c - 在 C 中交换双指针

Question

这是我的代码；我想交换下面的名字。我正在练习这个；

我收到的错误是

Segmentation fault

如有任何帮助，我们将不胜感激。

void nameSwap(char **wife[3],char **husband[3])
{
  int i;
  char **tmp[3];
for(i=0;i<3;i++)
 {

  *tmp[i]=*wife[i];
  *wife[i] = *husband[i];
  *husband[i] = *tmp[i];
 } 

int main(int argc,char  *argv[])
{
char *name1[3]={"Chicago","University","Computer"};
char *name2[3]={"I","Love","Uchicago"};

int k;
char **p1[3];
char **p2[3];

for(k=0;k<3;k++)
{
*p1[k]=name1[k];
*p2[k]=name2[k];
}

for(k=0;k<3;k++)
{
printf("%s %s\n",*p1[k],*p2[k]);
}

nameSwap(&p1[3],&p2[3]);

for(k=0;k<3;k++)
{
printf("%s %s\n",*p1[k],*p2[k]);
}
return 0;

}

Answer 1

双指针的意思

char ** 的情况

#include <stdio.h>

void nameSwap(char ***wife, char ***husband){
    char **tmp;

    tmp = *wife;
    *wife = *husband;
    *husband = tmp;
}

int main(void){
    char *name1[3]={"Chicago","University","Computer"};
    char *name2[3]={"I","Love","Uchicago"};

    int k;
    char **p1;
    char **p2;

    p1 = name1;
    p2 = name2;

    for(k=0;k<3;k++){
        printf("%s %s\n", p1[k], p2[k]);
    }

    nameSwap(&p1, &p2);

    for(k=0;k<3;k++){
        printf("%s %s\n", p1[k], p2[k]);
    }
    return 0;
}

<小时/>

指向char *var_name[3]的指针的情况

#include <stdio.h>

void nameSwap(char *(**wife)[3], char *(**husband)[3]){
    char *(*tmp)[3];

    tmp = *wife;
    *wife = *husband;
    *husband = tmp;
}

int main(void){
    char *name1[3]={"Chicago","University","Computer"};
    char *name2[3]={"I","Love","Uchicago"};

    int k;
    char *(*p1)[3];
    char *(*p2)[3];

    p1 = &name1;
    p2 = &name2;

    for(k=0;k<3;k++){
        printf("%s %s\n", (*p1)[k], (*p2)[k]);
    }

    nameSwap(&p1, &p2);

    for(k=0;k<3;k++){
        printf("%s %s\n", (*p1)[k], (*p2)[k]);
    }
    return 0;
}