是否有更简单的方法来使用指针（包括示例代码） [英] Is there an easier way to work with pointers (sample code included)

问题描述

是否有一种更简单的方法来编写cmp程序而不必通过所有指针操作？


#include< stdlib.h>

#include< string.h>


int cmp（const void * i，const void * j）

{

void * p1，* p2;

char ** s1，** s2;

p1 =（void *）i;

p2 =（void *）j;

s1 =（void *）p1;

s2 =（void *）p2;

返回strcmp（* s1，* s2）;

}


int main（无效）

{

char string0 [] =" zebra" ;;

char string1 [] =" hello";

char string2 [] =" goodbye" ;;


char * base [3];


base [0] =（char *）string0;

base [1] =（char *）string1;

base [2] =（char *）string2;


printf（" \ nbase [0]％s"，base [0]）;

printf（" \ base [1]％s"，base [1]）;

printf（" \ base [2]％s \ n"，base [2]）;


qsort（& base，3，sizeof（base [0]），（void *）cmp）;


printf（" \\\
base [0 ]％s"，base [0]）;

printf（" \ base [1]％s"，base [1]）;

printf（" ; \ nbase [2]％s \ n"，base [2]）;

}

TIA，


Walter

Is there an easier way to code the cmp procedure without going thru all
the pointer manipulations?

#include <stdlib.h>
#include <string.h>

int cmp(const void *i, const void *j)
{
void *p1, *p2;
char **s1, **s2;
p1=(void *) i;
p2=(void *) j;
s1=(void *) p1;
s2=(void *) p2;
return strcmp(*s1,*s2);
}

int main(void)
{
char string0[]="zebra";
char string1[]="hello";
char string2[]="goodbye";

char *base[3];

base[0]=(char *)string0;
base[1]=(char *)string1;
base[2]=(char *)string2;

printf("\nbase[0] %s",base[0]);
printf("\nbase[1] %s",base[1]);
printf("\nbase[2] %s\n",base[2]);

qsort(&base,3,sizeof(base[0]),(void *)cmp);

printf("\nbase[0] %s",base[0]);
printf("\nbase[1] %s",base[1]);
printf("\nbase[2] %s\n",base[2]);
}
TIA,

Walter

推荐答案

2006年3月8日10:22:53 -0800， wa *************** @ gmail.com 写道：

On 8 Mar 2006 10:22:53 -0800, wa***************@gmail.com wrote:

有没有一种更简单的方法来编写cmp程序而不必通过所有指针操作？

#include< stdlib.h>
#include < string.h>

int cmp（const void * i，const void * j）
{
void * p1，* p2;
char ** s1，** s2;
p1 =（void *）i;
p2 =（void *）j;
s1 =（void *）p1;
s2 =（void *）p2;
返回strcmp（* s1，* s2）;
}

Is there an easier way to code the cmp procedure without going thru all
the pointer manipulations?

#include <stdlib.h>
#include <string.h>

int cmp(const void *i, const void *j)
{
void *p1, *p2;
char **s1, **s2;
p1=(void *) i;
p2=(void *) j;
s1=(void *) p1;
s2=(void *) p2;
return strcmp(*s1,*s2);
}

int cmp（const void * i，const void * j）

{

char * s1 = i;

char * s2 = p;


返回strcmp（s1，s2）;

}

这就是你要做的事情吗？


BTW，请妥善缩进发布的代码。您的编译器可能不在乎，

但是人们在阅读它。


-

Al Balmer

Sun City，AZ

int cmp(const void *i, const void *j)
{
char *s1 = i;
char *s2 = p;

return strcmp(s1, s2);
}
Is that what you''re trying to do?

BTW, please properly indent code you post. Your compiler may not care,
but people reading it do.

--
Al Balmer
Sun City, AZ

wa***************@gmail.com 写道：

是否有一种更简单的方法来编写cmp程序而不必通过所有

除了过多无用的演员阵容外，你还遗漏了一个

的关键标题。

是的，你是太努力了。将下面的代码与您的代码进行比较，

跟随它：


#include< stdio.h>

#include < stdlib.h>

#include< string.h>


int compare_strings（const void * i，const void * j）

{

返回strcmp（*（char **）i，*（char **）j）;

}

void showstrs（size_t n，char * s [n]）

{

size_t i;

for（i = 0; i< n; i ++）

printf（" s [％u]％s \ n"，（unsigned）i，s [i]）;

}


int main（无效）

{

char * base [] = {" zebra"，" ;你好，再见 };

size_t n = sizeof base / sizeof * base;

showstrs（n，base）;

putchar（''\ n ''）;

qsort（base，n，sizeof * base，compare_strings）;

showstrs（n，base）;

返回0 ;

}


[输出]

s [0]斑马

s [1]你好< br $>
s [2]再见


s [0]再见

s [1]你好

s [2] zebra

[OP'代码]

#include< stdlib.h>
#include< string.h>

int cmp（const void * i，const void * j）
{
void * p1，* p2;
char ** s1，** s2;
p1 = （void *）i;
p2 =（void *）j;
s1 =（void *）p1;
s2 =（void *）p2;
返回strcmp（* s1，* s2）;
}
int main（void）
{char string0 [] =" zebra" ;;
char string1 [ ] =" hello";
char string2 [] =" goodbye";

char * base [3];

base [0] =（ char *）string0;
base [1] =（char *）string1;
ba se [2] =（char *）string2;

printf（" \ base [0]％s"，base [0]）;
printf（" \ nbase） [1]％s"，base [1]）;
printf（" \ base [2]％s \ n"，base [2]）;

qsort（ & base，3，sizeof（base [0]），（void *）cmp）;

printf（" \ base [0]％s"，base [0]）;
printf（" \ base [1]％s"，base [1]）;
printf（" \ base [2]％s \ n"，base [2]）;
}

