从malloc（）覆盖分配的内存 [英] Overwriting Allocated Memory from malloc()
问题描述
我一直在搞乱缓冲区，并且
我发现下面的代码将会运行没有分段错误的特殊情况。
据我所知，从调用malloc（）中覆盖分配的空间
导致未定义
行为。
任何想法为什么要覆盖动态
分配的空间不会导致分段
错误？
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
#include< errno.h>
int
main（无效）
{
char * ptr = NULL;
char buffer [256]; / *故意没有初始化* /
int y = 0;
if（（ptr = malloc（1））== 0）{
perror（" malloc（）问题"）;
退出（1）;
}
for（int i; i< 255; i ++）
buffer [i] =''Z'';
while（（ptr [y ] = buffer [y]）！=''\ 0''）
y ++;
（void）printf（" ptr is：％s \ n" ;，ptr）;
免费（ptr）;
返回0;
}
I have been messing around with buffers, and
I found it peculiar that the code below will
run without a segmentation fault.
As far as I know, overwriting the allocated space
from a call to malloc() results in undefined
behavior.
Any ideas why overwriting the dynamically
assigned space doesn''t cause a segmentation
fault?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
int
main(void)
{
char *ptr = NULL;
char buffer[256]; /* intentionally did not initialize */
int y = 0;
if ( ( ptr = malloc(1) ) == 0 ) {
perror("malloc() problem");
exit(1);
}
for ( int i; i < 255; i++ )
buffer[i] = ''Z'';
while ( (ptr[y] = buffer[y]) != ''\0'' )
y++;
(void)printf("ptr is: %s\n", ptr);
free(ptr);
return 0;
}
推荐答案
2004年7月8日星期四18:22:16 -0700，tweak< xb ********** @ cox.net>写道：
On Thu, 08 Jul 2004 18:22:16 -0700, tweak <xb**********@cox.net> wrote:
据我所知，从调用malloc（）覆盖分配的空间
会导致未定义的行为。
为什么覆盖动态分配的空间的任何想法都不会导致分段错误？
As far as I know, overwriting the allocated space
from a call to malloc() results in undefined
behavior.
Any ideas why overwriting the dynamically
assigned space doesn''t cause a segmentation
fault?
因为定义的未定义行为可以是什么都没有。你好b / b
幸运，仅此而已。
-
Eric Amick
哥伦比亚，MD
Because undefined behavior by definition can be anything at all. You
were lucky, nothing more.
--
Eric Amick
Columbia, MD
tweak写道：
tweak wrote:
我一直在搞乱缓冲区，并且
我发现下面的代码在没有分段错误的情况下运行是很奇怪的。
据我所知，覆盖分配的空间
malloc（）导致未定义的行为。
为什么覆盖动态分配的空间的任何想法都不会导致分段错误？
I have been messing around with buffers, and
I found it peculiar that the code below will
run without a segmentation fault.
As far as I know, overwriting the allocated space
from a call to malloc() results in undefined
behavior.
Any ideas why overwriting the dynamically
assigned space doesn''t cause a segmentation
fault?
为什么你会期望未定义的行为（UB）导致seg错误？
对我而言，UB似乎暗示任何事情都可能发生;好，坏，预期
或意外。
Why would you expect undefined behavior(UB) to cause seg fault?
To me, UB seems to suggest that anything can happen; good, bad, expected
or unexpected.
2004年7月8日星期四18:22:16 -0700，调整< xb ********** @ cox.net>
写道：
On Thu, 08 Jul 2004 18:22:16 -0700, tweak <xb**********@cox.net>
wrote:
我一直在搞乱缓冲区，并且我发现下面的代码在没有分段错误的情况下运行是很奇怪的。
据我所知，覆盖分配的空间
调用malloc（）导致未定义的行为。
为什么覆盖动态分配的空间的任何想法都不会导致分段错误？
未定义未定义的行为会导致段错误。在其最有害的形式中，它似乎正常工作，直到它能够满足墨菲的法律之一。
#include< ; stdio.h>
#include< stdlib.h>
#include< string.h>
#include< errno.h>
int
main（void）
char * ptr = NULL;
char buffer [256]; / *故意没有初始化* /
int y = 0;
if（（ptr = malloc（1））== 0）{
perror（" malloc（ ）问题"）;
退出（1）;
}
for（int i; i< 255; i ++）
buffer [i] ='' Z'';
while（（ptr [y] = buffer [y]）！=''\ 0''）
y ++;
（void）printf（ ptr是：％s \ n，ptr）;
免费（ptr）;
返回0;
}
I have been messing around with buffers, and
I found it peculiar that the code below will
run without a segmentation fault.
As far as I know, overwriting the allocated space
from a call to malloc() results in undefined
behavior.
Any ideas why overwriting the dynamically
assigned space doesn''t cause a segmentation
fault?
Undefined behavior is not guaranteed to cause a segment fault. In its
most pernicious form, it appears to work properly until it can satisfy
one of Murphy''s laws.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
int
main(void)
{
char *ptr = NULL;
char buffer[256]; /* intentionally did not initialize */
int y = 0;
if ( ( ptr = malloc(1) ) == 0 ) {
perror("malloc() problem");
exit(1);
}
for ( int i; i < 255; i++ )
buffer[i] = ''Z'';
while ( (ptr[y] = buffer[y]) != ''\0'' )
y++;
(void)printf("ptr is: %s\n", ptr);
free(ptr);
return 0;
}
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