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指针笔试题解析

程序员文章站 2024-03-15 19:20:12
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笔试题1

int main() 
{    
    int a[5] = { 1, 2, 3, 4, 5 };    
    int *ptr = (int *)(&a + 1);
    //&a取地址数组名–>数组指针->数组指针+1->跳过整个数组    
    printf( "%d,%d", *(a + 1), *(ptr - 1)); //2  5   
    return 0; 
} 

指针笔试题解析
笔试题2

这里告知结构体的大小是20个字节 
struct Test 
{    
    int Num;    
    char *pcName;    
    short sDate;    
    char cha[2];    
    short sBa[4]; 
}*p;
//假设p的值为0x10 0000.
int main() 
{    
    printf("%p\n", p + 0x1);//0x0010 0014
    //结构体指针+1,加20个字节,20按十六进制是:14
    printf("%p\n", (unsigned long)p + 0x1);//0x0010 0001
    //p+0x1 外加一个(unsigned long)类型,强制转换为无符号长整形,
    //再加1,结果就是1;(p指针指向空相当于0)按十六进制打印就是0x00100001   
    printf("%p\n", (unsigned int*)p + 0x1);//0x00100004 
    //p+0x1外加一个 (unsigned int*)类型,转换为指针,类型是 int 型,
    //所以占4个字节,跳过4个字节,按十六进制打印就是0x00100004;   
    return 0; 
}

笔试题3

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);//4 02000000 
    //ptr1[-1]= *(ptr1-1),取到元素4   
    return 0; 
}

指针笔试题解析
笔试题4

int main() 
{     
    int a[3][2] = { (0, 1), (2, 3), (4, 5) };
	int *p;
	p = a[0];
	printf("%d", p[0]);//1
	//逗号表达式,它的值为最后一个表达式的值
	//所以二维数组排列为:
	//1  3  
	//5  0
	//0  0
	return 0}

笔试题5

int main()
{
    int a[5][5];
	int(*p)[4];
	p = a;
	printf("%p,%d\n", &a[4][2] - &p[4][2], &a[4][2] - &p[4][2]);
	//打印结果为:fffffffc   -4
}

指针笔试题解析
笔试题6

int main()
{
    int aa[2][5] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
	//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));//10  5
	return 0}

指针笔试题解析
笔试题7

int main() 
{    
    char *a[] = {"work","at","alibaba"};    
    char**pa = a;    
    pa++;    
    printf("%s\n", *pa); //at
    //二级指针指向了指针数组名,++就是指针的指向加一,指到 at   
    return 0; 
}

笔试题8

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; 
} 

指针笔试题解析

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