Pointer written test questions ~ approaching Dachang

Preface

author ： Little snail rushes forward

quotes ： I can accept failure , But I can't accept giving up

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Catalog

Pen test 1：

Pen test 2

  Pen test 3

  Pen test 4

  Pen test 5

Pen test 6 

  Pen test 7

Pen test 8

  summary

  Next, I will share some written questions about pointer , I hope you can get something .

Pen test 1：

#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;
}

What is the result of this problem ？

For pointer type topics , Our best way to solve the problem is to draw pictures , I will use pictures to solve all the following pointer questions . 

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a Distribution diagram of array elements

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  As we can see from the picture The pointer ptr Pointing to Array a Other elements .

a It's an array name --> The address of the first element .

Then we (a+1) Skip an element from the first address , Point to the address of the next element ,*(a+1) You find the element 2.

that （ptr-1） Move left by a reshaped size , Point to 5 This element ,*（ptr-1） Find this element .

So our The result is 2,5.

Pen test 2

struct Test
{
 int Num;
 char *pcName;
 short sDate;
 char cha[2];
 short sBa[4];
}*p;
// hypothesis p  The value of is 0x000000.  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;
}

First We need to understand p What is it? ,p It's a pointer to a structure ,%p Value is the address of the printed pointer , We can assume that p The address for 0x000000, that p+0x1 What does that mean ？ It's really just the p+20-->0x000014.

among (unsigned long)p, Put the pointer p Force the type to be converted to an integer , therefore (unsigned long)p+0x1-->0x000001.

secondly (unsigned int*)p, This affects The pointer +- Number of bytes skipped , among ungsing int* The size is 4 byte , therefore  (unsigned long)p + 0x1-->0x000004.

  Pen test 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);
    return 0;
}

 

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a Memory distribution diagram of array

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Here we first understand clearly ,ptr1 and ptr2 All are int * Pointer to type .

(&a + 1） Represents skipping the entire array , Point to the element behind the array .

(int)a take a It's plastic surgery ,(int )a+1 Express a The value of the add 1, Just The pointer jumps over 1 Bytes ( Position as shown in the figure ).

%x Refers to printing in hexadecimal .

Well, we understand the meaning of these variable names , Let's start solving problems .

ptr1[-1]- ->*((&a+1)-1), That is, dereference the pointer to the last position of the array , Just Found the element 0x4.

*ptr2, Is to find ptr2 The element that the pointer points to , According to the memory access rules , To pour this out -->02 00 00 00 , So the final value printed on the screen is 2 00 00 00.

  Pen test 4

#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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a Distribution diagram of array elements

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  This topic is actually a pit , People mistakenly think that initialization is

0 1

2 3

4 5

This form , Actually （） It's a , The sign expression , The last value in the right bracket of the output value . So according to the array element distribution diagram above , Our output p[0]-->*(p+0) The value of is .

  Pen test 5

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;
}

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a Element distribution diagram of array  

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Here we understand a What is it? ？p What is it again? ？&p[4][2] and &a[4][2] What does it mean ?

among a It's a two-dimensional array ,a The type of int(*)[4],p For array pointers ,p The type of int(*)[4].

&a[4][2], The yellow square in the figure is a[4][2] Pointing elements .

p[4][2]-->&*(*(p+4)+2), First P+4 Point to the position in the diagram , Why? ? People may have questions , because p Think you point to 4 individual int type , So every time +1 Will skip 4 A plastic element . secondly , stay (*(p+4)+2),*(p+4) Will visit 4 An integer array , Array name + I'll skip it 2 Elements , Point to 3 Elements , stay *(*(p+4)+2) You'll find number one 3 Elements ( The green square in the picture ).

therefore , &p[4][2] - &a[4][2]( The pointer - The pointer )= The number of elements between pointers 4, Because it is the pointer to the address - The pointer result of the high address is -4.

-4

Original code ：10000000 00000000 00000000 00000100

Inverse code ：11111111 11111111 11111111 11111011

Complement code ：1111 1111 1111 1111 1111 1111 1111 1100

Hexadecimal ：F   F     F        F     F      F      F       C

%p Print FFFFFFFC.

%d Print -4.

Pen test 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", *(ptr1 - 1), *(ptr2 - 1));
    return 0;
}

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aa Element distribution diagram of array

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  This question and the written examination question 1 And its similar , It's easy to solve the problem after drawing .

ptr1-1--> Point to aa[1][4],*(prt1-1) find 10.

ptr2-1--> Point to aa[0][4],*(ptr2-1)z find 5.

  Pen test 7

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

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 a And pa Memory distribution

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  Here we also draw good pictures ,pa It's a secondary pointer ,a Is a constant string , among a The variable name of represents the address of the first element , Start pa-->w, later pa++-->a, therefore &s Print out ''at''.

Now let's officially enter our final topic , Are you ready for the challenge ？ 

Pen test 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);
 printf("%s\n", *--*++cpp+3);
 printf("%s\n", *cpp[-2]+3);
 printf("%s\n", cpp[-1][-1]+1);
 return 0;
}

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Initial memory layout

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  After we draw the initial memory layout , Start solving problems .

1**ccp, We can clearly see from the figure that it points to a string FIRST(cpp-->c+2).

2*--*++cpp+3( Be careful ++-- Are the values of the change quantity ), Here we first operate close to the operator of the variable ,++cpp, This makes cpp Point to c+1,*(c+1) find c+1, stay --*++cpp refer to c+1-->c,*--*++cpp Find the string ENTER The first address , then ,*--*++cpp+3, find E The address of , Print ER.(cpp-->c+1,c+1-->c).

3*cpp[-2]+3-->*(cpp-2)+3, here cpp-2 bring cpp Point to c+3,*(c+3) find FIRST The first address ,*(cpp-2)+3 find S The first address , Print ST.(cpp-->c+3)

4cpp[-1][-1]+1-->*(*(cpp-1)-1)+1,cpp-1 Point to c+2,*(cp-1)-1-->( Point to )c+1,*(*(cpp-1)-1) find NEW The first address ,*(*(cpp-1)-1)+1 find E The address of , Print EW.

  summary

1 It is very important to draw and analyze when doing pointer type topics .

2p[1] Can be equivalent to *(p+1),p[1][1] It can also be equivalent to *(*(p+1)+1).

3 Remember that the array name is the address of the first element （ I won't say more about the two special cases ).

4 The pointer - Pointer means the number of elements between two pointers .

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