64位分段错误 [英] Segmentation fault on 64 bit
问题描述
大家好，
我正在粘贴一段在32位系统上执行正常的代码，但是当编译64位编译器时，
因分段错误而失败。我是
在PA-RISC系统上使用HP-UX C编译器。从提到32到64位
系统的可移植性问题的文件中获取此代码
。
但是当我包括系统文件< malloc.h，代码在两个系统上执行罚款
。
#include< stdio.h>
int main（int argc，char ** argv）
{
char mystring1 [10] =" foo";
char * mystring2;
mystring2 =（char *）malloc（sizeof（char）*（long）10）;
strcpy（mystring2，" bar \ n \ 0"）;
printf（"％s％s"，mystring1，mystring2）;
返回0;
}
32位输出
foo bar
我会尝试调试但是无法针对原因。粘贴gdb输出的
剪辑。
程序收到信号SIGSEGV，分段错误。
0x800003ffff743ac8 in strlen + 0x10（ ）from /usr/lib/pa20_64/libc.2
（gdb）bt
＃0 0x800003ffff743ac8 in strlen + 0x10（）来自/ usr / lib / pa20_64 / libc .2
＃1 0x4000000000001f08 in main（argc = 1，argv = 0x800003ffff7f07a0）
test1.c：8
我假设它与malloc参数10有关，这个参数被强制为
long但无法推理出来。任何人都可以解释一下细节吗？
Hi All,
I am pasting a piece of code which executes fine on 32 bit system but
fails with a segmentation fault when compiled 64 bit compiler.I am
using a HP-UX C compiler on PA-RISC system. This code was picked up
from a document mentioning portability issues from 32 to 64 bit
systems.
But when I include the system file <malloc.h, the code executes fine
on both the systems.
#include <stdio.h>
int main(int argc, char **argv)
{
char mystring1[10] = "foo";
char *mystring2;
mystring2 = (char *)malloc(sizeof(char)*(long)10);
strcpy(mystring2, "bar\n\0");
printf("%s%s", mystring1, mystring2);
return 0;
}
Output on 32 bit
foo bar
I''ll tried to debug it but was unable to target the cause. pasting a
clip of the gdb output.
Program received signal SIGSEGV, Segmentation fault.
0x800003ffff743ac8 in strlen+0x10 () from /usr/lib/pa20_64/libc.2
(gdb) bt
#0 0x800003ffff743ac8 in strlen+0x10 () from /usr/lib/pa20_64/libc.2
#1 0x4000000000001f08 in main (argc=1, argv=0x800003ffff7f07a0) at
test1.c:8
I assume it is to do with the malloc argument 10 which is typecasted to
long but unable to reason out. Can anyone explain the details?
推荐答案
Linny说：
Linny said:
大家好，
我正在粘贴一段代码，该代码在32位系统上执行正常但是当编译64位编译器时，
因分段错误而失败。我是
在PA-RISC系统上使用HP-UX C编译器。从提到32到64位
系统的可移植性问题的文档中获取此代码
。
Hi All,
I am pasting a piece of code which executes fine on 32 bit system but
fails with a segmentation fault when compiled 64 bit compiler.I am
using a HP-UX C compiler on PA-RISC system. This code was picked up
from a document mentioning portability issues from 32 to 64 bit
systems.
你没能为malloc提供原型，这不是一个函数
返回int，所以行为是未定义的。你的编译器会有
警告你，但你通过（毫无意义地）投射malloc'的结果来阻止它。
您的问题在一些细节中有详细解释关于铸造的论文，你可以找到 http://www.cpax.org.uk/prg/writings/casting.php
但是当我包含系统文件时< malloc.h，代码在两个系统上执行良好
。
But when I include the system file <malloc.h, the code executes fine
on both the systems.
你真正想要的是适用于malloc的原型，你可以通过
获得，包括标准系统头，< stdlib。 h-而且，当你关于
时，为什么不添加< string.hfor strcpy？
标题不具装饰性！它们是出于某种目的而提供的。确保你
使用你需要的那些。
What you really want is a proper prototype for malloc, which you can get by
including the standard system header, <stdlib.h- and, while you''re about
it, why not add <string.hfor strcpy?
Headers are not decorative! They are provided for a purpose. Make sure you
use the ones you need.
