char *和char [] [英] char* and char[]
问题描述
大家好，
我是C的新手，所以请耐心等我:)
我需要创建一个数组在函数（局部变量）中，大小为
为1MB。由于局部变量存储在堆栈中，在运行时，我得到一个错误。
我尝试使用mallac（）函数。这个函数只有在我为$ * b $ b为char *而不是char []分配内存时才有效。
我希望操作函数中的字符串，但是在使用char *时我听说过
，这个值并不意味着要改变。好像我使用char []，我可以根据需要多次更改值（我希望b $ b想要）。
有人可以告诉我如何正确地将内存分配给
本地变量（字符串），这样我就可以将字符串复制到这个变量，并且
以任何方式操纵它以及我希望多少次。
提前致谢，
mike79
Hello all,
I am a newbie to C, so please bear with me :)
I need to create an array in a function (a local variable) with size
of 1MB. Since local variables get stored on the stack, at run-time, I
get an error.
I tried using mallac() function. This function only works when I am
allocating memory to a char* and not a char[].
I wish to manipulate the string in my function, but I''ve heard that
when using char*, the value is not meant to be altered. Where as if I
use char[], I can change the value as many times as I like (which I
want to).
Could someone please tell me how to properly allocate memory to a
local variable (string), so I can copy a string to this variable, and
manipulate it any way and how ever many times I wish.
Thanks in advance,
mike79
推荐答案
Artie Gold写道：
Artie Gold wrote:
也许这会有所帮助：
void some_function（）{
/ *以下是指向一个字符串文字* /
char * cannot_be_altered ="这个文字不能改变;
/ *以下是一个'char'的数组，不能改变
^^^^^^
ITYM可以（但是固定大小）* /
char can_be_altered [] =此文本可以更改;
/ *下面为39个字符的stri分配了足够的空间ng
和地址（如果成功）在char * /
char * can_also_be_altered = malloc（40）的指针中的地方;
if（can_also_be_altered！= NULL）
strcpy（ can_also_be_altered，此文本也可以更改;
/ *以下调用未定义的行为* /
strcpy（cannot_be_altered，fooled ya！）;
/ *以下是完全正常* /
can_be_altered [0] =''T'';
/ *以下调用未定义的行为* /
strcpy （can_be_altered，某些字符串的任意长度超过
之前的内容）;
/ *以下内容完全正常* /
can_also_be_altered [0 ] ='''T'';
/ * ......依旧...... * /
}
Perhaps this will help:
void some_function() {
/* the following is a pointer to a string literal */
char * cannot_be_altered = "this text cannot be altered";
/* the following is an array of `char''s that cannot be altered ^^^^^^
ITYM can (but is of fixed size) */
char can_be_altered[] = "this text can be altered";
/* the following allocated enough space for a 39 character string
and places that address (if successful) in a pointer to char */
char * can_also_be_altered = malloc(40);
if (can_also_be_altered != NULL)
strcpy(can_also_be_altered, "this text can also be altered";
/* the following invokes undefined behavior */
strcpy(cannot_be_altered, "fooled ya!");
/* the following is perfectly OK */
can_be_altered[0] = ''T'';
/* the following invokes undefined behavior */
strcpy(can_be_altered, "some string of arbitrary length exceeding "
"what was there before");
/* the following is perfectly OK */
can_also_be_altered[0] = ''T'';
/* ... and so on ... */
}
-
Bertrand Mollinier Toublet
现实存在 - Richard Heathfield，2003年7月1日
--
Bertrand Mollinier Toublet
"Reality exists" - Richard Heathfield, 1 July 2003
Bertrand Mollinier Toublet写道：
Bertrand Mollinier Toublet wrote:
Artie Gold写道：
Artie Gold wrote:
也许这会有所帮助：
void some_function（）{
/ *以下是指向字符串文字的指针* /
char * cannot_be_altered ="这个文字不能改变;
/ *以下是一个'char'的数组，不能改变
^^^^^^
ITYM可以
Perhaps this will help:
void some_function() {
/* the following is a pointer to a string literal */
char * cannot_be_altered = "this text cannot be altered";
/* the following is an array of `char''s that cannot be altered
^^^^^^
ITYM can
Doh！
当然！
--ag
[当然这可以导致对代码中的评论进行长时间的讨论。
希望它不会;-)]
Doh!
Of course!
--ag
[Of course this _could_ lead to a long discussion on comments in code.
