如何修复警告“格式指定类型'void*'，但参数具有类型'char'警告”

Question

我正在获得我想要的输出，但不知道如何摆脱这些警告。任何帮助是值得赞赏的。

警告：

> < li>

Format指定类型“void *”，但该参数的类型为“char”[-w Format]printf(" \ n指针变量的值为%p\n "，* myString)；

" Format指定类型为' void * '，但该参数的类型为' 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

首先这些电话

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

和

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

没有意义，因为您正在尝试使用字符的值作为指针值。

至于其他警告，则只需将指针投射到类型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 ));
}