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 www.cpax.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 www.cpax.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 www.cpax.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]
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