您能否通过执行以下 C 代码来解释为什么会输出：'dcc d'？（指针）

Question

My output according to the following code is dcba , but it is wrong.Why?

char xc[4] = {'a', 'b', 'c', 'd'};
char *xp[4];

int i;
for (i = 0; i < 4; i++) {     
    xp[i] = &xc[i];
}

for (i = 0; i < 4; i++) {     
    *xp[i] = *xp[3-i];
}

printf("%c %c %c %c", xc[0], xc[1], xc[2], xc[3]);

Answer 1

Follow what you did step-by-step.

Your code is equivalent to

char xc[4] = {'a', 'b', 'c', 'd'};

int i;

for (i = 0; i < 4; i++) {
    xc[i] = xc[3-i];
}

printf("%c %c %c %c", xc[0], xc[1], xc[2], xc[3]);

Then, trace is

i statement   xc[0] xc[1] xc[2] xc[3]
(before loop) 'a'   'b'   'c'   'd'
0 xc[0]=xc[3] 'd'   'b'   'c'   'd'
1 xc[1]=xc[2] 'd'   'c'   'c'   'd'
2 xc[2]=xc[1] 'd'   'c'   'c'   'd'
3 xc[3]=xc[0] 'd'   'c'   'c'   'd'

You should use a temporaly variable and be careful to swap the same pair twice to reverse the array.

char xc[4] = {'a', 'b', 'c', 'd'};
char *xp[4];

int i;
for (i = 0; i < 4; i++) {
    xp[i] = &xc[i];
}

for (i = 0; i < 3-i; i++) {
    char t = *xp[i];
    *xp[i] = *xp[3-i];
    *xp[3-i] = t;
}

printf("%c %c %c %c", xc[0], xc[1], xc[2], xc[3]);

Answer 2

Not really complex... If you expand you second for it does:

*xp[0] = *xp[3]; // xc[0] receive 'd' value, so xc array is now {'d', 'b', 'c', 'd'}
*xp[1] = *xp[2]; // xc[1] receive 'c' value, so xc array is now {'d', 'c', 'c', 'd'}
*xp[2] = *xp[1]; // xc[2] receive 'c' value, no change
*xp[3] = *xp[0]; // xc[3] receive 'd' value, no change

Answer 3

For i = 0 , xc[0] is modified and have value d . For i = 1 , xc[1] is modified and have value c . By this time a and b is no longer anywhere in your xc array.
Now you are using the value of xc[0] and xc[1] to modify the variables xc[3] and xc[2] respectively. The output would be

d c c d 

Answer 4

In the second loop, you want to use the initial values of xp in order to invert them, but by the time you reach the middle of the array, you have already modified the following values.

Declare a temporary variable and also change your printings in order to print xp instead of xc :

char xc[4] = {'a', 'b', 'c', 'd'};
char *xp[4];
char temp;
int i;

for (i = 0; i < 4; i++) {     
    xp[i] = &xc[i];
}

for (i = 0; i < 4; i++) {  
    temp = *xp[i];   
    *xp[i] = *xp[3-i];
    *xp[3-i] = temp;
}

printf(“%c %c %c %c”, xp[0], xp[1], xp[2], xp[3]);

Answer 5

After the first two iterations of the loop

for (i = 0; i < 4; i++) {     
    *xp[i] = *xp[3-i];
}

the original values of xc[0] and xc[1] were overwritten by values of xc[3] and xc[2] correspondingly and the array became to look like

char xc[4] = {'d', 'c', 'c', 'd'};

If you want to reverse the array then the loop can look like

size_t i;
size_t n = sizeof( xp ) / sizeof( *xp );

for ( i = 0; i < n / 2; i++ ) 
{
    char c = *xp[i];      
    *xp[i] = *xp[n-i-1];
    *xp[n-i-1] = c;
}

As result you will get that the array will look like

char xc[4] = {'d', 'c', 'b', 'a'};

That is you need to swap characters of the first half of the array with characters of the second half of the array.

Using this approach you can write a general function. For example

#include <stdio.h>

void reverse( char s[], size_t n )
{
    size_t i;

    for ( i = 0; i < n / 2; i++ )
    {
        char c = s[i];
        s[i] = s[n-i-1];
        s[n-i-1] = c;
    }
}    

int main( void ) 
{
    char xc[] = { 'a', 'b', 'c', 'd' };

    printf( "%c %c %c %c\n", xc[0], xc[1], xc[2], xc[3] );

    reverse( xc, sizeof( xc ) / sizeof( *xc ) );

    printf( "%c %c %c %c\n", xc[0], xc[1], xc[2], xc[3] );
}    

The program output is

a b c d
d c b a