Re: Allocations that overflow

From:
"Doug Harrison [MVP]" <dsh@mvps.org>
Newsgroups:
microsoft.public.vc.language
Date:
Thu, 19 Jun 2008 13:50:25 -0500
Message-ID:
<7h9l54djd4o9fresktnh0vcobj9itcjrkq@4ax.com>
On Thu, 19 Jun 2008 00:24:57 +0300, "Angel Tsankov"
<fn42551@fmi.uni-sofia.bg> wrote:

Hello,

According to the standard, what must the following function do if it is
passed std::numeric_limits<std::size_t>::max()?

struct S
{
char a[64]; // Any size greater than 1 would do.
};

S* allocate(std::size_t size)
{
return new S[size];
}


I can't find any prescribed behavior for this in the standard. However, VC9
does detect overflow for the multiplication of the count and object size.
Consider the following fragment:

int* f(size_t n)
{
   return new int[n];
}

Compiled with cl -O2 -EHs -FAs -c a.cpp, I get:

?f@@YAPAHI@Z PROC ; f, COMDAT

; 3 : return new int[n];

    mov eax, DWORD PTR _n$[esp-4]
    xor ecx, ecx
    mov edx, 4
    mul edx
    seto cl
    neg ecx
    or ecx, eax
    mov DWORD PTR _n$[esp-4], ecx
    jmp ??2@YAPAXI@Z ; operator new
?f@@YAPAHI@Z ENDP ; f

If overflow occurs, the argument to the operator new function will be
size_t(-1), i.e. all bits set, and operator new will detect this error.
(The seto/neg/or sequence does this.) Here's a full program to test it:

#include <stdio.h>
#include <stdexcept>

int* f(size_t n)
{
   return new int[n];
}

int main()
{
   try
   {
      f(size_t(-1));
   }
   catch (std::bad_alloc)
   {
      puts("1");
   }
}

X>cl -O2 -EHs -W4 a.cpp
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 15.00.21022.08 for
80x86
Copyright (C) Microsoft Corporation. All rights reserved.

a.cpp
Microsoft (R) Incremental Linker Version 9.00.21022.08
Copyright (C) Microsoft Corporation. All rights reserved.

/out:a.exe
a.obj

X>a
1

--
Doug Harrison
Visual C++ MVP

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