c 指针 - 警告格式 '%x' 需要类型为 'unsigned int' 的参数

Question

我目前正在阅读 Hacking the art of exploitation，其中有一个我似乎无法正确理解的示例。尝试编译导致错误:

./addressof.c: In function ‘main’:
./addressof.c:8:4: warning: format ‘%x’ expects argument of type ‘unsigned int’,
but argument 2 has type ‘int *’ [-Wformat]


#include <stdio.h>
int main() {
   int int_var = 5;
   int *int_ptr;

   int_ptr = &int_var; // Put the address of int_var into int_ptr.

   printf("int_ptr = 0x%08x\n", int_ptr);
   printf("&int_ptr = 0x%08x\n", &int_ptr);
   printf("*int_ptr = 0x%08x\n\n", *int_ptr);

   printf("int_var is located at 0x%08x and contains %d\n", &int_var, int_var);
   printf("int_ptr is located at 0x%08x, contains 0x%08x, and points to %d\n\n",
      &int_ptr, int_ptr, *int_ptr);
}

我知道错误在哪里，我只是不确定如何解决这个问题。

Answer 1

指针的格式说明符是%p，而不是%x。 (参见 here)