PTR [英] ptr

问题描述

大家好。

我对这个程序有疑问

#include< stdio.h>

main（）

{

char str1 [] =" hello";

char * p = hello;

p =" bye" ;

}

根据这个程序，p是指向字符串hello的指针。当我们指向

指向再见时。只有3个字母应该擦除。但它是再见。

为什么？

Hi everyone
I have a doubt in this program
#include<stdio.h>
main()
{
char str1[]="hello";
char *p=hello;
p="bye";
}
According to this program p is pointer to the string hello.When we made
to point to bye.Only 3 letters should erase.But it is giving bye .
Why?

推荐答案

嗨！


考虑你的程序这个小修改：


#include< stdio.h>

main（）

{

char str1 [] =" hello";

char * p =" hello";

printf（" ;％u =％s"，p，p）;

p =" bye";

printf（"％u =％s"，p，p） ;

}


要观察的输出：


4202496 =你好

4202510 = bye


这意味着两个字符串hello和再见出现在2

不同的内存位置，通过更改指针

''p'指向你的地址，观察到的输出是合理的。


如果我的解释错误，请纠正我。

Hi!

Consider this small modificationt o your program:

#include<stdio.h>
main()
{
char str1[]="hello";
char *p="hello";
printf("%u = %s",p,p);
p="bye";
printf("%u = %s",p,p);
}

Output to be observed:

4202496 = hello
4202510 = bye

which means that the two strings "hello" and "bye" are present at 2
different memory locations and by changing the address to which pointer
''p'' points to you, the observed output is justified.

Please correct me if my interpretation is wrong.

vim写道：

vim wrote:

大家好
我对这个程序有疑问


我也有疑问，因为它根本没有编译：


gcc.exe -c main.c -pedantic -W -Wall -Wextra -ansi


main.c中包含的文件：1：

main.c：3：警告：返回类型默认为int'

main.c：在函数main中：

main.c：5：错误：`hello''未声明（在此函数中首次使用）

main.c：5：错误:(每个未声明的标识符仅报告一次

main.c：5：错误：对于它出现的每个函数。）

main.c：4：警告：未使用的变量`str1''

make。 exe：*** [m ain.o]错误1

执行终止

#include< stdio.h>
main（）
{
char str1 [ ] =" hello";
char * p = hello;
p =" bye";
}
根据这个程序，p是指向字符串hello的指针。当我们指示再见。只有3个字母应该擦除。但它是再见。
为什么？

Hi everyone
I have a doubt in this program
I have a doubt as well, since it doesn''t compile at all:

gcc.exe -c main.c -pedantic -W -Wall -Wextra -ansi

In file included from main.c:1:
main.c:3: warning: return type defaults to `int''
main.c: In function `main'':
main.c:5: error: `hello'' undeclared (first use in this function)
main.c:5: error: (Each undeclared identifier is reported only once
main.c:5: error: for each function it appears in.)
main.c:4: warning: unused variable `str1''
make.exe: *** [main.o] Error 1
Execution terminated
#include<stdio.h>
main()
{
char str1[]="hello";
char *p=hello;
p="bye";
}
According to this program p is pointer to the string hello.When we made
to point to bye.Only 3 letters should erase.But it is giving bye .
Why?

[如果你修复它，并添加一个`printf（）`因此它输出

的东西。]

因为你指定给`p`的不是字符串bye但指向

常量字符串文字bye的指针。在'str1`的情况下，你所写的

只是数组初始化的简写：


char str1 [] = {''h' '，'''，''''，'''，''o''，''\ 0''};


你可能意味着：


char * p =" hello";


再次向`p`指定一个指向常量的指针string literal

" hello" ;.


常量字符串文字保存在内存中（你不知道，

并且不小心），你不能修改它们，即：


* p =''Z'';

是非法的。

[Provided you fix it, and add a `printf()` so it does output
something.]
Because what you assign to `p` is not a string "bye" but a pointer to a
constant string literal "bye". In case of `str1`, what you''ve written
is just shorthand for array initialisation:

char str1[] = {''h'',''e'',''l'',''l'',''o'',''\0''};

In the second line you probably meant:

char *p = "hello";

which again assigns to `p` a pointer to a constant string literal
"hello".

Constant string literals are held somewhere in memory (you don''t know,
and shoudln''t care), and you''re not allowed to modify them, i.e.:

*p = ''Z'';

is illegal.

HandledException写道：

HandledException wrote:

嗨！

考虑一下你的程序的这个小修改：
＃include< stdio.h>
main（）


int main （无效）

{
char str1 [] =" hello"; char * p =" hello";
printf（"％u =％s"，p，p）;


输出指针的唯一正确方法是使用％p说明符。

p =" bye";
printf（"％u =％s"，p，p）;


如果你没有以''\ n''终止ouptut，你可能根本看不到它

（你可以使用` fflush（）`以及。）


此外，在C99中不需要，但仍然很好：


返回0;


（注意，C99不允许`main（）`的隐式类型，所以你的代码不是
C99，而`return`实际上是，需要。）

}

要观察的输出：

4202496 =你好
4202510 = bye

这意味着两个字符串你好和再见存在于2个不同的存储位置，通过改变指针
''p'指向你的地址，观察到的输出是合理的。

请纠正我，如果我的解释是错误的。

Hi!

Consider this small modificationt o your program:

#include<stdio.h>
main()
int main(void)
{
char str1[]="hello";
char *p="hello";
printf("%u = %s",p,p);
The only correct way to output pointers is using "%p" specifier.
p="bye";
printf("%u = %s",p,p);
If you don''t terminate ouptut with a ''\n'', you may not see it at all
(you can use `fflush()` as well).

Also, not required in C99, but nice still:

return 0;

(NB, C99 wouldn''t allow implicit type for `main()`, so your code is not
C99, and the `return` is, in fact, required.)
}

Output to be observed:

4202496 = hello
4202510 = bye

which means that the two strings "hello" and "bye" are present at 2
different memory locations and by changing the address to which pointer
''p'' points to you, the observed output is justified.

Please correct me if my interpretation is wrong.

这不是解释，而是实验。你解释了*什么*

发生了，而不是*如何*或*为什么*。您还在我上面指出的corected

代码中出错。


引用上下文。阅读< http://cfaj.freeshell.org/google/>。

It''s not interpretation, it''s experimentation. You explained *what*
happens, not *how* or *why*. You also made errors in your "corected"
code which I pointed out above.

Quote context. Read <http://cfaj.freeshell.org/google/>.

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