TIA，

Is there an easier way to code the cmp procedure without going thru all
the pointer manipulations?
In addition to an excessive number of useless casts, you have left out a
key header.
Yes, you are trying too hard. Compare the code below to your code,
which follows it:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int compare_strings(const void *i, const void *j)
{
return strcmp(*(char **) i, *(char **) j);
}

void showstrs(size_t n, char *s[n])
{
size_t i;
for (i = 0; i < n; i++)
printf("s[%u] %s\n", (unsigned) i, s[i]);
}

int main(void)
{
char *base[] = { "zebra", "hello", "goodbye" };
size_t n = sizeof base / sizeof *base;
showstrs(n, base);
putchar(''\n'');
qsort(base, n, sizeof *base, compare_strings);
showstrs(n, base);
return 0;
}

[output]
s[0] zebra
s[1] hello
s[2] goodbye

s[0] goodbye
s[1] hello
s[2] zebra
[OP''s code]
#include <stdlib.h>
#include <string.h>

int cmp(const void *i, const void *j)
{
void *p1, *p2;
char **s1, **s2;
p1=(void *) i;
p2=(void *) j;
s1=(void *) p1;
s2=(void *) p2;
return strcmp(*s1,*s2);
}

int main(void)
{
char string0[]="zebra";
char string1[]="hello";
char string2[]="goodbye";

char *base[3];

base[0]=(char *)string0;
base[1]=(char *)string1;
base[2]=(char *)string2;

printf("\nbase[0] %s",base[0]);
printf("\nbase[1] %s",base[1]);
printf("\nbase[2] %s\n",base[2]);

qsort(&base,3,sizeof(base[0]),(void *)cmp);

printf("\nbase[0] %s",base[0]);
printf("\nbase[1] %s",base[1]);
printf("\nbase[2] %s\n",base[2]);
}
TIA,

^^^

你是票据收藏家还是剧本小子？

^^^
Are you a bill-collector or a script-kiddie?

文章< 11 ********************** @ z34g2000cwc.googlegroups .com>，

< wa *************** @ gmail.com>写道：

In article <11**********************@z34g2000cwc.googlegroups .com>,
<wa***************@gmail.com> wrote:

有没有一种更简单的方法来编写cmp程序而不必通过所有指针操作？

#include< stdlib.h>
#include< string.h>

int cmp（const void * i，const void * j）
{
void * p1，* p2;
char ** s1，** s2;
p1 =（void *）i;
p2 =（void *）j;
s1 =（void *）p1;
s2 =（void *）p2;
返回strcmp（* s1，* s2）;
}

Is there an easier way to code the cmp procedure without going thru all
the pointer manipulations?

#include <stdlib.h>
#include <string.h>

int cmp(const void *i, const void *j)
{
void *p1, *p2;
char **s1, **s2;
p1=(void *) i;
p2=(void *) j;
s1=(void *) p1;
s2=(void *) p2;
return strcmp(*s1,*s2);
}

[...其余代码被剪断。 ..]


通常，在代码中使用强制转换表示

出错了。这条规则有例外，但在

一般情况下，演员应该在你的脑海中引起一个红旗。这是

代码的修改版本，没有强制转换。它是短的

，应该更容易理解和调试：


#include< stdlib.h>

#include < stdio.h>

#include< string.h>


int cmp（const void * a，const void * b）

{

char * const * aa = a;

char * const * bb = b;

返回strcmp（* aa，* bb）;

}


int main（无效）

{

char string0 [] =" zebra";

char string1 [] =" hello";

char string2 [] =" goodbye";

char * base [3];


base [0] = string0;

base [1] = string1;

base [2] = string2;


qsort（base，（sizeof base）/（sizeof base [0]），sizeof base [0]，cmp）;


printf（" base [0] =％s \ n"，base [0]）;

printf（" base [1] =％s \ n"，base [1]）;

printf（" base [2] =％s \ n"，base [2]）;


返回EXIT_SUCCESS;

}


-

Rouben Rostamian

[... the rest of the code snipped ...]

As a rule, the use of casts in a code is an indication that
something is wrong. There are exceptions to this rule, but in
general a cast should raise a red flag in your mind. Here is
a modified version of your code without casts. It is shorte
and should be easier to understand and debug:

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

int cmp(const void *a, const void *b)
{
char * const *aa = a;
char * const *bb = b;
return strcmp(*aa, *bb);
}

int main(void)
{
char string0[] = "zebra";
char string1[] = "hello";
char string2[] = "goodbye";
char *base[3];

base[0] = string0;
base[1] = string1;
base[2] = string2;

qsort(base, (sizeof base)/(sizeof base[0]), sizeof base[0], cmp);

printf("base[0] = %s\n",base[0]);
printf("base[1] = %s\n",base[1]);
printf("base[2] = %s\n",base[2]);

return EXIT_SUCCESS;
}

--
Rouben Rostamian

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