C language pointer interview question - the second bullet

Catalog

The first question is

The second question is  

  Third question

  Fourth question

  Fifth question

Sixth question  

  Question seven

  The eighth question

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The first question is

#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 The type is int(*)[5], there (int*) It is mandatory to cast &a Type conversion to int*

ptr-1 Then it will point to 5, Then dereference , final result *（ptr-1） be equal to 5

a Point to first element ,a+1 Point to the second element

The second question is  

#include<stdio.h>
struct Test
{
	int Num;
	char* pcName;
	short sDate;
	char cha[2];
	short sBa[4];
}*p=(struct Test*)0x100000;
// hypothesis p  The value of is 0x100000.  What are the values of the expressions in the following table ？
// It is known that , Structure Test The variable size of type is 20 Bytes 
int main()
{
	printf("%p\n", p + 0x1);
	printf("%p\n", (unsigned long)p + 0x1);
	printf("%p\n", (unsigned int*)p + 0x1);
	return 0;
}

 

because Test The size of the type is 20 Bytes , and p Is precisely test type  ,0x1 yes 16 Number in hexadecimal 1, yes 1*16^0,

p+0x1: It is equivalent to giving p Add twenty bytes , and p The content is 0x1000000, This is a 16 Number in hexadecimal , We should put this 20 Convert to 16 Number in hexadecimal , The result of the conversion is 14, So the answer is 0x100000+14=0x100014

(unsigned long)p+0x1: Is to put p Cast to integer （ Unsigned long shaping ）, After plastic surgery, the result is 1048576, Then add 1（16 It's binary 1 and 10 It's binary 1 identical ）, Turn into 1048577, To 16 Into the system for 0x100001

（unsigned int*）p： hold p Cast to （unsigned int *） type , The size of this type of permission is four bytes , When p+0x1=p+1 after , Because the permission size is 4 Bytes , therefore p Across 4 Bytes, so the result is 0x100004

  Third question

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 The original type of is int(*)[4], Cast type to （int*）ptr[-1] How to get , Please look at the picture ,ptr[-1]=*(ptr1+(-1))=*(ptr1-1)

*ptr2, The end result is because *ptr It is an integer dereference , So the next four bytes are accessed , The final result is related to small end storage

 

  Fourth question

#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;
}

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  Fifth question

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; 
}

 

 

 

Here is the -4 Print it as an address  

Sixth question  

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; 
}

 

 

  Question seven

#include <stdio.h>
int main()
{
 char *a[] = {"work","at","alibaba"};
 char**pa = a;
 pa++;
 printf("%s\n", *pa);
 return 0; 
}

 

 

because pa The object is char * Of , So every time 1, Add one more char * Type size  

  The eighth question

 

#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;
}