将指针数组分配给字符串

Question

我试过在网上搜索，但没有找到与我想做的一样的东西。希望你能帮帮我。

问题：我有 2 个字符数组，即 string 和 string2。两者最初都是空的。然后我给这些字符串 2 个输入。我想使用指向 string2 的指针在字符串中分配值。

这是我的代码：

int main(void) {
    char string[2][MAXLEN];
    char string2[2][MAXLEN];
    char *pointer[2];
    pointer[0] = &string2[0];
    pointer[1] = &string2[1];
    scanf("%s", string[0]); //Assume i scan in "ab"
    scanf("%s", string[1]); //assume i scan in "cd"

    //Now string[0] contains "ab" and string[1] contains "cd"
    //I want to deferences the pointer and assign "cd" to it, so string2[0] will contain "cd"

    *pointer[0] = string[0];
    printf("%s", string2[0]); //but this does not print "cd"
}

*编辑我知道我可以使用 strcpy，但我正在尝试学习使用指针来做到这一点。在这方面的任何帮助都会很棒。

Answer 1

因为您没有复制编译器可以理解的整个信息结构，所以您需要单独复制数组的每个元素。通常这是通过 for 循环检查 NUL 或大小来完成的，但我在作弊，只是向您展示了执行所需副本的语法：

#define MAXLEN 10

int main(void) 
{
    char string[2][MAXLEN];
    char string2[2][MAXLEN];
    char *pointer[2];
    pointer[0] = &string2[0];
    pointer[1] = &string2[1];

    // Replace scanf for simplicity
    string[0][0] = 'a'; string[0][1] = 'b'; string[0][2] = '[=10=]';
    string[1][0] = 'c'; string[1][1] = 'b'; string[1][2] = '[=10=]';

    // For loop or strcpy/strncpy/etc. are better, but showing array method of copying
    pointer[0][0] = string[1][0];
    pointer[0][1] = string[1][1];
    pointer[0][2] = string[1][2];

    printf("%s", string2[0]);

    return 0;
}

对于指针，你可以这样做：

#define MAXLEN 10

int main(void) {
  char string[2][MAXLEN];
  char string2[2][MAXLEN];
  char *pointer[2];
  pointer[0] = &string2[0];
  pointer[1] = &string[1];  // changed this

  string[0][0] = 'a'; string[0][1] = 'b'; string[0][2] = '[=11=]';
  string[1][0] = 'c'; string[1][1] = 'd'; string[1][2] = '[=11=]';

  *pointer[0]++ = *pointer[1]++;
  *pointer[0]++ = *pointer[1]++;
  *pointer[0]++ = *pointer[1]++;

  printf("%s", string2[0]);

  return 0;
}

上面的指针魔法变成：

  char temp = *pointer[1];  // Read the char from dest.
  pointer[1]++;  // Increment the source pointer to the next char.
  *pointer[0] = temp;  // Store the read char.
  pointer[0]++;   // Increment the dest pointer to the next location.

我做了 3 次 - 每个输入字符一次。用 while() 围绕它检查 sourcePtr == '\0' 基本上把它变成 strcpy().

另一个有趣的示例，其中取消引用可能会达到您的预期：

typedef struct foo
{
    char mystring[16];
} FOO;

FOO a,b;

// This does a copy
a = b;

// This also does a copy
FOO *p1 = &a, *p2=&b;
*p1 = *p2;

// As does this
*p1 = a;

// But this does not and will not compile:
a.mystring = b.mystring;

// Because arrays in 'C' are treated different than other types.
// The above says: Take the address of b.mystring and assign it (illegally because the array's location in memory cannot be changed like this) to a.mystring.

将指针数组分配给字符串

Assigning array of pointers to string

c

arrays

pointers

2d