char指针数组 [英] array of char pointers
问题描述
我认为我终于明白了。唉。
给出：
char * s [] = {" Jan"，" Feb"，" Mar"，" April"};
是否可以将char * p指向s？
* p = s ...不会这样做，正如我所料。
* p = s [0] ...我相信它会指向s中的第一个元素。但是
这不是我想要的。我想要的是一个指针，当
递增时，它将产生Jan，Feb，Mar，April和4月。等等
我相信我想要的是指针相当于：
printf（"％s"，s [i]） ，（其中i = 0-> 3）
谢谢。
And I thought I understood it, finally. Alas.
given:
char *s[]={"Jan","Feb","Mar","April"};
is it possible to have char *p point at s?
*p = s...does not do it, as I expect.
*p=s[0]...I believe sets it to point to the first element in s. But
this is not what I want. What I want is a pointer which, when
incremented will produce "Jan","Feb","Mar","April" etc.
I believe what I want is a pointer equivalent of:
printf("%s", s[i]), (where i = 0->3)
Thanks.
推荐答案
mdh写道：
mdh wrote:
最后，我想我明白了。唉。
给出：
char * s [] = {" Jan"，" Feb"，" Mar"" April" };
是否可以将char * p指向s？
And I thought I understood it, finally. Alas.
given:
char *s[]={"Jan","Feb","Mar","April"};
is it possible to have char *p point at s?
概念上可能更容易想到s作为char **。
所以给定char * p;
p = s [0];
p现在指向字符串文字Jan。
It might be conceptually easier to think of s as a char**.
so given char* p;
p = s[0];
p now points to the string literal "Jan".
* p = s ...没有按照我的预期这样做。 />
* p = s [0] ...我相信它设置为指向s中的第一个元素。但是
这不是我想要的。我想要的是一个指针，当
递增时，它将产生Jan，Feb，Mar，April和4月。等等
*p = s...does not do it, as I expect.
*p=s[0]...I believe sets it to point to the first element in s. But
this is not what I want. What I want is a pointer which, when
incremented will produce "Jan","Feb","Mar","April" etc.
然后你希望p成为另一个char **。
Then you want p to be another char**.
我相信我想要的是一个指针相当于：
printf（"％s"，s [i]），（其中i = 0-> 3）
I believe what I want is a pointer equivalent of:
printf("%s", s[i]), (where i = 0->3)
char ** p = s;
for（unsigned n = 0; n< 4; ++ n）
{
printf（"％s"，* p）;
++ p;
}
-
Ian Collins。
char** p = s;
for( unsigned n = 0; n < 4; ++n )
{
printf("%s", *p );
++p;
}
--
Ian Collins.
4月17日下午6:59，Ian Collins< ian-n。 .. @ hotmail.comwrote：
On Apr 17, 6:59 pm, Ian Collins <ian-n...@hotmail.comwrote:
mdh写道：
mdh wrote:
给出：
char * s [] = {" Jan"，" Feb"，" Mar"，" April"};
given:
char *s[]={"Jan","Feb","Mar","April"};
>
从概念上讲，将s视为char **可能更容易。
那么你希望p成为另一个char **。
>
It might be conceptually easier to think of s as a char**.
Then you want p to be another char**.
char ** p = s;
for（unsigned n = 0; n< 4 ; ++ n）
{
printf（"％s"，* p）;
++ p;
char** p = s;
for( unsigned n = 0; n < 4; ++n )
{
printf("%s", *p );
++p;
Ian ...得到JanJanJan
如果我用* p ++取代* p得到了预期。
这有意义吗？
Ian...got "JanJanJan"
if I replaced *p with *p++ got as expected.
Does this make sense?
mdh< m ... @ comcast.netwrote：
mdh <m...@comcast.netwrote:
我认为我终于明白了。唉。
给出：
char * s [] = {" Jan"，" Feb"，" Mar"，" April" };
是否可以将char * p指向s？
And I thought I understood it, finally. Alas.
given:
char *s[]={"Jan","Feb","Mar","April"};
is it possible to have char *p point at s?
No.
No.
* p = s ...没有按照我的预期这样做。
* p = s [0] ...我相信它会指向s中的第一个元素。但是
这不是我想要的。我想要的是一个指针，当
递增时，它将产生Jan，Feb，Mar，April和4月。等等
*p = s...does not do it, as I expect.
*p=s[0]...I believe sets it to point to the first element in s. But
this is not what I want. What I want is a pointer which, when
incremented will produce "Jan","Feb","Mar","April" etc.
要指向一个元素，你需要一个指向元素类型的指针。
指向一个元素对于一个元素数组，你需要一个
指向元素类型的指针。
所以，要指向一个X数组的元素，你需要一个指针
到X.
因此，要指向一个char *数组的元素，你需要
a指针到char *。换句话说，你需要一个char **。
To point to an element, you need a pointer to the element type.
To point to an element of an array of elements, you need a
pointer to the element type.
So, to point to an element of an array of X, you need a pointer
to X.
Thus, to point to an element of an array of char *, you need
a pointer to char *. In other words, you need a char **.
我相信我想要的是指针相当于：
printf（"％s"，s [i]），（其中i = 0-> 3）
I believe what I want is a pointer equivalent of:
printf("%s", s[i]), (where i = 0->3)
#include< stdio.h>
#define countof（X）（（size_t）（sizeof（X）/ sizeof *（X）））
int main（void）
{
const char * s [] = {" Jan"，Feb，Mar，April};
const char ** p;
size_t i;
puts（" Loop 1："）;
for（i = 0; i< countof（s）; i ++）
puts（s [i]）;
puts（" \ nLoop 2："）;
for（p = s，i = 0; i< countof（s）; i ++）
puts（p [i ]）;
put（" \\\
Loop 3："）;
for（p = s; p<& s [countof] （s）]; p ++）
put（* p）;
返回0;
}
#include <stdio.h>
#define countof(X) ( (size_t) ( sizeof(X)/sizeof*(X) ) )
int main(void)
{
const char *s[]={"Jan","Feb","Mar","April"};
const char **p;
size_t i;
puts("Loop 1:");
for (i = 0; i < countof(s); i++)
puts(s[i]);
puts("\nLoop 2:");
for (p = s, i = 0; i < countof(s); i++)
puts(p[i]);
puts("\nLoop 3:");
for (p = s; p < &s[countof(s)]; p++)
puts(*p);
return 0;
}
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