目录
- #include<stdio.h>
- int main()
- {
- int a[5] = { 1, 2, 3, 4, 5 };
- int* ptr = (int*)(&a + 1);
- printf("%d,%d", *(a + 1), *(ptr - 1));
- return 0;
- }
&a的类型是int(*)[5],这里的(int*)是强制类型转换把&a类型转换为int*
ptr-1之后会指到5,然后再解引用,最终结果*(ptr-1)等于5
a指向首元素,a+1指向第二个元素
- #include<stdio.h>
- struct Test
- {
- int Num;
- char* pcName;
- short sDate;
- char cha[2];
- short sBa[4];
- }*p=(struct Test*)0x100000;
- //假设p 的值为0x100000。 如下表表达式的值分别为多少?
- //已知,结构体Test类型的变量大小是20个字节
- int main()
- {
- printf("%p\n", p + 0x1);
- printf("%p\n", (unsigned long)p + 0x1);
- printf("%p\n", (unsigned int*)p + 0x1);
- return 0;
- }
由于Test类型的大小是20个字节,而p正好是test类型 ,0x1是16进制下的数字1,是1*16^0,
p+0x1:相当于给p加了二十个字节,而p的内容是0x1000000,这是16进制下的数字,我们应把这20转换为16进制下的数字,转换结果为14,所以答案是0x100000+14=0x100014
(unsigned long)p+0x1:就是把p强制转换为整形(无符号长整形),转为整形后结果是1048576,之后再加1(16进制的1和10进制的1相同),变为1048577,转为16进制为0x100001
(unsigned int*)p:把p强制转换为(unsigned int *)类型,这个类型的权限大小是四个字节,当p+0x1=p+1之后,由于权限大小为4个字节,所以p跨过4个字节因此结果为0x100004
- int main()
- {
- int a[4] = { 1, 2, 3, 4 };
- int *ptr1 = (int *)(&a + 1);
- int *ptr2 = (int *)((int)a + 1);
- printf( "%x,%x", ptr1[-1], *ptr2);
- return 0;
- }
ptr1的类型原来是int(*)[4],强制类型转换为(int*)ptr[-1]如何得来,请看上图,ptr[-1]=*(ptr1+(-1))=*(ptr1-1)
*ptr2,最终结果是因为*ptr是整形解引用,所以访问了后面的四个字节,最终结果和小端存储有关
- #include <stdio.h>
- int main()
- {
- int a[3][2] = { (0, 1), (2, 3), (4, 5) };
- int* p;
- p = a[0];
- printf("%d", p[0]);
- return 0;
- }
- int main()
- {
- int a[5][5];
- int(*p)[4];
- p = a;
- printf( "%p,%d\n", &p[4][2] - &a[4][2], &p[4][2] - &a[4][2]);
- return 0;
- }
这里把-4当作地址去打印了
- int main()
- {
- int aa[2][5] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
- int *ptr1 = (int *)(&aa + 1);
- int *ptr2 = (int *)(*(aa + 1));
- printf( "%d,%d", *(ptr1 - 1), *(ptr2 - 1));
- return 0;
- }
- #include <stdio.h>
- int main()
- {
- char *a[] = {"work","at","alibaba"};
- char**pa = a;
- pa++;
- printf("%s\n", *pa);
- return 0;
- }
因为pa指向的对象是char *的,所以每次加1,加一个char *类型大小
- #include <stdio.h>
- int main()
- {
- char* c[] = { "ENTER","NEW","POINT","FIRST" };
- char** cp[] = { c + 3,c + 2,c + 1,c };
- char*** cpp = cp;
- printf("%s\n", **++cpp);
- printf("%s\n", *-- * ++cpp + 3);
- printf("%s\n", *cpp[-2] + 3);
- printf("%s\n", cpp[-1][-1] + 1);
- return 0;
- }