[dish of learning notes, dog learning C] Dachang written test, is that it?

Topic 1 ：

// What is the result of the program ？
int main(){
    

    int a[5] = {
     1, 2, 3, 4, 5 };
    int *ptr = (int *)(&a + 1);
    printf( "%d,%d", *(a + 1), *(ptr - 1));
    
    return 0; 
}

 

int main(){
    

    int a[5] = {
     1, 2, 3, 4, 5 };
    int *ptr = (int *)(&a + 1);
    //&a Get the address of the entire array ,+1 Skip this array and point to the next address 
    //(int *) Strong transition type 
    printf( "%d,%d", *(a + 1), *(ptr - 1));
    //a+1 Represents the address of the second element , Dereference to get the second element , That is to say 2
    //ptr - 1 Get the address of the fifth element , Dereference to get the fifth element , That is to say 5
    
    return 0; 
}

---

Topic two ：

struct Test
{
    
	int Num;
	char* pcName;
	short sDate;
	char cha[2];
	short sBa[4];
}*p;
// hypothesis p  The value of is 0x100000.  What are the values of the expressions in the following table ？
int main()
{
    
	printf("%p\n", p + 0x1);
	printf("%p\n", (unsigned long)p + 0x1);
	printf("%p\n", (unsigned int*)p + 0x1);
	return 0;
}

 

struct Test
{
    
	int Num;
	char* pcName;
	short sDate;
	char cha[2];
	short sBa[4];
}*p;
//1 individual int4 byte ,1 individual char*4 byte ,1 individual short2 byte 
//1 Two elements char Array 2 byte ,1 Four elements short Array 8 byte 
// So the size of this structure is 4+4+2+2+8=20 byte 
int main()
{
    
	printf("%p\n", p + 0x1);
	//0x1 Lower hexadecimal representation 1
	//p It's a pointer to a structure ,p The value of is 0x100000
	// Adding one is equivalent to adding 20,20 In hexadecimal, it is 14
	// So the result is 0x100014
	printf("%p\n", (unsigned long)p + 0x1);
	//(unsigned long) hold p The type of , Plus one is plus one 
	// The result is 0x100001
	printf("%p\n", (unsigned int*)p + 0x1);
	//(unsigned int*) hold p Strong into a int Pointer to type 
	// Plus one is equal to plus four 
	// The result is 0x100004
	
	return 0;
}

---

Question three ：

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;
}

 

int main()
{
    
	int a[4] = {
     1, 2, 3, 4 };
	int* ptr1 = (int*)(&a + 1);
	//&a Get the address of the entire array ,+1 Skip the entire array 
	//int* Strong transition type ,ptr1 Only four bytes are skipped for each addition and subtraction 
	int* ptr2 = (int*)((int)a + 1);
	//a Represents the address of the first element , Be forced to int type , It's a string of binary numbers 
	//+1, This is the string of binary numbers +1, It is equivalent to changing one byte 
	//int* Strong transition type , Length programming 4 Bytes 
	printf("%x,%x", ptr1[-1], *ptr2);
	//ptr[-1] amount to *(ptr1-1), The result is 4, Print in hexadecimal or 4
	//*ptr2 The resulting Hex is 2000000
	
	return 0;
}

---

Question 4 ：

int main()
{
    
	int a[3][2] = {
     (0, 1), (2, 3), (4, 5) };
	int* p;
	p = a[0];
	printf("%d", p[0]);

	return 0;
}

 

int main()
{
    
	int a[3][2] = {
     (0, 1), (2, 3), (4, 5) };
	// Be careful ！！！ Here is () No {}, Who is deceived? I don't speak （doge）
	// therefore a[3][2] What is actually stored is { 1, 3, 5}
	int* p;
	p = a[0];
	//p Stored is an array a[3][2] The first line of the address 
	printf("%d", p[0]);
	//p[0] amount to *p, namely *a[0], Get the first element in the first line , That is to say 1

	return 0;
}

---

Question five ：

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;
}

 

