const char * = new char [6] [英] const char* = new char[6]

问题描述

大家好


我有


const char * p =" Hello";


所以，这里的内存不是由C ++编译器为p分配的，因此我不能访问p [0]将内容修改为Kello

p [ 0] ='''K''; //运行时错误


所以我做了


const char * p = new char [6];


但是我如何将其初始化为Hello？我的要求是 - 我想要

a const char *初始化后来想要修改内容。


我知道如下所示的方式
< br $>
const char p [] =" hello";

const_cast< char&>（p [0]）='''K''; //好吧


但是如何用指针实现这个？


我所知道的是 - 当我做的时候C编译器分配内存
const char * =" hello";


但C ++编译器不这样做。欢迎任何帮助。


谢谢

SS

解决方案

< blockquote> SS写道：

我有


const char * p =" Hello" ;;


所以，这里的内存不是由C ++编译器为p分配的，因此我不能访问p [0]将内容修改为Kello

p [0] =''K''; //运行时错误


所以我做了


const char * p = new char [6];


但是我如何将其初始化为Hello？我的要求是 - 我想要

一个const char *初始化，后来想要修改内容。

无论你尝试什么，这都将是未定义的行为。

修改一个声明为const的对象只是一个禁忌。

我知道如下所示的方式


const char p [] =" hello" ;

const_cast< char&>（p [0]）='''K''; // OK

不，不行。这是未定义的行为。它可以在你的平台上运行 - 现在为

。

但如何用指针实现这一点？


我所知道的是 - C编译器在我执行时分配内存

const char * =" hello";


但C ++编译器不这样做。欢迎任何帮助。

Best


Kai-Uwe Bux

< blockquote> SS写道：

大家好


我有


const char * p =" Hello";


所以，这里的内存不是由C ++编译器为p
分配的

p是只有一个指针。编译器确实为指针分配存储器

本身。

因此我无法访问p [0]将内容修改为 Kello

p [0] =''K''; //运行时错误

编译器为字符串文字分配存储空间，但作为常量。所以

你可以访问它，但你不能写它。

所以我做了


const char * p = new char [6];


但是如何将其初始化为Hello？

你不能直接初始化它。您可以使用strcpy将文字

复制到您的数组中，例如：


strcpy（p，" Hello"）;


但是，除非你删除''const''，否则没有演员就不会工作。


顺便说一句：别忘了删除你的一旦你不再需要它就行了。

我的要求是 - 我想要一个const char *初始化然后想要

修改内容。

如果你想修改它，不要把它变成常量。

我知道一个方式如下所示


const char p [] =" hello";

const_cast< char&>（p [0]）='' K ''; // OK

如果要写入数组，为什么要使数组为const？只做：


char p [] =" hello";

p [0] =''K'';

但如何用指针实现这一点？

char p [] =" hello";

char * ptr = p;

ptr [ 0] ='''K'';


但为什么你真的想在这里有指针？

什么我知道是 - C编译器在我这样做时分配内存

const char * =" hello" ;;


但是C ++编译器不这样做。

C处理上述代码行的方式与C ++的处理方式没有区别。


另一个问题：为什么你使用原始指针来代替

std :: string？

SS写道：

>

const char * p =" Hello";


所以，这里内存不是由C ++编译器为p分配的，因此我不能访问p [0]将内容修改为Kello

p [0] ='' K ''; //运行时错误

不，这不正确。首先，你需要了解两件事。

P是一个指针。它是一个保存char的地址的变量。

p本身的内存是在声明的上下文中分配的。

（很可能是函数的本地函数）包含它。


你好在这种情况下，实现在一些

未指定的位置分配为（const）字符数组，这个值是

转换为指针并存储在p中。


首先，您的错误应该发生在运行时。编译器

应该在编译时拒绝存储到

a const char的访问冲突。


即使你不是在这里使用const：

char * p =" Hello";

p [0] =''K'';

是未定义的行为。不允许使用字符串文字内存

要更改（编译器允许此单据是历史记录）。

>

所以我做了


const char * p = new char [6 ];


但是我如何将其初始化为Hello？我的要求是 - 我想要

一个const char *初始化，后来想要修改内容。

Yoj不能将它初始化，但你可以复制到它。

strcpy（p，hello）;

>

我知道如下所示的方式


const char p [] = " hello";

const_cast< char&>（p [0]）='''K''; //好吧


但是如何用指针实现这个？


我所知道的是 - 当我做的时候C编译器分配内存
const char * =" hello";


但C ++编译器不这样做。欢迎任何帮助。

Nope，C和C ++在这方面实际上表现相同。

Hi Everyone

I have

const char *p = "Hello";

So, here memory is not allocated by C++ compiler for p and hence I
cannot access p[0] to modify the contents to "Kello"
p[0] = ''K''; // error at runtime

So I did

const char *p = new char[6];

But then how do I initialize it to "Hello"? My requirement is - I want
a const char* initialised and later want to modify the contents.

