Dachang classic pointer written test questions

This part is the classic pointer question , Practice hard , If you refuse, you will , As soon as I write fei

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Catalog ：

Pointer classic question

1. This topic is the epitome of classic pointer pen test questions

2. drawing 、 drawing 、 drawing

3. Classic questions ：

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Classic questions ：

Pen test 1：

  Particular attention ：

①int* ptr = (int*)(&a + 1);

//&a It takes out the address of the entire array , therefore &a The type is int(*)[5], This type of +-1 What I'm skipping is 5 individual int type

// Cast , The type changes , And the value remains the same

②printf("%d,%d", *(a + 1), *(ptr - 1));

//*(ptr-1) Why just skip 1 individual int type , because ptr The type is int*, This type of +-1 skip 1 individual int type

if int* p=&a;* explain p It's a pointer variable ,int explain p The object is int type .p The type is int*

Answer to this question ：2,5

The code is as follows ：

#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;
}

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Pen test 2：

Particular attention ：

there 0x1 It's hexadecimal , Convert to decimal to 1

It is known that p The value of is 0x100000,

①printf("%p\n", p + 0x1);

//p Is a structure type pointer variable ,p+-1 It means skipping a structure type size

if p Is an integer pointer variable ,p+-1 It means skipping an integer size

②printf("%p\n", (unsigned long)p + 0x1);

(unsigned long)p,p The type of changes but the value does not change ,

p The value of is 0x100000, convert to unsigned int Type followed by 1048576,+1 by 1048577, The address of this value is 0x100001

③printf("%p\n", (unsigned int*)p + 0x1);
// if p It's an integer pointer ,p+-1 Represents skipping an unsigned integer size

Answer to this question ：

    0x100014,0x100001,0x100004

The code is as follows ：

#include <stdio.h>

struct Test
{
	int Num;
	char* pcName;
	short sDate;
	char cha[2];
	short sBa[4];
}*p;

int main()
{
	printf("%p\n", p + 0x1);

	printf("%p\n", (unsigned long)p + 0x1);

	printf("%p\n", (unsigned int*)p + 0x1);

	return 0;
}

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Pen test 3：

Particular attention ：

①int* ptr2 = (int*)((int)a + 1);

//a Is the address of the first element of the array , Force type to int type , The value remains the same , The type changes

// if a=0x0012ff40, be a+1=0x0012ff44,(int)a+1=0x0012ff41

The storage of the array in memory ,vs Adopt small end storage mode ,0x00 00 00 01 Store in memory as 01 00 00 00

②printf("%x,%x", ptr1[-1], *ptr2);

//ptr2 The type is int*, Dereference operation access 1 Integer sizes , Then print in hexadecimal

Answer to this question ：4,2000000

The code is as follows ：

#include <stdio.h>

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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Pen test 4：

  Particular attention ：

①int a[3][2] = { (0,1),(2,3),(4,5) };

// Here is the comma expression , Note that （）, No { } , See the figure below to see the difference ！

Answer to this question ：1

The code is as follows ：

#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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Pen test 5：

  Particular attention ：

①int a[5][5];

// In a two-dimensional array int a[5][5] in ,a Is the address of the first element of the array , The type is int (*)[5];

②int(*p)[4];

//p The array pointed to has 4 Elements , Each for int type ,+1 skip 1 individual （4 individual int An array of types ）

③p = a;

   int (*)[4]=int (*)[5];

   printf("%p\n", &p[4][2] - &a[4][2]);
//&p[4][2] and &*(*(p+4)+2) equally .p Is an array pointer ,+4 skip 4 individual （4 individual int An array of types ）

//p++ It is the green icon in the above figure , Orange is p[4] Quoting

④&p[4][2] - &a[4][2]=-4;

//-4 Press %p Print , Yes, it will -4 The complement of is converted to hexadecimal

Original code ：

10000000 00000000 00000000 00000100

Inverse code ：

11111111 11111111 11111111 11111011

Complement code ：

11111111 11111111 11111111 11111100

obtain 0xFF FF FF FC

Answer to this question ：FF FF FF FC,-4

The code is as follows ：

#include <stdio.h>

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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Pen test 6：

Particular attention ：

①int* ptr1 = (int*)(&aa + 1);

//&aa What you get is the address of the entire array ,+1 Skip the entire 2D array

②int* ptr2 = (int*)(*(aa + 1));

//aa Is the address of the first row of the two-dimensional array ,

*(aa + 1)=aa[1],aa[1] Is the address of the first element in the second line , This type is int*, So here's (int*) redundant

Answer to this question ：10,5

The code is as follows ：

#include <stdio.h>

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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Pen test 7：

  Particular attention ：

①char* a[ ] = { "work","at","alibaba" };

// This array stores work、at、alibaba The address of the first element

②char** pa = a;

//pa Pointing to a The second element of the array , After dereferencing, we get at The address of the first element

③printf("%s\n", *pa);

//printf When printing ,%s According to the address of the first element , Print backwards , encounter \0 stop it

Answer to this question ：at

The code is as follows ：

#include <stdio.h>

int main()
{
	char* a[] = { "work","at","alibaba" };

	char** pa = a;

	pa++;

	printf("%s\n", *pa);

	return 0;
}

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Pen test 8：

This topic is the epitome of pointer pen test questions

Particular attention ：

Code analysis ：

++cpp after cpp The address pointed to will change

①printf("%s\n", **++cpp);

//++cpp after ,cpp Point to c+2, After the first dereference, we get c+2, and c+2 Pointing to the array c The third element , The array is obtained after the second dereference c The third element “POINT” The address of the first character ,printf When printing ,%s According to the first character P Print the address backwards , encounter \0 stop it

②printf("%s\n", *--*++cpp+3);

//++cpp after ,cpp Point to c+1, After the first dereference, we get c+1,c+1 Again -1 Pointing to the array c First element , The array is obtained after the second dereference c First element "ENTER" The address of the first character ,+3 obtain T The address of ,printf When printing ,%s According to the characters T Print the address backwards , encounter \0 stop it

③printf("%s\n", *cpp[-2]+3);

//**(cpp-2)+3;cpp-2 Point to c+3, After the first dereference, we get c+3, The address points to the array c Fourth element , The array is obtained after the second dereference c Fourth element "FIRST" The address of the first character ,+3 obtain R The address of ,printf When printing ,%s According to the characters R Print the address backwards , encounter \0 stop it

④printf("%s\n", cpp[-1][-1]+1);

//cpp[-1][-1]+1 and *(*(cpp-1)-1)+1;cpp-1 Point to c+2, After the first dereference, we get c+2, The address -1 Pointing to the array c The second element , The array is obtained after the second dereference c The second element "NEW" The address of the first character ,+1 obtain E The address of ,printf When printing ,%s According to the characters E Print the address backwards , encounter \0 stop it

Answer to this question ：POINT,ER,ST,EW

The code is as follows ：

#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;
}

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Conclusion ：

If it helps you ,

Don't forget it give the thumbs-up + Focus on Oh , Crab

If it helps you ,

Don't forget it give the thumbs-up + Focus on Oh , Crab

If it helps you ,

Don't forget it give the thumbs-up + Focus on Oh , Crab