为什么这个 C 程序不给出给定字符串的反转？

Question

虽然 length() 函数工作正常（当我评论其他代码并且只评论运行那部分时），但为什么这个程序不给出给定字符串 computer 的反转并给出输出正确，但第二个 reverse() 函数没有给出任何输出。

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

 int length(char *);
 char *reverse(char *, int);

int main()
{
char word[] = "COMPUTER";
int count;

count = length("COMPUTER");

printf("%s", reverse(word, count));

}

int length(char *p)
{
int count;
for (count = 0; *(p + count) != '[=10=]'; count++);
    
 return (count);
 }

 char *reverse(char *p, int count)
 {
char temp;
for (int i = 0; i < count / 2; i++)
{
    temp = *(p + i);
    *(p + i) = *(p - (count - 1) - i);
    *(p - (count - 1) - i) = temp;
}
return (p);
 }

Answer 1

这些表达式语句

*(p + i) = *(p - (count - 1) - i);
*(p - (count - 1) - i) = temp;

不正确，

你的意思好像是

*(p + i) = *(p + ( count - 1 ) - i);
*(p + (count - 1) - i) = temp;

也代替这个说法

count = length("COMPUTER");

写成

逻辑上会更一致

count = length( word );

这是一个演示程序。

#include <stdio.h>

size_t length( const char * );
char * reverse( char *, size_t );
 
int main(void) 
{
    char word[] = "COMPUTER";
    size_t count = length( word );

    puts( reverse( word, count ) );

}

size_t length( const char *p )
{
    size_t count = 0;

    while ( *( p + count ) != '[=14=]' ) ++count;
    
    return count;
}

char * reverse( char *p, size_t count )
{
    for ( size_t i = 0; i < count / 2; i++ )
    {
        char temp = *( p + i );
        *( p + i ) = *( p + count - 1 - i );
        *( p + count - 1 - i ) = temp;
    }
    
    return p;
}

程序输出为

RETUPMOC

为什么这个 C 程序不给出给定字符串的反转？

Why this C-program doesn't gives the reverse of given string?

c

reverse

c-strings

pointer-arithmetic

function-definition