有关堆栈溢出（in vs 2005）的读书笔记堆栈中 申请大数组

部分原创

 

66下一步

Vs 堆栈 理论修改的上限是多少 ？ 问题2： 全局变量去 最大的上限是多少？ 堆栈（休养版本测试） 2012年3月31日17:31:27 3月

 



 



 



从 10^5到 10^6数组，导致vs…

当前：搭好脚手架。

 

改进： 修改vs系数。（byte 为单位i）

改进二： 细化 时间粒度。

 

 

以上是我在项目汇总的修改。下面是阅读笔记

 

 

 

 

 

test dword ptr [eax],eax ; probe page.

2010年05月18日 星期二 20:19

局部数组变量定义超过所分配的最大空间

-----------------------------------

Posts Tagged 变量

局部数组变量定义所分配的最大空间为多少?如何设置大小

有两个程序

A：

1. #include "stdafx.h"
2. int _tmain(int argc, _TCHAR* argv[])
3. {
4. int nArray[256000] = {0};
5. nArray[1] = 5;
6. printf("array 1 is %d",nArray[1]);
7. return 0;
8. }

B：

1. #include "stdafx.h"
2. int _tmain(int argc, _TCHAR* argv[])
3. {

4. int nArray[260000] = {0};

   //66 大于 1m了

5. nArray[1] = 5;
6. printf("array 1 is %d",nArray[1]);
7. return 0;
8. }

 

大家通过运行可以发现，A是可以正常运行的，B虽然编译通过了，可是当运行时就会弹出错误

错误的原因，就是栈溢出

 

局部变量的申请空间是存放于栈中，windows里默认栈内存是1M

所以当申请空间大于1M时就会出现溢出错误

通过debug就会进入以下文件chkask.asm

page ,132

title chkstk – C stack checking routine

;***

;chkstk.asm – C stack checking routine

;

; Copyright (c) Microsoft Corporation. All rights reserved.

;

;Purpose:

; Provides support for automatic stack checking in C procedures

; when stack checking is enabled.

;

;*******************************************************************************

.xlist

include cruntime.inc

.list

; size of a page of memory

_PAGESIZE_ equ 1000h

CODESEG

page

;***

;_chkstk – check stack upon procedure entry

;

;Purpose:

; Provide stack checking on procedure entry. Method is to simply probe

; each page of memory required for the stack in descending order. This

; causes the necessary pages of memory to be allocated via the guard

; page scheme, if possible. In the event of failure, the OS raises the

; _XCPT_UNABLE_TO_GROW_STACK exception.

;

; NOTE: Currently, the (EAX < _PAGESIZE_) code path falls through

; to the "lastpage" label of the (EAX >= _PAGESIZE_) code path. This

; is small; a minor speed optimization would be to special case

; this up top. This would avoid the painful save/restore of

; ecx and would shorten the code path by 4-6 instructions.

;

;Entry:

; EAX = size of local frame

;

;Exit:

; ESP = new stackframe, if successful

;

;Uses:

; EAX

;

;Exceptions:

; _XCPT_GUARD_PAGE_VIOLATION – May be raised on a page probe. NEVER TRAP

; THIS!!!! It is used by the OS to grow the

; stack on demand.

; _XCPT_UNABLE_TO_GROW_STACK – The stack cannot be grown. More precisely,

; the attempt by the OS memory manager to

; allocate another guard page in response

; to a _XCPT_GUARD_PAGE_VIOLATION has

; failed.

;

;*******************************************************************************

public _alloca_probe

_chkstk proc

_alloca_probe = _chkstk

push ecx

; Calculate new TOS.

lea ecx, [esp] + 8 – 4 ; TOS before entering function + size for ret value

sub ecx, eax ; new TOS

; Handle allocation size that results in wraparound.

; Wraparound will result in StackOverflow exception.

sbb eax, eax ; 0 if CF==0, ~0 if CF==1

not eax ; ~0 if TOS did not wrapped around, 0 otherwise

and ecx, eax ; set to 0 if wraparound

mov eax, esp ; current TOS

and eax, not ( _PAGESIZE_ – 1) ; Round down to current page boundary

cs10:

cmp ecx, eax ; Is new TOS

jb short cs20 ; in probed page?

mov eax, ecx ; yes.

pop ecx

xchg esp, eax ; update esp

mov eax, dword ptr [eax] ; get return address

mov dword ptr [esp], eax ; and put it at new TOS

ret

; Find next lower page and probe

cs20:

sub eax, _PAGESIZE_ ; decrease by PAGESIZE

test dword ptr [eax],eax ; probe page.

jmp short cs10

_chkstk endp

end

 

提示栈溢出

所以解决此问题的方法就是扩大栈空间的大小

方法为

项目->属性->链接器->系统->堆栈保留大小

注：这里填的是字节数

如果你想把他扩大为2M的话，

1024*1024*2 = 2097152

然后再编译运行的话A,B就都可以通过了

 

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