走过一系列的char指针 [英] walking through an array of char pointers
问题描述
我似乎无法让它工作：
#include< stdio.h>
#include< stdlib.h> ;
#include< string.h>
int main（）
{
char * names [3];
char ** np;
names [0] =" jack";
names [1] =" jill";
names [2] =" zack";
while（** np！=''\\ \\ 0''）{
printf（"％s\ n"，* np）;
np ++;
}
返回0;
}
打印3个名字后，它会打印垃圾然后有时会出现故障。
我知道我可以使用for循环轻松地完成它，但那不是我想要的东西
for。我的印象实际上是名字：
[] - > " jack \0"
[] - > jill \0
[] - > " zack \0"
[] - > \ 0
如果是这样的话，我不能执行上述循环吗？
这样有效：
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int main（）
{
char * names [3];
char ** np;
names [0] =" jack";
names [1] =" jill";
names [ 2] =" zack";
while（** np！=''\'''）
printf（" \％ s \ n"，*（* np）++）;
np ++;
while（** np！=''\ 0''）
printf（" \％s\ n"，*（* np）++）;
np ++;
while（** np！=''\'''）
printf（" \％s\\\
"，*（* np）++）;
返回0;
{
但是，我需要3个循环来完成我想要做的一个。
我不想使用我知道有n $ $
str的事实数组中的数据。我想使用这样一个事实：每个字符串都是
null终止然后有一个最终的null终止符。
任何建议都会受到赞赏。
- gaga
I can''t seem to get this to work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *names[3];
char **np;
names[0] = "jack";
names[1] = "jill";
names[2] = "zack";
while (**np != ''\0'') {
printf("%s\n",*np);
np++;
}
return 0;
}
after printing the 3 names, it prints garbage then sometimes seg faults.
I know i can do it easily using a for loop, but that''s not what I am looking
for. I was under the impression names is actually:
[] --> "jack\0"
[] --> "jill\0"
[] --> "zack\0"
[] --> \0
If, that is so, shouldn''t I be able to perform the above loop?
This works:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *names[3];
char **np;
names[0] = "jack";
names[1] = "jill";
names[2] = "zack";
while (**np != ''\0'')
printf("\%s\n",*(*np)++);
np++;
while (**np != ''\0'')
printf("\%s\n",*(*np)++);
np++;
while (**np != ''\0'')
printf("\%s\n",*(*np)++);
return 0;
{
But, I need 3 loops for what I''d like to do in one.
I do not want to use the fact that I know there are n
strings in the array. I want to use the fact that each string is
null terminated then there is a final null terminator.
any suggestions would be appreciated.
- gaga
推荐答案
gaga< zi ******* @ aol.com>写道：
gaga <zi*******@aol.com> wrote:
我似乎无法让这个工作：
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int main（）
{char / names [3];
char ** np;
names [0] =" jack";
names [1] =" jill";
names [2] =" zack";
虽然（** np！=''\'''）{
np此时尚未初始化。我想你的意思是在循环开始之前设置
设置
np =名字;
。但即便如此，在打印zack之后，没有理由将循环停止。因为之后np
会增加到__后第三个数组元素，
这可能是一些随机位模式，当解释为
as一个char指针，并不指向存储''\0'''charac-
ter的东西。所以你需要你的
''名称''数组的第四个元素，它指向一个空字符串。
printf（"％s \ n"， * np）;
np ++;
}
返回0;
}
你需要这样的东西代替让它工作：
int main（无效）
{
char * names [] = {" jack" ，jill，zack，" };
char ** np =姓名;
while（** np！=''\'''）
printf（"％s\ n"，* np ++）;
返回EXIT_SUCCESS;
}
这个有效：
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int main（）
{char / names [3];
char ** np;
names [0] =" jack";
names [1] =" jill";
names [2] =" zack";
while（** np！=''\ 0''）
printf（" \％s\\\
"，*（* np）++）;
np ++;
while（** np！=''\ 0''）
printf（" \％s \ n"，*（* np）++）;
np ++;
while（** np！=''\'''）
printf（" \％s\\\
"，*（* np）++）;
返回0;
{
I can''t seem to get this to work: #include <stdio.h>
#include <stdlib.h>
#include <string.h> int main()
{
char *names[3];
char **np; names[0] = "jack";
names[1] = "jill";
names[2] = "zack";
while (**np != ''\0'') {
np hasn''t been in initialized at this point. I guess you meant to
set
np = names;
before the start of the loop. But even then there''s no reason why
the loop should stop after printing "zack" because afterwards np
would get incremented to point _after_ the third array element,
which probably is some random bit pattern which, when interpreted
as a char pointer, doesn''t point to something where a ''\0'' charac-
ter would be stored. So you would need a fourth element of your
''names'' array that would point to an empty string.
