Newbee数组指针问题 [英] Newbee array pointer question
问题描述
为什么下面的代码不起作用？
我想在函数中初始化一个字符串，但是我获得了那些STATUS_ACCESS_VIOLATION异常的
。
#include< stdio.h>
#include< stdlib.h>
void init（char * buffer）
{
buffer =（char *）malloc（6 * sizeof（char））;
buffer [0] =''H'';
buffer [1] =''e'';
buffer [2] =''l'';
buffer [3] ='' l'';
buffer [4] =''o'';
buffer [5] =''\''';
}
int main（）
{
char * buffer;
init（缓冲区）;
printf（"％s"，buffer）;
}
任何解决方法或如何解决此问题？
TIA，
Martin
Why ist the following code not working?
I want to initialize a string in function, but I am
getting those STATUS_ACCESS_VIOLATION exceptions.
#include <stdio.h>
#include <stdlib.h>
void init(char* buffer)
{
buffer = (char*) malloc(6 * sizeof(char));
buffer[0] = ''H'';
buffer[1] = ''e'';
buffer[2] = ''l'';
buffer[3] = ''l'';
buffer[4] = ''o'';
buffer[5] = ''\0'';
}
int main()
{
char* buffer;
init(buffer);
printf("%s", buffer);
}
Any workaround or how to for this?
TIA,
Martin
推荐答案
Martin Andert< ma ******** ***@gmx.de>潦草地写道：
Martin Andert <ma***********@gmx.de> scribbled the following:
为什么以下代码不起作用？
我想在函数中初始化一个字符串，但是我正在获取那些STATUS_ACCESS_VIOLATION异常。
#include< stdio.h>
#include< stdlib.h>
void init（char * buffer）
{
buffer =（char *）malloc（6 * sizeof（char））;
Why ist the following code not working?
I want to initialize a string in function, but I am
getting those STATUS_ACCESS_VIOLATION exceptions.
#include <stdio.h>
#include <stdlib.h> void init(char* buffer)
{
buffer = (char*) malloc(6 * sizeof(char));
这是一个非常常见的新手问题。即使参数的类型是指针，C参数传递也是严格按值计算的。因此，你只需要在这里修改init（）函数的缓冲区本地副本。尝试
将指针传递给缓冲区。
-
/ - Joona Palaste（pa ***** @ cc.helsinki.fi）-------------芬兰-------- \
\ --------- -----------------------------------------------规则！ -------- /
不，玛吉，不是阿兹台克人，奥尔梅克！ Ol-mec！
- Lisa Simpson
This is a very common newbie problem. C parameter passing is strictly
by value, even if the parameter''s type is a pointer. Therefore you''re
only modifying the init() function''s local copy of buffer here. Try
passing a pointer to buffer instead.
--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"No, Maggie, not Aztec, Olmec! Ol-mec!"
- Lisa Simpson
Martin Andert写道：
Martin Andert wrote:
为什么会这样以下代码不起作用？
我想在函数中初始化一个字符串，但是我正在获取那些STATUS_ACCESS_VIOLATION异常。
#include< stdio.h>
#include< stdlib.h>
void init（char * buffer）
char * init（void）
{
char * buffer;
所以你不会造成内存泄漏，并且可以返回指针。
{
buffer =（char *）malloc（6 * sizeof（char））;
buffer = malloc（6）;
足够并且更好。根据定义，sizeof char是1.
buffer [0] =''H'';
buffer [1] =''e'';
buffer [2] = ''l'';
缓冲区[3] =''l'';
缓冲区[4] =''o'';
缓冲区[5] =''\ 0 '';
返回缓冲区; }
int main（）
{char * buffer;
init（缓冲区）;
buffer = init（）;
printf（"％s"，buffer）;
}
任何变通方法或如何做到这一点？
Why ist the following code not working?
I want to initialize a string in function, but I am
getting those STATUS_ACCESS_VIOLATION exceptions.
#include <stdio.h>
#include <stdlib.h>
void init(char* buffer)
char *init(void)
{
char *buffer;
so you don''t create a memory leak, and can return the pointer.
{
buffer = (char*) malloc(6 * sizeof(char));
buffer = malloc(6);
suffices and is better. By definition sizeof char is 1.
buffer[0] = ''H'';
buffer[1] = ''e'';
buffer[2] = ''l'';
buffer[3] = ''l'';
buffer[4] = ''o'';
buffer[5] = ''\0'';
return buffer; }
int main()
{
char* buffer;
init(buffer);
buffer = init();
printf("%s", buffer);
}
Any workaround or how to for this?
见上文。
-
查克F（cb********@yahoo.com）（cb********@worldnet.att.net）
可用于咨询/临时嵌入式和系统。
< http://cbfalconer.home.att.net>使用worldnet地址！
See above.
--
Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
Martin Andert写道：
Martin Andert wrote:
为什么以下代码不起作用？
我想在函数中初始化一个字符串，但是我正在获取那些STATUS_ACCESS_VIOLATION异常。
＃include< stdio.h>
#include< stdlib.h>
void init（char * buffer）
{/> buffer =（char *）malloc（6 * sizeof（char））;
buffer [0] =' 'H'';
buffer [1] =''e'';
buffer [2] =''l'';
buffer [3] =''l'';
buffer [4] =''o'';
缓冲区[5] =''\ 0'';
}
int main（）
{
char * buffer;
init（缓冲区）;
printf（"％s"，buffer）;
}
任何解决方法或如何做到这一点？
Why ist the following code not working?
I want to initialize a string in function, but I am
getting those STATUS_ACCESS_VIOLATION exceptions.
#include <stdio.h>
#include <stdlib.h>
void init(char* buffer)
{
buffer = (char*) malloc(6 * sizeof(char));
buffer[0] = ''H'';
buffer[1] = ''e'';
buffer[2] = ''l'';
buffer[3] = ''l'';
buffer[4] = ''o'';
buffer[5] = ''\0'';
}
int main()
{
char* buffer;
init(buffer);
printf("%s", buffer);
}
Any workaround or how to for this?
#include< stdio.h>
#include< stdlib.h>
void init（char ** buffer）
{
* b uffer = malloc（6 * sizeof ** buffer）;
if（* buffer！= NULL）{
buffer [0] [0] =''H'' ;
buffer [0] [1] =''e'';
buffer [0] [2] =''l'';
buffer [0] [3] =''l'';
buffer [0] [4] =''o'';
buffer [0 ] [5] =''\ 0'';
}
}
int main（void）
{
char * buffer;
init（& buffer）;
if（buffer！ = NULL）{
printf（"％s \ n"，buffer）;
免费（缓冲区）;
}
返回0;
}
-
pete
#include <stdio.h>
#include <stdlib.h>
void init(char **buffer)
{
*buffer = malloc(6 * sizeof **buffer);
if (*buffer != NULL) {
buffer[0][0] = ''H'';
buffer[0][1] = ''e'';
buffer[0][2] = ''l'';
buffer[0][3] = ''l'';
buffer[0][4] = ''o'';
buffer[0][5] = ''\0'';
}
}
int main(void)
{
char *buffer;
init(&buffer);
if (buffer != NULL) {
printf("%s\n", buffer);
free(buffer);
}
return 0;
}
--
pete
这篇关于Newbee数组指针问题的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
