在Unix中工作但不在Windows中工作 [英] Working in Unix but not in windows
问题描述
这段代码不能在windows中工作但在Unix上工作不是
得到原因..帮帮我
#include< iostream .h>
#include< stdio.h>
#include< string.h>
int addDynamicMemory（ char ** ptr，int size）;
int addDynamicMemory（char ** ptr，int size）
{
int currSize;
currSize = strlen（* ptr）;
size = currSize + size;
printf（"在重新分配大小为％之前） d \ n"，size）;
* ptr =（char *）realloc（* ptr，size * sizeof（char））;
printf（" memory已分配\ n"）;
返回0;
}
int main（int argc，char * argv [] ）
{
char ** test;
char * test1;
ws1_c =" san jay" ;
test =& test1;
* test =（char *）malloc（1 * sizeof（char））; //如果U会成功* test =
（char *）malloc（10 * sizeof（char））它会起作用.... :(很棒..
strcpy（* test，" TEST_11"）;
printf（" Curr val of one％s\ n，* test）;
* test =（char *）realloc（* test，20 * sizeof（char））;
strcat（* test，" 3333333333"）;
printf（ 在分配val之后是％s \ n，* test）;
addDynamicMemory（test，40）;
strcat（* test，& ; 444444444"）;
printf（" After allocation val is％s \ nn"，* test）;
addDynamicMemory（test，20） ;
strcat（* test，" 7777777"）;
printf（" After allocation val is％s \ n"，* test）;
addDynamicMemory（test，20）;
strcat（* test，" 888888888888888"）;
printf（" After allocation val是％s \ n，*测试）;
}
Chk这段代码让我知道这是不是你想要的......否则我可能会改变错误的东西。如果那就是你想要的那样我可以用b $ b解释你的代码在windows上出了什么问题
#include< iostream.h>
#include< stdio.h>
#include< string.h>
#include< stdlib.h>
int addDynamicMemory（char ** ptr，int size）;
int addDynamicMemory（char ** ptr，int size）
{
int currSize;
currSize = strlen（* ptr）;
size = currSize + size;
printf（ 在重新分配大小为％d \ n之前，大小）;
* ptr =（char *）realloc（* ptr，size * sizeof（char））;
printf（内存已分配\ n）;
返回0;
}
// int main（int argc，char * argv []）
void main（）
{
char ** test;
char * test1;
// ws1_c =" san jay" ;;
test =& test1;
* test =（char *）malloc（1 * sizeof（char））; //如果U会成功* test =
（char *）malloc（10 * sizeof（char））它会起作用.... :(很棒..
strcpy（* test，" TEST_11"）;
printf（" Curr val of one％s\ n，* test）;
* test =（char *）realloc（* test，20 * sizeof（char））;
strcat（* test，" 3333333333"）;
printf（ 在分配val之后是％s \ n，* test）;
addDynamicMemory（test，40）;
strcat（* test，& ; 444444444"）;
printf（" After allocation val is％s \ nn"，* test）;
addDynamicMemory（test，20） ;
strcat（* test，" 7777777"）;
printf（" After allocation val is％s \ n"，* test）;
addDynamicMemory（test，20）;
strcat（* test，" 888888888888888"）;
printf（" After allocation val是％s \ n，*测试）;
}
Ashwin
您好Ashwin
在更改main（）签名后仍然面临同样的问题
....我在MS VC ++中编译它并面临这个问题。 。
回复我
Sanjay
sa ********* @ gmail.com 写道：这个一段代码不能在Windows中工作，但在Unix中工作没有
得到原因..帮助我
#include< iostream.h>
这不是标准的C标题（它也不是C ++标题，但是我/ b $ b离开......）。幸运的是（在这种情况下）你不会使用它。
#include< stdio.h>
#include< string.h>
#include< stdlib.h> / * for malloc（）/ realloc（）prototypes * /
int addDynamicMemory（char ** ptr，int size）;
int addDynamicMemory（char ** ptr，int size）
定义将很好地作为原型，谢谢。 {
int currSize;
currSize = strlen（* ptr）;
size = currSize + size;
printf（重新分配大小为％d \ n之前， size）;
* ptr =（char *）realloc（* ptr，size * sizeof（char））;
不要从* alloc函数转换返回值。这是不必要的
并且可以掩盖错误（因为它在这里......见上文）。
`sizeof（char）的定义是1。 br />
*从不直接realloc（） - 如果失败就会造成内存泄漏。
（总是* realloc（）到temp。）
printf（" memory allocated \ n"）;
返回0;
}
int main（int argc，char * argv [] ）
{
char ** test;
char * test1;
ws1_c =" san jay" ;;
test =& test1;
* test =（char *）malloc（1 * sizeof（char））; //如果U会成功* test =
（char *）malloc（10 * sizeof（char））它会起作用.... :(很棒。
为什么不直接使用`test1''？
[...以及关于转换返回值的相同评论和
sizeof（char）]
.. strcpy（* test，" TEST_11"）;
你只为一个'char'分配了空间。你有什么期望？
printf（" Curr val of one％s\ n，* test）;
* test =（char *）realloc（* test，20 * sizeof（char） ）;
strcat（* test，" 3333333333"）;
printf（" After allocation val is％s \ nn"，* test）;
addDynamicMemory（ test，40）;
strcat（* test，" 444444444"）;
printf（" After allocation val is％s \ nn"，* test）;
addDynamicMemory（test，20）;
strcat（* test，" 7777777"）;
printf（" After allocation val is％s \ nn"，* test）;
addDynamicMemory（test，20）;
s trcat（* test，" 888888888888888"）;
printf（分配后val为％s \ n，* test）;
}
显然，你缺乏线索。
幸运的是，线索很容易获得。阅读一本好书C
（ http：// www / accu / org 是一个很好的建议来源; K& R II很难获得b
）。阅读C-FAQ。浏览新闻：comp.lang.c。
完成所有这些后，如果遇到问题，请回复。
HTH，
- g
-
Artie Gold - 德克萨斯州奥斯汀
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
如果你没有什么可隐瞒的，你没有尝试！
This piece of code is not working in windows but working in Unix Not
getting the reason .. help me out
#include <iostream.h>
#include <stdio.h>
#include <string.h>
int addDynamicMemory(char **ptr, int size);
int addDynamicMemory(char **ptr, int size)
{
int currSize;
currSize = strlen(*ptr);
size = currSize + size;
printf("before re Allocation size is %d\n",size);
*ptr = (char*) realloc(*ptr, size*sizeof(char));
printf("memory allocated \n");
return 0;
}
int main(int argc, char* argv[])
{
char **test;
char *test1;
ws1_c = "san jay";
test = &test1;
*test = (char*) malloc(1*sizeof(char)); // If U will make it *test =
(char*) malloc(10*sizeof(char)) It will work .... :( amazing ..
