如何修复警告 "Format specifies type 'void *' but the argument has type 'char' warning"

Question

我得到了我想要的输出，但不知道如何消除这些警告。感谢任何帮助。

警告：

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

"格式指定类型 '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，因为您正试图将字符的值用作指针值。

至于其他警告，则只需将指针转换为类型 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 ));
}

如何修复警告 "Format specifies type 'void *' but the argument has type 'char' warning"

How to fix warning "Format specifies type 'void *' but the argument has type 'char' warning"

c

pointers

printf

c-strings

conversion-specifier