C语言指针笔试题

1.

int main()
{
    int a[5] = { 1,2,3,4,5 };
    int* ptr = (int*)(&a + 1);//&a为一整个a数组的地址
    printf("%d,%d\n",*(a+1),*(ptr-1));//2,5
    return 0;
}

 2.指针加减整数

struct Test
{
    int Num;
    char* pcName;
    short sDate;
    char cha[2];
    short sBa[4];
}*p;
//假设p的值为0x100000。如下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()
{
    p = (struct Test*)0x100000;
    //下面所用知识解释：16进制的0x1就是1；(unsigned long)p将p强制转换成无符号整数；(unsigned int*)p将p转换成int类型的指针
    printf("%p\n", p + 0x1);
    //p + 0x1将p结构体整体的地址加1即将p的地址向后加20个字节，因为Test类型的变量大小是20个字节
    printf("%p\n", (unsigned long)p + 0x1);
    //(unsigned int*)p+1将(unsigned int*)p所代表的十进制值加1，它的地址为0x100000+1即(重点）0x100001
    printf("%p\n", (unsigned int*)p + 0x1);
    //(unsigned int*)p + 0x1即p的地址加上int*类型的大小（4个字节）
    return 0;
}

3.

int main()
{
	int a[4] = { 1,2,3,4 };
	int* ptr1 = (int*)(&a + 1);
	int* ptr2 = (int*)((int)a + 1);
    //(int)a将a的首元素'1'的地址转换为int，再加上1然后再转换为int*，此时((int)a + 1)的地址为'1'的地址向后移动(重点)1个字节
	printf("%x,%x", ptr1[-1], *ptr2);
	return 0;
}

4.

int main()
{
	int a[3][2] = { (0,1),(2,3),(4,5) };//此处有逗号表达式，所以实际为a[3][2]={1,3,5}
	int* p;
	p = a[0];//将a[0]首元素的地址赋给p
	printf("%d\n", p[0]);//1
	return 0;
}

 5.

int main()
{
	int a[5][5];
	int(*p)[4];
	p = a;//a将首元素地址即二维数组a第一行传给p
	printf("%p,%d", &p[4][2] - &a[4][2], &p[4][2] - &a[4][2]);//（重点）地址相减的差为两地址之间的元素个数
	return 0;
}

6. 

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\n", *(ptr1 - 1), *(ptr2 - 1));//10     5
	return 0;
}

7.

int main()
{
	char* a[] = { "work","at","alibaba" };
	char** pa = a;
	pa++;
	printf("%s\n", *pa);//at
	return 0;
}

8.（本次内容的难点）

int main()
{
	char *c[] = { "ENTER","NEW","POINT","FIRST" };
	char* *cp[] = { c + 3,c + 2,c + 1,c };
	char*** cpp = cp;
	printf("%s\n", **++cpp);      //POINT
	printf("%s\n", *--*++cpp+3);  //ER
	printf("%s\n", *cpp[-2]+3);   //ST
	printf("%s\n", cpp[-1][-1]+1);//EW
	return 0;
}