Question

#include <stdio.h>

int main() {
   int i;

   char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
   int int_array[5] = {1, 2, 3, 4, 5};

   char *char_pointer;
   int *int_pointer;

   char_pointer = int_array;
   int_pointer = char_array;

   for(i=0; i < 5; i++) {
      printf("[integer pointer] points to %p, which contains the integr %d\n", int_pointer, *int_pointer);
      int_pointer = int_pointer + 1;
   }

   for(i=0; i < 5; i++) {
      printf("[char pointer] points to %p, which contains the char '%c'\n", char_pointer, *char_pointer);
      char_pointer = char_pointer + 1;
   }
}

显示以下错误：

pointer_types2.c: In function ‘main’:
pointer_types2.c:13:17: warning: assignment from incompatible pointer type [enabled by default]
    char_pointer = int_array;
                 ^
pointer_types2.c:14:16: warning: assignment from incompatible pointer type [enabled by default]
    int_pointer = char_array;
                ^

Answer 1

您无法执行此操作，因为char和int的存储空间大小不同4。具体来说，您可以尝试：

char_pointer = (char *)&int_array;
int_pointer = (int *)&char_array;

或

char_pointer = (char *)int_array;
int_pointer = (int *)char_array;

哪个会在没有警告的情况下编译，但结果是：

<强>输出：

$ ./bin/juxt
[integer pointer] points to 0x7fff0e89d7d0, which contains the integr 1684234849
[integer pointer] points to 0x7fff0e89d7d4, which contains the integr 101
[integer pointer] points to 0x7fff0e89d7d8, which contains the integr 243914712
[integer pointer] points to 0x7fff0e89d7dc, which contains the integr 32767
[integer pointer] points to 0x7fff0e89d7e0, which contains the integr 243914672
[char pointer] points to 0x7fff0e89d7b0, which contains the char ''
[char pointer] points to 0x7fff0e89d7b1, which contains the char ''
[char pointer] points to 0x7fff0e89d7b2, which contains the char ''
[char pointer] points to 0x7fff0e89d7b3, which contains the char ''
[char pointer] points to 0x7fff0e89d7b4, which contains the char ''

注意： int指针d7d0，d7d4，......的增量，并与字符指针值d7b0进行比较， d7b1，...以这种方式交换指针最终会导致它们指向内存中的错误位置。

Answer 2

它们是不同的类型，你不能这样做

指针只能指向与自身类型相同的对象。因为不同的类型在记忆中有不同的长度。例如，这里，int的长度为4，而char的长度为1.当我们对指针进行一些操作时，这里，+ 1，指针移动的步骤不同，当你想访问存储的数据时可能会导致一些错误那里。

然而在某些情况下，使用指针指向不同类型是安全的，它的高级技能，小心

Answer 3

您可以通过类型转换修复这些警告：

   char_pointer = (char *)int_array;
   int_pointer = (int *)char_array;

指针并置使得int_pointer指向字符数据，反之亦然

3 个答案: