Make learning pointer easier (3)

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Preface

Pointer interview questions , The understanding of pointer no longer stays at the simple level of knowledge , Instead, you can know how the pointer in the interview question is examined ;

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One 、 Pointer and array written test question analysis

1.1 One dimensional array

First, let's talk about the knowledge points ： Very important ！！！
The meaning of array names ：

1. sizeof( Array name ), The array name here represents the entire array , It calculates the size of the entire array .
2. & Array name , The array name here represents the entire array , It takes out the address of the entire array .
3. In addition, all array names represent the address of the first element .

int a[ ] = {1,2,3,4};
printf( “%d\n”,sizeof(a) );
printf( “%d\n”,sizeof(a+0) );
printf( “%d\n”,sizeof(a) );
printf( “%d\n”,sizeof(a+1) );
printf( “%d\n”,sizeof(a[1]) );
printf( “%d\n”,sizeof(&a) );
printf( “%d\n”,sizeof(&a) ;
printf( “%d\n”,sizeof(&a+1) );
printf( “%d\n”,sizeof(&a[0]) );
printf( “%d\n”,sizeof(&a[0]+1) );

1.2 A character array

char arr[] = {‘a’,‘b’,‘c’,‘d’,‘e’,‘f’};
printf(“%d\n”, sizeof(arr));
printf(“%d\n”, sizeof(arr+0));
printf(“%d\n”, sizeof(*arr));
printf(“%d\n”, sizeof(arr[1]));
printf(“%d\n”, sizeof(&arr));
printf(“%d\n”, sizeof(&arr+1));
printf(“%d\n”, sizeof(&arr[0]+1));
printf(“%d\n”, strlen(arr));
printf(“%d\n”, strlen(arr+0));
printf(“%d\n”, strlen(*arr));
printf(“%d\n”, strlen(arr[1]));
printf(“%d\n”, strlen(&arr));
printf(“%d\n”, strlen(&arr+1));
printf(“%d\n”, strlen(&arr[0]+1));

char arr[] = “abcdef”;
printf(“%d\n”, sizeof(arr));
printf(“%d\n”, sizeof(arr+0));
printf(“%d\n”, sizeof(*arr));
printf(“%d\n”, sizeof(arr[1]));
printf(“%d\n”, sizeof(&arr));
printf(“%d\n”, sizeof(&arr+1));
printf(“%d\n”, sizeof(&arr[0]+1));
printf(“%d\n”, strlen(arr));
printf(“%d\n”, strlen(arr+0));
printf(“%d\n”, strlen(*arr));
printf(“%d\n”, strlen(arr[1]));
printf(“%d\n”, strlen(&arr));
printf(“%d\n”, strlen(&arr+1));
printf(“%d\n”, strlen(&arr[0]+1));

char *p = “abcdef”;
printf(“%d\n”, sizeof§);
printf(“%d\n”, sizeof(p+1));
printf(“%d\n”, sizeof(*p));
printf(“%d\n”, sizeof(p[0]));
printf(“%d\n”, sizeof(&p));
printf(“%d\n”, sizeof(&p+1));
printf(“%d\n”, sizeof(&p[0]+1));
printf(“%d\n”, strlen§);
printf(“%d\n”, strlen(p+1));
printf(“%d\n”, strlen(*p));
printf(“%d\n”, strlen(p[0]));
printf(“%d\n”, strlen(&p));
printf(“%d\n”, strlen(&p+1));
printf(“%d\n”, strlen(&p[0]+1));

1.3 Two dimensional array

int a[3][4] = {0};
printf( “%d\n”,sizeof(a) );
printf( “%d\n",sizeof(a[0][0]) );
printf( “%d\n”,sizeof(a[0]) );
printf( ”%d\n",sizeof(a[0]+1) );
printf( “%d\n”,sizeof(* (a[0]+1)) );
printf( “%d\n”,sizeof(a+1) );
printf( “%d\n”,sizeof(* ( a+1) ) );
printf( “%d\n”,sizeof(&a[0]+1) );
printf( “%d\n”,sizeof( * (&a[0]+1) ) );
printf( “%d\n”,sizeof(*a) );
printf( “%d\n”,sizeof(a[3]) );

2. Pointer written test questions

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 the program ？

// Because I haven't learned the structure yet , The size of the structure is 20 Bytes
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 ？
// 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;
}

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;
}

#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;
}

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 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;
}

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

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;
}

summary

Pointer's blog is over , Please look forward to the next blog ！！！