiza at gmail dot com>]

2007/4/9, J.C. Pizarro <jcpiza@gmail.com> wrote:
> _Z29__allocate_array_of_RossRidgejjj:
> [ gcc v3.4.6 : 9 instructions ]
>         movl    4(%esp), %edx
>         cmpl    12(%esp), %edx   # comparing and ?????? i lose me
>         movl    8(%esp), %eax
>         orl     $-1, %eax
>         imull   %edx, %eax   # signed multiply!!! 1 bit signed + unsigned 31x31!!!
>         pushl   %eax
>         call    _Znaj
>         popl    %edx
>         ret
>
> _Z29__allocate_array_of_RossRidgejjj:
> [ gcc 4.1.3 20070326 (prerelease) : 8 instructions ]
>         movl    4(%esp), %eax
>         orl     $-1, %ecx
>         cmpl    12(%esp), %eax   # comparing and ?????? i lose me
>         movl    8(%esp), %edx
>         movl    %edx, %ecx
>         imull   %eax, %ecx   # signed multiply!!! 1 bit signed + unsigned 31x31!!!
>         movl    %ecx, 4(%esp)
>         jmp     _Znaj
>
> _Z29__allocate_array_of_JCPizarrojjj:
> [ gcc 4.1.3 20070326 (prerelease) and gcc 3.4.6 : 7 instructions ]
>         movl    4(%esp), %edx
>         cmpl    12(%esp), %edx   # comparing and ?????? i lose me
>         movl    8(%esp), %eax
>         movl    $-1, 4(%esp)
>         imull   %edx, %eax   # signed multiply!!! 1 bit signed + unsigned 31x31!!!
>         movl    %eax, 4(%esp)
>         jmp     _Znaj
>
> -----------------------------------------------------------------------------
>
> J.C. Pizarro

I don't see a conditional jump or a test of the zero flag. Am i confuse?

The multiply is signed. It is need more researching a little bit.

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