gcc编译器

Question

这是C中的一个简单程序，用于说明C如何为变量分配内存。我的想法是每次重新编译程序时都会显示一个新的内存地址。

以下是该计划：

#include<stdio.h>

main()
{
        int a;
        int b = 10;
        float c = 5.8;

        printf("the address is %u \n", &a);
        printf("the address is %u \n", &b);
        printf("the address is &u \n", &c);

        getch();
}

当我运行此gcc program_name.c时，我收到一系列错误：

program_name.c:3:1: warning: type specifier missing, defaults to 'int'
      [-Wimplicit-int]
main()
^
program_name.c:9:33: warning: format specifies type 'unsigned int' but the
      argument has type 'int *' [-Wformat]
        printf("the address is %u \n", &a);
                               ~~      ^~
program_name.c:10:33: warning: format specifies type 'unsigned int' but the
      argument has type 'int *' [-Wformat]
        printf("the address is %u \n", &b);
                               ~~      ^~
program_name.c:11:33: warning: format specifies type 'unsigned int' but the
      argument has type 'float *' [-Wformat]
        printf("the address is %u \n", &c);
                               ~~      ^~
program_name.c:13:2: warning: implicit declaration of function 'getch' is
      invalid in C99 [-Wimplicit-function-declaration]
        getch();
        ^
5 warnings generated.
Undefined symbols for architecture x86_64:

我无法理解为什么会抛出这些错误。当然%u是一个完全合适的格式说明符。我错过了什么？如何重新格式化代码以阻止这些错误？

编辑：根据回复，似乎正确的代码应该是

#include<stdio.h>

int main()
{
        int a;
        int b = 10;
        float c = 5.8;

        printf("the address is %p \n", &a);
        printf("the address is %p \n", &b);
        printf("the address is %p \n", &c);

        return 0;
}

这确实运行没有错误。