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]);
	// Subtracting two pointers yields the number of elements between them ,&p[4][2] And &a[4][2] The number of elements between is 4
	// But the address is from low to high , So the result is -4
	// Negative numbers are stored in memory as complements 
	//-4 Complement code ：11111111 11111111 11111111 11111100
	//%p Print as a pointer , It doesn't matter what complement the original code will 
	// Print this string of binary directly in hexadecimal form , The result is FF FF FF FC
	//%d The printed value is -4
	
	return 0;
}

---

Question six ：

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;
}

 

int main()
{
    
	int aa[2][5] = {
     1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
	int* ptr1 = (int*)(&aa + 1);
	//&aa Get the address of the entire array ,+1 Skip this array 
	//int* Strong transition type ,ptr1 The length is 4 Bytes 
	int* ptr2 = (int*)(*(aa + 1));
	//aa Indicates the address of the first line ,+1 Get the second line of address 
	// Dereference to get the address of the first element in the second line 
	//(int*) It's a smoke bomb , The result is the same with or without it 
	printf("%d,%d", *(ptr1 - 1), *(ptr2 - 1));
	//ptr1-1 Get the address of the fifth element in the second line , Dereference to get 10
	//ptr2-1 Get the address of the fifth element in the first line , Dereference to get 5
	
	return 0;
}

---

Question seven ：

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

 

int main()
{
    
	char* a[] = {
     "work","at","alibaba" };
	//"alibaba" Focus on , It shows that this is Ali's problem （doge）
	//char* type , So the address of the first letter of three words is stored 
	char** pa = a;
	//a Represents the address of the first element , One char* The address of , Store with secondary pointer 
	pa++;
	//pa++, Point to the second element 
	printf("%s\n", *pa);
	//pa Dereference to get string “at” The first letter “a” The address of 
	//%s, from “a” Start printing , encounter “\0” stop it 
	// The string will automatically add a “\0”
	// So the result is “at”
	// Ali ？ Is this ？ I , Would like to be doing ！ Tomorrow, ,offer！ understand ？
	
	return 0;
}

---

Question eight ：

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;
}

 

int main()
{
    
	char* c[] = {
     "ENTER","NEW","POINT","FIRST" };
	//char* type , So the address of the first letter of three words is stored 
	char** cp[] = {
     c + 3,c + 2,c + 1,c };
	char*** cpp = cp;
	//cpp Deposit cp First element address 
	printf("%s\n", **++cpp);
	//++cpp Deposit cp The address of the second element ,cpp The stored value has changed 
	// Dereference to get c+2 namely c The address of the third element 
	// Dereference again to get ‘P’ The address of 
	// be %s Print out "POINT"
	printf("%s\n", *-- * ++cpp + 3);
	//cpp Deposit is cp[1] The address of ,++cpp Point to cp[2]
	// Dereference to get “c+1”,--(c+1) obtain c, Point to ‘E’ The address of 
	//‘E’+3 Get a second ‘E’ The address of 
	//%s From the second ‘E’ Start printing , The result is “ER”
	printf("%s\n", *cpp[-2] + 3);
	//cpp Deposit is cp[2] The address of ,cpp[-2] namely *(cpp-2), obtain “c+3”
	// Dereference again , obtain ‘F’ The address of 
	//‘F’+3 obtain ‘S’ The address of 
	//%s from ‘S’ Start printing , The result is “ST”
	printf("%s\n", cpp[-1][-1] + 1);
	//cpp[-1][-1] amount to *(*(cpp-1)-1)
	//*(cpp-1) obtain “c+2”,c+2-1 == c+1
	//*(c+1) obtain ‘N’ The address of 
	//‘N’+1 obtain ‘E’ The address of 
	//%s from ‘E’ Start printing , The result is ‘EW’

	return 0;
}