>
#include< stdio.h> ;
int main（int argc，char ** argv）
{
char mystring1 [10] =" foo";
char * mystring2;
mystring2 =（char *）malloc（sizeof（char）*（long）10）;
strcpy（mystring2，" ; bar\\\
\0"）;
printf（"％s％s"，mystring1，mystring2）;
返回0;
}
>
#include <stdio.h>
int main(int argc, char **argv)
{
char mystring1[10] = "foo";
char *mystring2;
mystring2 = (char *)malloc(sizeof(char)*(long)10);
strcpy(mystring2, "bar\n\0");
printf("%s%s", mystring1, mystring2);
return 0;
}
#include< stdlib.h>
#include< string.h>
#include< stdio.h>
int main（int argc，char ** argv）
{
char mystring1 [10] =" foo";
char * mystring2;
mystring2 = malloc（sizeof * mystring2 * 10）;
if（mystring2！= NULL）
{
strcpy（mystring2，" bar \ n"）;
printf （"％s％s"，mystring1，mystring2）;
fr ee（mystring2）;
}
返回0;
}
我在这里：〜/ scratchmake
gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align
-Wpointer-arith -Wbad-function-cast -Wmissing-prototypes
-Wstrict-prototypes -Wmissing-declarations -Winline -Wundef
-Wnested-externs -Wcast-qual -Wshadow -Wconversion -Wwrite-strings
-Wno-conversion -ffloat-store -O2 -g -pg -c -o foo.o foo.c
foo.c：在函数`main''：
foo.c：5：警告：未使用的参数`argc''
foo.c：5：警告：未使用的参数`argv''
gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align
-Wpointer-arith -Wbad-function-cast -Wmissing-prototypes
-Wstrict -prototypes -Wmissing-declarations -Winline -Wundef
-Wnested-externs -Wcast-qual -Wshadow -Wconversion -Wwrite-strings
-Wno-conversion -ffloat-store -O2 -g -pg -o foo foo.o -lm
我@here：〜/ scratch。/ foo
foobar
-
Richard Heathfield
Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上面的域名（但显然放弃了www）
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
int main(int argc, char **argv)
{
char mystring1[10] = "foo";
char *mystring2;
mystring2 = malloc(sizeof *mystring2 * 10);
if(mystring2 != NULL)
{
strcpy(mystring2, "bar\n");
printf("%s%s", mystring1, mystring2);
free(mystring2);
}
return 0;
}
me@here:~/scratchmake
gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align
-Wpointer-arith -Wbad-function-cast -Wmissing-prototypes
-Wstrict-prototypes -Wmissing-declarations -Winline -Wundef
-Wnested-externs -Wcast-qual -Wshadow -Wconversion -Wwrite-strings
-Wno-conversion -ffloat-store -O2 -g -pg -c -o foo.o foo.c
foo.c: In function `main'':
foo.c:5: warning: unused parameter `argc''
foo.c:5: warning: unused parameter `argv''
gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align
-Wpointer-arith -Wbad-function-cast -Wmissing-prototypes
-Wstrict-prototypes -Wmissing-declarations -Winline -Wundef
-Wnested-externs -Wcast-qual -Wshadow -Wconversion -Wwrite-strings
-Wno-conversion -ffloat-store -O2 -g -pg -o foo foo.o -lm
me@here:~/scratch./foo
foobar
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
Linny写道：
=
Linny wrote:
=
我粘贴了一段代码，该代码在32位系统上执行正常，但
在编译时因分段错误而失败64位编译器。我是在PA-RISC系统上使用HP-UX C编译器的
。从提到32到64位
系统的可移植性问题的文件中获取此代码
。
但是当我包括系统文件< malloc.h，
I am pasting a piece of code which executes fine on 32 bit system but
fails with a segmentation fault when compiled 64 bit compiler.I am
using a HP-UX C compiler on PA-RISC system. This code was picked up
from a document mentioning portability issues from 32 to 64 bit
systems.
But when I include the system file <malloc.h,
当然你的意思是< stdlib.h>？没有< malloc.hin
标准C.
Surely you mean <stdlib.h>? There''s no such thing as <malloc.hin
standard C.
代码执行正常
$ b两个系统都是$ b。
#include< stdio.h>
int main（int argc，char ** argv）
{
char mystring1 [10] =" foo";
char * mystring2;
mystring2 =（char *）的malloc（的sizeof（char）的*（长）10）;
the code executes fine
on both the systems.
