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指针面试题,对指针的理解不再停留在简单的知识层面上,而是可以知道面试题中指针的考察是怎样的;
1. Analysis of pointers and arrays written test questions
1.1 One-dimensional arrays
First of all, let me talk about the knowledge point: it is very important! ! !
The meaning of the array name:
- sizeof (array name), where the array name represents the entire array, and the size of the entire array is calculated.
- & array name, where the array name represents the entire array, and the address of the entire array is taken out.
- All array names other than that 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 Character arrays
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 arrays
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 the structure has not yet been learned, the size of the structure is told here to be 20 bytes
struct Test
{ int Num; char *pcName; short sDate; char cha[2]; short sBa[4]; }*p; / /Assume the value of p is 0x100000. What are the values of the expressions in the following table? //It is known that the variable size of the structure Test 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;
}
Summarize
The pointer's blog has come to an end, so stay tuned for the next blog! ! !