I know a way as written below

const char p[] = "hello";
const_cast<char&>(p[0]) = ''K''; //OK

But how to acheive this with pointers?

What I know is - C compiler allocates memory when I do
const char* = "hello";

But C++ compiler does not do that. Any help is welcome.

Thanks
SS

解决方案

S S wrote:

I have

const char *p = "Hello";

So, here memory is not allocated by C++ compiler for p and hence I
cannot access p[0] to modify the contents to "Kello"
p[0] = ''K''; // error at runtime

So I did

const char *p = new char[6];

But then how do I initialize it to "Hello"? My requirement is - I want
a const char* initialised and later want to modify the contents.

That will always be undefined behavior, regardless of what you try.
Modifying an object that was declared const is simply a no-no.

I know a way as written below

const char p[] = "hello";
const_cast<char&>(p[0]) = ''K''; //OK

Nope, not OK. It''s undefined behavior. It may work on your platform -- for
now.

But how to acheive this with pointers?

What I know is - C compiler allocates memory when I do
const char* = "hello";

But C++ compiler does not do that. Any help is welcome.

Best

Kai-Uwe Bux

S S wrote:

Hi Everyone

I have

const char *p = "Hello";

So, here memory is not allocated by C++ compiler for p

p is only a pointer. The compiler does allocate storage for the pointer
itself.

and hence I cannot access p[0] to modify the contents to "Kello"
p[0] = ''K''; // error at runtime

The compiler allocates storage for the string literal, but as a constant. So
you can access it, but you''re not allowed to write to it.

So I did

const char *p = new char[6];

But then how do I initialize it to "Hello"?

You can''t directly initialize it. You can use strcpy to copy the literal
over to your array, like:

strcpy(p, "Hello");

However, that won''t work without a cast unless you remove the ''const''.

Btw: Don''t forget to delete your array as soon as you don''t need it anymore.

My requirement is - I want a const char* initialised and later want to
modify the contents.

If you want to modify it, don''t make it const.

I know a way as written below

const char p[] = "hello";
const_cast<char&>(p[0]) = ''K''; //OK

Why do you make the array const if you want to write to it? Just do:

char p[] = "hello";
p[0] = ''K'';

But how to acheive this with pointers?

char p[] = "hello";
char* ptr = p;
ptr[0] = ''K'';

But why do you actually want to have a pointer here?

What I know is - C compiler allocates memory when I do
const char* = "hello";

But C++ compiler does not do that.

There is no difference between the way C treats the above line of code and
the way C++ does.

Another question: Why do you use raw pointers to char instead of
std::string?

S S wrote:

>
const char *p = "Hello";

So, here memory is not allocated by C++ compiler for p and hence I
cannot access p[0] to modify the contents to "Kello"
p[0] = ''K''; // error at runtime

No, that is not correct. First, you need to understand two things.
P is a pointer. It is a variable that holds the address of a char.
The memory for p itself is allocated in the context it was declared
(most likely local to a the function that contains it).

"Hello" in this context is allocated by the implementation at some
unspecified location as a array of (const) chars, this value is
converted to a pointer and stored in p.

First, your error should have happened NOT at runtime. The compiler
should reject at compile time the access violation of storing into
a const char.

Even if you were not to use const here:
char* p = "Hello";
p[0] = ''K'';
is undefined behavior. The string literal memory is not allowed
to be changed (the fact the compiler lets this slip is historical).

>
So I did

const char *p = new char[6];

But then how do I initialize it to "Hello"? My requirement is - I want
a const char* initialised and later want to modify the contents.

Yoj can''t initalialize it, but you can copy into it.
strcpy(p, "hello");

>
I know a way as written below

const char p[] = "hello";
const_cast<char&>(p[0]) = ''K''; //OK

But how to acheive this with pointers?

What I know is - C compiler allocates memory when I do
const char* = "hello";

But C++ compiler does not do that. Any help is welcome.

Nope, C and C++ actually behave identically in this aspect.

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