基本程序,有趣的输出？说明

我正在学习指针,我正在从教程中运行这个程序,它提供了一些有趣的输出.

#include 

int main () {

    int var = 20; /*actual variable declaration */
    int *ip; /*pointer variable declaration*/

    ip = &var; /*store addres of var in ip variable*/

    printf("Address of var variable: %x\n", &var);

    /*address stored in ip variable*/
    printf("Address of stored in ip variable: %x\n", ip);
    /*access the value using pointer */
    printf("value of *ip variable: %x\n", *ip);

    return 0;

}

根据我的理解,该网站说我的输出应该是两个匹配的内存位置,而值为20的"**ip变量"部分.我对内存位置没有任何问题,它们都像人们想象的一样,最后一部分,它告诉我*ip变量的值是14？

Address of var variable: 6d73b8cc
Address of stored in ip variable: 6d73b8cc
value of *ip variable: 14

这里发生了什么？用gcc编译,它给出一个关于格式%x的警告,期望一个无符号整数,但参数2的类型为....

MorePointers.c:10:9: warning: format ‘%x’ expects argument of type ‘unsigned int’, but argument 2 has type ‘int *’ [-Wformat=]
  printf("Address of var variable: %x\n", &var);
         ^
MorePointers.c:13:9: warning: format ‘%x’ expects argument of type ‘unsigned int’, but argument 2 has type ‘int *’ [-Wformat=]
  printf("Address of stored in ip variable: %x\n", ip);

请纠正我,如果我错了,但可能是由于格式,当期望无符号整数时可能导致数字不是20？作为无符号整数的值在0到65,535或0到4,294,967,295之间,而整数在-32,768到32,767或-2,147,483,648到2,147,483,647之间？

发生的事情是您将整数值打印20为十六进制值,正确显示为14.我怀疑你在期待20十进制.:)

一些修复: -

printf("Address of var variable: %x\n", &var);

这是不正确的,应该将指针打印为指针值

printf("Address of var variable: %p\n", &var); /* << ensure pointer value */

当你打电话时printf,它会将固定文本和参数按顺序混合......

printf( "hello %d world %d today %d\n", 1,2,3 );  

今天打印你好1世界2 3

显示参数的方式基于格式说明符.

%s /* displays as a string */
%d /* displays as an integer in normal decimal */
%x /* displays an integer in hexa-decimal. */
%f /* displays as a floating point */
%p /* displays a pointer address */

Decimal goes      0,1,2,..,8,9,10,11,...,14,15,16,17,...,20
Hexadecimal goes  0,1,2,..,8,9, A, B,..., E, F,10,11,...,14