指针综合练习

#include <stdio.h>
int main(int argc, char* argv[])
{
printf("sizeof(argv):%d\n", sizeof(argv));
int var_i = 5;
char var_c = 'p';
char* p = &var_c;
printf("*p = %c\n", *p);

//测试大端小端
//short var_s = 0x0001;
///char* ps = &var_s;

//printf("ps:%d\n", ps);

//p = &var_i;//就取了1个字节，有的编译器报错，有的编译器警告
//printf("*p = %d\n", *p);//超过255的数字不能正常打印
//var_i=400时，会打印符号位，结果为-112


int a[2][3] = { 1, 2, 3, 4, 5, 6 };
int (*pa)[3] = a;
printf("**a = %d\n", **a);//取出1
printf("(*a)[2] = %d\n", (*a)[0]);//取1


printf("*(*(pa+1)+1) = %d\n", *(*(pa + 1) + 1));//取出5
printf("*(*(a+1)+1) = %d\n", *(*(a+1)+1));//取出5
printf("(*(a+1))[2] = %d\n", (*(a+1))[1]);//取5
printf("*(*a+4) = %d\n", (*(*a + 4)));//取5


printf("(*a+1) = %d\n", (*a + 1));
printf("(*a) = %d\n", (*a));


printf("(a+1) = %d\n", (a + 1));
printf("(a) = %d\n", a);


printf("sizeof(a):%d\n", sizeof(a));
printf("sizeof(*a):%d\n", sizeof(*a));
printf("sizeof(pa):%d\n", sizeof(pa));
//myCalc();
return 0;

}

运行结果：