c - 如何修复警告“格式指定类型'void *'但参数具有类型'char'警告”

Question

我得到了我想要的输出，但不知道如何摆脱这些警告。任何帮助表示赞赏。

警告：

1. Format 指定类型 'void *' 但参数类型为 'char' [-Wformat] printf("\n指针变量的值为 %p\n", *myString);

2. "Format 指定类型 'void *' 但参数的类型为 'char' [-Wformat] printf("%p\n", myString[x]);

    #include <stdio.h>
    #include <stdlib.h>
   
    int main() {
   
      char *myString = "Daniel";
      int x;
   
      printf("\nThe pointer variable's value is %p\n", *myString);
      printf("\nThe pointer variable points to %s\n", myString);
      printf("\nThe memory location for each character are: \n");
   
   
      for (x = 0;x < 7;x++){
       printf("%p\n", myString[x]);
      }
   
    return 0;
    }
   

输出：

     The pointer variable's value is 0x44

     The pointer variable points to Daniel

     The memory location for each character are: 
     0x44
     0x61
     0x6e
     0x69
     0x65
     0x6c
     (nil)

Answer 1

对于初学者来说，这些电话

printf("\nThe pointer variable's value is %p\n", *myString);

和

printf("%p\n", myString[x]);

没有意义，因为您试图将字符的值用作指针值。

至于其他警告，只需将指针转换为 type void *。例如

printf("\nThe pointer variable's value is %p\n", ( void * )myString);
printf("\nThe pointer variable points to %s\n", myString);
printf("\nThe memory location for each character are: \n");


for (x = 0;x < 7;x++){
 printf("%p\n", ( void * )( myString + x ));
}