[C language] advanced pointer exercise 1

One 、 Pointer advanced key knowledge summary

Two 、 Advanced pointer exercises 1

Please think about the running results of the following programs , Give the right answer .

// 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));
// 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(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));
// 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]));   

3、 ... and 、 Advanced pointer exercises 1 Answer and analysis

explain , So here's what I chose x86 Environment is 32 Bit platform , So the results of pointer size calculation are 4. But I didn't give the premise when analyzing , So two results are given 4/8.

（ One ） answer

1. The first group

2. The second group

（1） The first group

（2） The second group

It's obvious from the above picture that it's hanging

（3） The third group

It's obvious from the above picture that it's hanging

3. The third group

（ Two ） analysis

Four 、 Focus on

1.sizeof Only focus on type, not content and meaning ;strlen Pay attention to the end of string \0.

2. Array names have only two meanings , Most of the time, it represents the address of the first element of the array , There are only two cases that represent the entire array ——sizeof( Array name )【 Calculate the memory size of the entire array 】,& Array name 【 The address of the entire array 】

3. The result of pointer operation has a lot to do with the meaning of pointer type . Pointer types have two meanings , One is pointer execution +1/-1 How far is the address skipped by the operation of step , Another is the permission of pointer dereference , for example int The dereference permission of type is 4 Bytes ,char Yes. 1 Bytes ,double* Yes. 8 Bytes, etc .（ We can adopt different types according to our own needs ）