strcpy(*test,"TEST_11");
printf("Curr val of one %s\n",*test);
*test = (char*) realloc(*test, 20*sizeof(char));
strcat(*test,"3333333333");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 40) ;
strcat(*test,"444444444");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 20) ;
strcat(*test,"7777777");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 20) ;
strcat(*test,"888888888888888");
printf("After allocation val is %s\n",*test);
}
Hi,
Chk this code and let me know if this is what you want...otherwise I
might have changed the wrong things. If that is what you wanted I can
explain what is wrong with your code on windows
#include <iostream.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int addDynamicMemory(char **ptr, int size);
int addDynamicMemory(char **ptr, int size)
{
int currSize;
currSize = strlen(*ptr);
size = currSize + size;
printf("before re Allocation size is %d\n",size);
*ptr = (char*) realloc(*ptr, size*sizeof(char));
printf("memory allocated \n");
return 0;
}
//int main(int argc, char* argv[])
void main()
{
char **test;
char *test1;
//ws1_c = "san jay";
test = &test1;
*test = (char*) malloc(1*sizeof(char)); // If U will make it *test =
(char*) malloc(10*sizeof(char)) It will work .... :( amazing ..
strcpy(*test,"TEST_11");
printf("Curr val of one %s\n",*test);
*test = (char*) realloc(*test, 20*sizeof(char));
strcat(*test,"3333333333");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 40) ;
strcat(*test,"444444444");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 20) ;
strcat(*test,"7777777");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 20) ;
strcat(*test,"888888888888888");
printf("After allocation val is %s\n",*test);
}
Ashwin
Hi Ashwin
Still Facing the same problem after changing the main() signature also
.... I am compiling this in MS VC++ and facing this problem..
Reply me back
Sanjay
sa*********@gmail.com wrote:This piece of code is not working in windows but working in Unix Not
getting the reason .. help me out
#include <iostream.h> This is not a standard C header (it''s not a C++ header either, but I
digress...). Fortunately (in this context) you don''t use it.
#include <stdio.h>
#include <string.h> #include <stdlib.h> /* for malloc()/realloc() prototypes */
int addDynamicMemory(char **ptr, int size);
int addDynamicMemory(char **ptr, int size) The definition will serve quite well as a prototype, thank you. {
int currSize;
currSize = strlen(*ptr);
size = currSize + size;
printf("before re Allocation size is %d\n",size);
*ptr = (char*) realloc(*ptr, size*sizeof(char));
Do not cast the return value from *alloc functions. It is unnecessary
and can mask errors (as it has here...see above).
`sizeof(char) is 1 by definition.
*Never* directly realloc() -- it can create a memory leak if it fails.
(Always* realloc() to a temp.)
printf("memory allocated \n");
return 0;
}
int main(int argc, char* argv[])
{
char **test;
char *test1;
ws1_c = "san jay";
test = &test1;
*test = (char*) malloc(1*sizeof(char)); // If U will make it *test =
(char*) malloc(10*sizeof(char)) It will work .... :( amazing .
Why not just use `test1''?
[...as well as same comments about casting the return value and
sizeof(char)]
.. strcpy(*test,"TEST_11");
You''ve only allocated space for one `char''. What do you expect?
printf("Curr val of one %s\n",*test);
*test = (char*) realloc(*test, 20*sizeof(char));
strcat(*test,"3333333333");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 40) ;
strcat(*test,"444444444");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 20) ;
strcat(*test,"7777777");
printf("After allocation val is %s\n",*test);
addDynamicMemory(test, 20) ;
strcat(*test,"888888888888888");
printf("After allocation val is %s\n",*test);
}
Obviously, you lack a clue.
Fortunately, clues are easy to obtain. Read a good book on C
(http://www/accu/org is a good source for suggestions; K&R II is hard to
beat). Read the C-FAQ. Browse news:comp.lang.c.
Once you''ve done all that, please post back if you run into problems.
HTH,
--ag
--
Artie Gold -- Austin, Texas
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
"If you have nothing to hide, you''re not trying!"
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