Hopefully it won''t ;-)]
（但是具有固定的大小）* /
char can_be_altered [] ="此文本可以更改;
/ *以下分配足够的空间容纳39个字符的字符串
和地址（如果成功）放在指向char * /
char * can_also_be_altered = malloc（40）的指针中;
if（can_also_be_altered！= NULL）
strcpy（can_also_be_altered，此文本也可以更改;
/ *以下调用未定义的行为* /
strcpy（cannot_be_altered，" fooled ya！" ）;
/ *以下是完全正常* /
can_be_altered [0] =''T'';
/ *以下调用未定义的行为* /
strcpy（can_be_altered，一些任意长度的字符串超过
之前的内容）;
/ *以下内容完全正常* /
can_also_be_altered [0] =''T'';
/ * ......依旧...... * /
}
(but is of fixed size) */
char can_be_altered[] = "this text can be altered";
/* the following allocated enough space for a 39 character string
and places that address (if successful) in a pointer to char */
char * can_also_be_altered = malloc(40);
if (can_also_be_altered != NULL)
strcpy(can_also_be_altered, "this text can also be altered";
/* the following invokes undefined behavior */
strcpy(cannot_be_altered, "fooled ya!");
/* the following is perfectly OK */
can_be_altered[0] = ''T'';
/* the following invokes undefined behavior */
strcpy(can_be_altered, "some string of arbitrary length exceeding "
"what was there before");
/* the following is perfectly OK */
can_also_be_altered[0] = ''T'';
/* ... and so on ... */
}
-
Artie Gold - 德克萨斯州奥斯汀
--
Artie Gold -- Austin, Texas
A rtie Gold< ar ******* @ austin.rr.com>在消息新闻中写道：< 3F ************** @ austin.rr.com> ...
Artie Gold <ar*******@austin.rr.com> wrote in message news:<3F**************@austin.rr.com>...
mike79写道：
大家好，
我是C的新手，所以请耐心等待我:)
我需要在函数（局部变量）中创建一个大小的数组
1MB。由于局部变量存储在堆栈中，在运行时，我收到错误。
我尝试使用mallac（）函数。这个函数只有在我为char *而不是char []分配内存时才有效。
我希望在我的函数中操作字符串，但我听说过
使用char *时，该值不应更改。好像我使用char []，我可以随意更改值（我想要的）。
Hello all,
I am a newbie to C, so please bear with me :)
I need to create an array in a function (a local variable) with size
of 1MB. Since local variables get stored on the stack, at run-time, I
get an error.
I tried using mallac() function. This function only works when I am
allocating memory to a char* and not a char[].
I wish to manipulate the string in my function, but I''ve heard that
when using char*, the value is not meant to be altered. Where as if I
use char[], I can change the value as many times as I like (which I
want to).
啊，我看看你困惑的地方！
也许这会有所帮助：
void some_function（）{
/ *以下是指向字符串文字的指针* /
char * cannot_be_altered ="此文本无法更改" ;;
/ *以下是一个无法更改的`char'数组
（但是固定大小）* /
char can_be_altered [] ="此文本可以更改;
/ *下面为39个字符的字符串分配了足够的空间
和地方指向char * /
char * can_also_be_altered = malloc（40）的指针（如果成功）;
if（can_also_be_altered！= NULL）
strcpy（can_also_be_altered，" this text can也可以改变;
/ *以下调用未定义的行为* /
strcpy（cannot_be _altered，fooled ya！）;
/ *以下内容非常好* /
can_be_altered [0] =''T'';
/ *以下调用未定义的行为* /
strcpy（can_be_altered，某些字符串的任意长度超过
之前的内容）;
/ *以下完全没问题* /
can_also_be_altered [0] =''T'';
Ah, I see where you''re confused!
Perhaps this will help:
void some_function() {
/* the following is a pointer to a string literal */
char * cannot_be_altered = "this text cannot be altered";
/* the following is an array of `char''s that cannot be altered
(but is of fixed size) */
char can_be_altered[] = "this text can be altered";
/* the following allocated enough space for a 39 character string
and places that address (if successful) in a pointer to char */
char * can_also_be_altered = malloc(40);
if (can_also_be_altered != NULL)
strcpy(can_also_be_altered, "this text can also be altered";
/* the following invokes undefined behavior */
strcpy(cannot_be_altered, "fooled ya!");
/* the following is perfectly OK */
can_be_altered[0] = ''T'';
/* the following invokes undefined behavior */
strcpy(can_be_altered, "some string of arbitrary length exceeding "
"what was there before");
/* the following is perfectly OK */
can_also_be_altered[0] = ''T'';
怎么样
int main（）{
char * p =" hello";
char c [5] =" hello";
c [1 ] =''y''; / *不确定这个！* /
p = c; / *也不确定这个！* /
返回0 ;
}
不是var''c''内存是只读的！
另外p = c;不应该是一个问题吗？
- 拉维
}
how about
int main (){
char *p = "hello";
char c[5] = "hello";
c[1] = ''y'';/* Not sure on this one !*/
p = c;/* Not sure on this one too !*/
return 0;
}
Isnt it that var ''c'' memory is read only !
Also p=c; shouldnt be a problem right ?
- Ravi
}
这篇关于char *和char []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！