#include <stdio.h>
int main(int argc, char **argv)
{
char mystring1[10] = "foo";
char *mystring2;
mystring2 = (char *)malloc(sizeof(char)*(long)10);
BOOOOOOOOOOOOM。
你做了什么？你没有在范围内声明`malloc`
。编译器必须为您声明它。它给了
它是一个返回类型的`int`，因为它必须。演员到char *
隐瞒了这一点。
我的赌注是调用约定不同于
64位实现和32位实现，
，会发生什么是64位地址获得
传回一个不同于int的寄存器，
所以代码将垃圾转换为char *。 （或者
结果被截断为32位，或者其他什么。谁知道
知道什么？）
BOOOOOOOOOOOOM.
What did you do? You did not have a declaration of `malloc`
in scope. The compiler had to declare it for you. It gave
it a return-type of `int`, as it had to. The cast to char*
conceals this.
My bet is that the calling conventions are different on
the 64-bit implementation and the 32-bit implementation,
and what happens is that the 64-bit address gets
passed back in a different register than an int would,
so the code casts garbage to char*. (Or maybe the
result gets truncated to 32 bits, or something. Who
knows?)
我假设它与malloc参数10有关，这个参数被强制为
long但无法推理出来。任何人都可以解释细节吗？
I assume it is to do with the malloc argument 10 which is typecasted to
long but unable to reason out. Can anyone explain the details?
不，这是因为编译器首先被误导，然后
被禁止投诉。开头的代码是错的，
你所看到的是/原因/为什么它错了。
#include< ; stdlib.h>
...
char * mystring2 = malloc（10 * sizeof（* mystring2））;
...
会解决它。
-
Chris" Essen -6并且计算 ; Dollin
得分，哇。如果我想得分，我会去玩/年龄的蒸汽/。
No, it''s because the compiler was first misled and then
prevented from complaining. The code was wrong to start with,
and what you''re seeing is the /reason/ why it''s wrong.
#include <stdlib.h>
...
char *mystring2 = malloc( 10 * sizeof( *mystring2 ) );
...
would fix it.
--
Chris "Essen -6 and counting" Dollin
Scoring, bah. If I want scoring I''ll go play /Age of Steam/.
Linny写道：
Linny wrote:
大家好，
我粘贴了一段在32位系统上执行正常的代码，但是当编译64位编译器时，
因分段错误而失败.I我在PA-RISC系统上使用HP-UX C编译器获得
。从提到32到64位
系统的可移植性问题的文件中获取此代码
。
但是当我包括系统文件< malloc.h，代码在两个系统上执行罚款
。
#include< stdio.h>
Hi All,
I am pasting a piece of code which executes fine on 32 bit system but
fails with a segmentation fault when compiled 64 bit compiler.I am
using a HP-UX C compiler on PA-RISC system. This code was picked up
from a document mentioning portability issues from 32 to 64 bit
systems.
But when I include the system file <malloc.h, the code executes fine
on both the systems.
#include <stdio.h>
您忘了包含malloc的头文件。
我会留给您找到您需要的头文件 - 它虽然
不是malloc.h。
You forgot to include the header file for malloc.
I''ll leave it up to you to find which one you need - it
is not malloc.h though.
int main（int argc，char ** argv）
{
char mystring1 [10] =" foo";
char * mystring2;
mystring2 =（char *）malloc（sizeof （字符）*（长）10）;
int main(int argc, char **argv)
{
char mystring1[10] = "foo";
char *mystring2;
mystring2 = (char *)malloc(sizeof(char)*(long)10);
不需要这个转换为char *，并且可能会发出警告
你应该关心。
sizeof（char）总是一个，并且没有任何理由
对于长时间的演员来说。除非你知道
究竟是什么原因，否则不要四处乱扔。
This cast to a char *is not needed, and might supress a warning
you should care about.
sizeof(char) is always one, and there isn''t any justification
for the cast to long. Don''t throw around casts unless you know
exactly why.
strcpy（mystring2，" bar\\\
\0" ）;
strcpy(mystring2, "bar\n\0");
字符串文字已经空终止，在这种情况下无需再添加另一个。
String literals are already null terminated, no need
to add another one in this case.
printf（"％s％s"，mystring1，mystring2）;
返回0;
}
输出32位
printf("%s%s", mystring1, mystring2);
return 0;
}
Output on 32 bit
[snip]
[snip]
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