帮助* x ++ = * y ++; [英] Help for *x++=*y++;

问题描述

这个表达式到底是做什么的。

* x ++ = * y ++;

x和y都是char指针。

x指向p_ch它指向C字符串。
$ b $确实直接指向C-Strings。


我知道后增量运算符

的优先级高于

解除引用运算符。


....

p_ch = cstring;

x = p_ch;

* x =''\''';

* x ++ = * y ++;

....


''\ 0''会发生什么（* x =''\''';？

有人知道吗？


thx

What does this expression exactly do.
*x++=*y++;
x and y are both char pointers.
x points on p_ch which points on a C-string.
y indeed points directly on a C-Strings.

I know that the post-increment operator
has got a higher priority than the
dereference operator.

....
p_ch=cstring;
x=p_ch;
*x=''\0'';
*x++=*y++;
....

What happens with the ''\0'' (*x=''\0''; ?
Does somebody know?

thx

推荐答案

" Marco Stauder"< go ***** @ gmx.de>写道......

"Marco Stauder" <go*****@gmx.de> wrote...

这个表达式到底是做什么的。
* x ++ = * y ++;
x和y都是char指针。
x点在p_ch上指向一个C弦。
确实直接指向一个C弦。

我知道这个位置t-increment运算符
比
解引用运算符具有更高的优先级。


正确。


事件链类似于


// - 开始表达式


char c = * y; //暂时的


y = y + 1; //不一定在这里，但肯定是

//在原始expr结束之前


* x = c;


x = x + 1;


// - 结束表达

...
p_ch = cstring;
x = p_ch;
* x =''\''';


这似乎没必要。

* x ++ = * y ++;
...

什么发生在''\ 0''（* x =''\''';？


你是什么意思，会发生什么事情？

有人知道吗？

What does this expression exactly do.
*x++=*y++;
x and y are both char pointers.
x points on p_ch which points on a C-string.
y indeed points directly on a C-Strings.

I know that the post-increment operator
has got a higher priority than the
dereference operator.
Correct.

The chain of events is similar to

// -- beginning of the expression

char c = *y; // a temporary

y = y + 1; // not necessarily here, but certainly
// before the end of the original expr

*x = c;

x = x + 1;

// -- end of the expression

...
p_ch=cstring;
x=p_ch;
*x=''\0'';
This doesn''t seem necessary.
*x++=*y++;
...

What happens with the ''\0'' (*x=''\0''; ?
What do you mean "what happens"?
Does somebody know?

无论是谁写的都可能知道。


Victor

Whoever wrote it probably knows.

Victor

" Marco Stauder"< go ***** @ gmx.de>在消息新闻中写道：rg *********** ********************* @ 4ax.com ...

\

"Marco Stauder" <go*****@gmx.de> wrote in message news:rg********************************@4ax.com...
\

我知道后增量运算符
比
解引用运算符具有更高的优先级。


这个词是优先的。是的你是对的。

* x ++表示*（x ++）。

（不是（* x）++）。

...
p_ch = cstring;
x = p_ch;
* x =''\''';
上面被下一行删除。* x ++ = * y ++;

I know that the post-increment operator
has got a higher priority than the
dereference operator.
The word is precedence. Yes you are right.
*x++ means *(x++).
(not (*x)++).

...
p_ch=cstring;
x=p_ch;
*x=''\0''; The above obliterated by the next line. *x++=*y++;

表达式x ++和y ++都产生了原始值x和y

（在增量之前），因此从最初指向x最初指向的位置y

分配一个字符。此外，x和

y都会增加到指向下一个字符。


char str [] =" 0123456789";

x = str;

y =" abcdefghij";


* x ++ = * y ++;


将''a''移动到str中的''0'。

x和y递增（所以x现在是str + 1，指向''1''和y指着b）。

The expressions x++ and y++ both yield the original values of x and y
(before the increment), so one character is assigned from the where y
originally pointed to where x originally pointed. In addition, both x and
y are incremented to point at the next character.

char str[] = "0123456789";
x = str;
y = "abcdefghij";

*x++ = *y++;

moves the ''a''in y to the ''0'' in str.
x and y are incremented (so x is now str+1, pointing at the ''1'' and y is pointing at the b).

在文章< rg ******************* *************@4ax.com>，

Marco Stauder< go ***** @ gmx.de>写道：

In article <rg********************************@4ax.com>,
Marco Stauder <go*****@gmx.de> wrote:

这个表达式到底是做什么的。
* x ++ = * y ++;



#include< iostream>

使用命名空间std;


void foo（char * x，const char * y）

{

char * start_of_x = x;

while（* y）{

cout<< start_of_x<< " " << （int）x<< " " << （int）y<< endl;

* x ++ = * y ++;

}

}


int main（）

{

const char * b =" abcdefg" ;;

char * a = new char [8];

//以下是严格必要的吗？我不这么认为，但我不确定。

for（int i = 0; i< 8; ++ i）

a [i] = 0;

foo（a，b）;

cout<< a;

}


它是什么样的* x ++ = * y ++在上面做了什么？


x和y都是char指针。
x点在p_ch上指向一个C字符串。
y确实直接指向一个C字符串。

我知道后增量运算符
比
解引用运算符具有更高的优先级。


我不得不猜测下面的类型。你已经声明

x和y是char *所以我假设p_ch和cstring也是char *'。

...
p_ch = cstring;
现在p_ch和cstring都指向同一个地方。 X = P_CH;
，因此是x * x =''\''';
现在p_ch，cstring和x指向的一个地方包含一个

''\'''* x ++ = * y ++;
现在该地方包含y指向的地方

所包含的相同值。 p_ch和cstring仍然指向那个地方，但是

x指向下一个地方。

''\ 0''会发生什么（* x =''\\ \\ 0'';？
有人知道吗？

What does this expression exactly do.
*x++=*y++;

#include <iostream>
using namespace std;

void foo( char* x, const char* y )
{
char* start_of_x = x;
while ( *y ) {
cout << start_of_x << " " << (int)x << " " << (int) y << endl;
*x++ = *y++;
}
}

int main()
{
const char* b = "abcdefg";
char* a = new char[8];
// is the below strictly necessary? I don''t think so but am not sure.
for ( int i = 0; i < 8; ++i )
a[i] = 0;
foo( a, b );
cout << a;
}

What does it look like "*x++ = *y++" is doing in the above?

x and y are both char pointers.
x points on p_ch which points on a C-string.
y indeed points directly on a C-Strings.

I know that the post-increment operator
has got a higher priority than the
dereference operator.

I''m having to guess the types on the below. You have already stated that
x and y are char* so I''m assuming that p_ch and cstring are also char*''s.
...
p_ch=cstring; now p_ch and cstring are both pointing to the same place. x=p_ch; and so is x *x=''\0''; now that one place that p_ch, cstring, and x is pointing to contains a
''\0'' *x++=*y++; now that place contains the same value that was contained by the place
that y was pointing to. p_ch and cstring still point to that place, but
x points to the next place.
What happens with the ''\0'' (*x=''\0''; ?
Does somebody know?

\ 0没有任何反应。有些事情发生在x是的地方

指向，但它设置为''\0''。

Nothing happens to the ''\0''. Something happens to the place that x is
pointing to though, it is set to ''\0''.

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