[written examination question analysis] | | get [sizeof and strlen] [pointer and array] graphic explanation + code analysis

    Keep going.

---

Catalog

sizeof And strlen

brief introduction ：

subject ：

1. One dimensional array  

2. A character array

 3. Two dimensional array

Pointer written test questions

---

sizeof And strlen

brief introduction ：

sizeof yes C/C++ One of them The operator （operator）, Simply put, its function is to return the proportion of an object or type Number of memory bytes .

notes ：sizeof It's the operator , It's not a function , You don't need to include any header files when using .

strlen yes c/c++ One of them function , Its function is calculation The length of a specified string .strlen What is done is the work of a counter , It comes from somewhere in memory （ It can start with a string , Somewhere in the middle , Even some uncertain memory area ） Start scanning , Until I met the first character string Terminator '\0' until , Then return the counter value ( Length does not contain '\0').

notes ：strlen Is the function , It is necessary to introduce The header file .string.h or cstring.

The function prototype ：

size_t strlen(const char *string);

（typedef unsigned int size_t ;）

sizeof And strlen The difference between ：

sizeof The empty character \0 Count in , and strlen Will not \0 Count in .

subject ：

Keep four tenets in mind before doing questions ：

The meaning of array names ：

***1.sizeof( Array name ）, The array name here represents the whole array , It calculates the size of the entire array .

***2.& Array name , The array here represents the whole array , It takes out the address of the entire array .

3. In addition, all array names represent the address of the first element .

4. A few bytes of a pointer is language independent , It's about the addressing capability of the system .

A pointer in 32 Bit operating system , Occupy 4 Bytes

A pointer in 64 Bit operating system , Occupy 8 Bytes

1. One dimensional array  

    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));

answer （32 Bit platform ）：

analysis :

2. A character array

1）

    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));
	

analysis ：

2） 

	char arr[] = { 'a', 'b', 'c', 'd', 'e', 'f' };
	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));

  answer ：

3）

    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));

answer ：

  analysis ：

4）

    char *p = "abcdef";
	printf("%d\n", sizeof(p));
	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(p));
	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));

analysis ：

 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]));

answer ：

analysis ： 

---

Pointer written test questions

1） 

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

2） 

// 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 ？
int main()
{
 printf("%p\n", p + 0x1);
 printf("%p\n", (unsigned long)p + 0x1);
 printf("%p\n", (unsigned int*)p + 0x1);
 return 0;
}

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;
}

answer ： 

 

  analysis ：

 4）

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

answer ： 

 

analysis ： 

 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;
}

answer ： 

 

 

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;
}

answer ：

 

 7）

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

answer ： 

analysis ：

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;
}

answer ： 

analysis ：