字符串问题？ [英] String Problem?

问题描述

大家好，


我发现

char x [4] = {" my"};

can编译。


但是

char x [4];

x = {" my"};

无法编译。


为什么？

如有任何建议，我们将不胜感激！


Best问候，

Davy

解决方案

文章< 11 ********* *************@g44g2000cwa.googlegroups .com> ;,

Davy< zh ******* @ gmail.com>写道：

我发现
char x [4] = {" my"};
可以编译。
但是
char x [4];
x = {" my"};
无法编译。
为什么？

因为它是这样的。


变量声明的初始化器有语法可用

在其他地方不可用。

-

相机制造商暂时推迟推出

亚毫巴分辨率生物超维等离子体空间聚合成像，

但指示仍然指日可待。

Ok davy，要了解这个，

你需要一些C / C ++处理数组名称和数组的方式背景

指针。


char * p1 =" hello";

char p2 [] =你好;


现在.. p1和p2可用于大多数地方可互换..喜欢

1. p1 [0];

2. *（p2 + 2）

等等..

但是很少有例外。

1. p2 ++

2. p2 - = 10;

3. p2 = p1

4. char * ch =" world " ;

p2 = ch;

无效


拇指规则是你不能对其进行任何修改操作地址

由数组名称指向。你将得到一个LValue所需的错误。


Davy < ZH ******* @ gmail.com>在消息中写道

news：11 ********************** @ g44g2000cwa.googlegr oups.com ...

大家好，

我发现
char x [4] = {" my"};
可以编译。

但是
char x [4];
x = {" my"};
无法编译。

为什么？
任何建议都会非常感谢！

致以诚挚的问候，
Davy

Davy写道：< blockquote class =post_quotes>大家好，

我发现
char x [4] = {" my"};
可以编译。


是的，但最好这样做：


char x [] =" my";


这种方式只有足够的空间用于我的提供。

但是
char x [4];
x = {" my"};
无法编译。


因为x的类型...


char x [4];


....实际上是......


char * const x;


....这意味着指针的值英寸×"是不变的 - 可能不是

修改，而这样做......


x = {" my"}; // - 实际上x =" my" ...


....尝试修改指针x，这就是你得到编译器的原因

错误。


建议如下：


char x [100] = {''\ 0''};

std :: ostrstream os（x，sizeof（x））;

os<< Hooh，hah <<的std ::结束; //不要忘记std :: ends;


或...


std :: string x（" my" ）;


或......


char x [4] = {''\ 0''};

strcpy（x，" my"）;


请记住，数据的内容存在于地址

位置所在的x点，是可以修改的。地址本身（或ptr的

值）不是。


问候，


W


为什么？
任何建议都将不胜感激！

此致，Davy

Hi all,

I found
char x[4]={"my"};
can be compiled.

But
char x[4];
x={"my"};
can not be compiled.

Why?
Any suggestions will be appreciated!

Best regards,
Davy

解决方案

In article <11**********************@g44g2000cwa.googlegroups .com>,
Davy <zh*******@gmail.com> wrote:

I found
char x[4]={"my"};
can be compiled. But
char x[4];
x={"my"};
can not be compiled. Why?

Because that''s how it is.

Initializers on a variable declaration have syntaxes available
that are not available in other places.
--
Camera manufacturers have temporarily delayed introduction of
sub-millibarn resolution bio-hyperdimensional plasmatic space polyimaging,
but indications are that is still just around the corner.

Ok davy, to understand this,
you need some background on the way C / C++ deals with array names and array
pointers.

char* p1 = "hello";
char p2[] = hello;

now.. p1 and p2 can be used in most places interchangeable.. like
1. p1[0];
2. *(p2+2)
etc etc..
but there are few exceptioins for this.
1. p2++
2. p2 - =10;
3. p2 = p1
4. char* ch = "world" ;
p2 = ch;
ARE NOT VALID

the thumb rule is "you cant do any modication operations to the address
pointed by the array name". You will get an LValue required error.


"Davy" <zh*******@gmail.com> wrote in message
news:11**********************@g44g2000cwa.googlegr oups.com...

Hi all,

I found
char x[4]={"my"};
can be compiled.

But
char x[4];
x={"my"};
can not be compiled.

Why?
Any suggestions will be appreciated!

Best regards,
Davy

Davy wrote:

Hi all,

I found
char x[4]={"my"};
can be compiled.
Yes, but it is better to do this:

char x[] = "my";

This way just enough space for "my" is provided.
But
char x[4];
x={"my"};
can not be compiled.
Because the type of x in...

char x[4];

....is effectively...

char* const x;

....which means the value of the pointer "x" is constant - may not be
modified, whereas doing this...

x={"my"}; //- actually x = "my"...

....attemps to modify the pointer x, which is why you get a compiler
error.

Suggestions as follow:

char x[100] = { ''\0'' };
std::ostrstream os( x, sizeof(x) );
os << "Hooh, hah" << std::ends; //don''t forget std::ends;

or...

std::string x("my");

or...

char x[4] = { ''\0'' };
strcpy( x, "my" );

Remember, the contents of the array which exists at the address
location whereto x points, is modifiable. The address itself (or the
value of the ptr) is not.

Regards,

W


Why?
Any suggestions will be appreciated!

Best regards,
Davy

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