printf("%s\n",*np);
np++;
}
return 0;
}
You would need something like this instead to get it working:
int main( void )
{
char *names[ ] = { "jack", "jill", "zack", "" };
char **np = names;
while ( **np != ''\0'' )
printf( "%s\n", *np++ );
return EXIT_SUCCESS;
}
This works: #include <stdio.h>
#include <stdlib.h>
#include <string.h> int main()
{
char *names[3];
char **np; names[0] = "jack";
names[1] = "jill";
names[2] = "zack";
while (**np != ''\0'')
printf("\%s\n",*(*np)++);
np++;
while (**np != ''\0'')
printf("\%s\n",*(*np)++);
np++;
while (**np != ''\0'')
printf("\%s\n",*(*np)++);
return 0;
{
如果这个版本适合您，那只是偶然的 - 在我的机器上它
立即崩溃。什么是'\\'％s"和*（* np）++应该这样做吗？
问候，Jens
-
\ Jens Thoms Toerring ___ Je *********** @ physik.fu-berlin.de
\ __________________________ http://www.toerring.de
gaga写道：
我似乎无法让这个工作：
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int main（）
{
char *名称[3];
4 / *需要一个最终指针* / char ** np;
np =& names; / * initialize * / names [0] =" jack";
names [1] =" jill";
names [2] =" zack" ;;
names [3] ="" ;; / *结束标记* /
while（** np！=''\'''）{
printf（"％s \ n"，* np）;
np ++ ;
打印3个名字后，它会打印垃圾然后有时会发生故障。我知道我可以使用for循环轻松完成，但那不是我想要的。我的印象名称是
I can''t seem to get this to work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *names[3]; 4 /* need a final pointer */ char **np;
np = &names; /* initialize */ names[0] = "jack";
names[1] = "jill";
names[2] = "zack"; names[3] = ""; /* end marker */
while (**np != ''\0'') {
printf("%s\n",*np);
np++;
}
return 0;
}
after printing the 3 names, it prints garbage then sometimes seg
faults. I know i can do it easily using a for loop, but that''s
not what I am looking for. I was under the impression names is
尝试上面注释的更改/添加。更简单的是：
char * names [] = {" jack"，" jill"，" zack"，""};
char * * np;
for（np =& names; ** np; np ++）printf（"％s \ n，* np）;
-
fix（vb。）：1。纸张覆盖，模糊，隐藏在公众视野之外; 2.
以一种产生意想不到的后果的方式来解决，比原来的问题更糟糕。用法：Windows ME
修复了Windows 98 SE的许多缺点。 - Hutchison
Try the commented changes/additions above. Simpler is:
char *names[] = {"jack", "jill", "zack", ""};
char* *np;
for (np = &names; **np; np++) printf("%s\n, *np);
--
fix (vb.): 1. to paper over, obscure, hide from public view; 2.
to work around, in a way that produces unintended consequences
that are worse than the original problem. Usage: "Windows ME
fixes many of the shortcomings of Windows 98 SE". - Hutchison
CBFalconer写道：
CBFalconer wrote:
gaga写道：
gaga wrote:
我可以'似乎让这个工作：
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int main（）
{char * names [3];
I can''t seem to get this to work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *names[3];
4 / *需要一个最终指针* /
4 /* need a final pointer */
char ** np;
char **np;
np =& names; / * initialize * /
np = &names; /* initialize */
names [0] =" jack";
names [1] =" jill";
names [2] = " zack";
names[0] = "jack";
names[1] = "jill";
names[2] = "zack";
names [3] ="" ;; / *结束标记* /
names[3] = ""; /* end marker */
while（** np！=''\ 0''）{
printf（"％s \ n，* np）;
np ++;
}
返回0;
打印3个名字后，打印垃圾然后有时候seg
错误。我知道我可以使用for循环轻松完成，但那不是我想要的。我的印象名称是
while (**np != ''\0'') {
printf("%s\n",*np);
np++;
}
return 0;
}
after printing the 3 names, it prints garbage then sometimes seg
faults. I know i can do it easily using a for loop, but that''s
not what I am looking for. I was under the impression names is
尝试上面注释的更改/添加。更简单的是：
char * names [] = {" jack"，" jill"，" zack"，""};
char * * np;
for（np =& names; ** np; np ++）printf（"％s \ n，* np）;
Try the commented changes/additions above. Simpler is:
char *names[] = {"jack", "jill", "zack", ""};
char* *np;
for (np = &names; **np; np++) printf("%s\n, *np);
char * names [] = {" jack"，" jill"，" zack"，NULL};
char ** np = names;
while（ * np）printf（"％s\ n"，* np ++）;
请注意，names是指向char的指针数组。 ''names''衰减到指向数组第一个元素的
指针。指向char的指针。所以
赋值是''np = names;''，而不是''np =& names;''。
-
Joe Wright mailto：jo ******** @ comcast.net
所有内容都应尽可能简单，但并不简单。
---阿尔伯特爱因斯坦---
char *names[] = {"jack", "jill", "zack", NULL };
char **np = names;
while (*np) printf("%s\n", *np++);
Note that names is an array of pointers to char. ''names'' decays to a
pointer to the array''s first element. A pointer to char. So the
assignment is ''np = names;'', not ''np = &names;''.
--
Joe Wright mailto:jo********@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
这篇关于走过一系列的char指针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！
