catalogue
2.1 print string pointer array
2.2 printing integer pointer array
3.1 definition of array pointer
3.2 & array name VS array name
Definition of pointer:
- A pointer is a variable used to store an address, which uniquely identifies a piece of memory space.
- The size of the pointer is fixed at 4 / 8 bytes (32-bit platform / 64 bit platform).
- Pointers are typed. The type of pointer determines the step size of the + - integer of the pointer and the permission of pointer dereference operation.
- Pointer operation.
When defining multiple variables in succession:
#include <stdio.h>
int main()
{
int a, b; //int a; int b;
int* pa, pb; //int* pa; int pb;
int* pa, *pb; //int* pa; int* pb;
return 0;
}You can also use typedef or #define. Refer to: Comparison of typedef and #define differences
1. Character pointer
1.1 method of use
General use:
#include <stdio.h>
int main()
{
char ch = 'w';
char *pc = &ch;
*pc = 'w';
return 0;
}
There is another way to use it:
#include <stdio.h>
int main()
{
const char* pstr = "hello bit."; //Here, a string is stored in the read-only data area, and the content cannot be changed
printf("%s\n", pstr);
return 0;
}
1.2 testing
#include <stdio.h>
int main()
{
char str1[] = "hello bit.";
char str2[] = "hello bit.";
const char *str3 = "hello bit.";
const char *str4 = "hello bit.";
if(str1 == str2)
printf("str1 == str2\n");
else
printf("str1 != str2\n");
if(str3 == str4)
printf("str3 == str4\n");
else
printf("str3 != str4\n");
return 0;
}
The running result is: STR1= str2 str3 == str4
The array name represents the address of its first element, and the other two pointers point to the address of the first element stored in the read-only area (the address is the same)
2. Pointer array
- Pointer array: an array of pointers
- Integer array: an array that holds integers
- Character array: an array of characters
int* arr1[10]; / / array of integer pointers
char *arr2[4]; / / array of first level character pointers
char **arr3[5]; / / array of L2 character pointers
2.1 print string pointer array
#include <stdio.h>
int main()
{
char* arr[] = { "abcdef", "qwer", "zhangsan" };
int i = 0;
int sz = sizeof(arr) / sizeof(arr[0]);
for (i = 0; i < sz; i++)
{
printf("%s ", arr[i]); //Dereference is not required because the string is an array of char, that is, the nesting of the first element address
}
return 0;
}
2.2 printing integer pointer array
#include <stdio.h>
int main()
{
int arr1[] = { 1,2,3,4,5 };
int arr2[] = { 2,3,4,5,6 };
int arr3[] = { 3,4,5,6,7 };
int* arr[] = { arr1,arr2,arr3 };
int i = 0;
for (i = 0; i < 3; i++)
{
int j = 0;
for (j = 0; j < 5; j++)
{
printf("%d ", arr[i][j]); //*(*(arr+i)+j)
}
printf("\n");
}
return 0;
}3. Array pointer
3.1 definition of array pointer
Array pointer: pointer to an array
int* p1[10]; / / pointer array
int (*p2)[10]; / / array pointer (explanation: P is first combined with * to indicate that P is a pointer variable, and then points to an array of 10 integers. Therefore, P is a pointer to an array, which is called array pointer)
3.2 & array name VS array name
Analyze the difference between arr and & arr
#include <stdio.h>
int main()
{
int arr[10] = { 0 };
printf("%p\n", arr);
printf("%p\n", arr+1);
printf("%p\n", &(arr[0]));
printf("%p\n", &(arr[0])+1);
printf("%p\n", &arr); //int(*p)[10] = &arr;
printf("%p\n", &arr+1);
return 0;
}In fact: & arr represents the address of the array, not the address of the first element of the array.
In this example, the type of & arr is int(*)[10], which is an array pointer type
expand
#include <stdio.h>
int main()
{
char arr[5];
char(*pa)[5] = &arr;
int* parr[6];
int* (*pp)[6] = &parr;
return 0;
}At this time, the type of pp is int* (*)[6]
3.3 use of array pointer
Two dimensional array: arr[i] = = * (arr+i) = = p[i] = * (p+i)
#include <stdio.h>
void print(int(*p)[5], int r, int c)
{
int i = 0;
for (i = 0; i < r; i++)
{
int j = 0;
for (j = 0; j < c; j++)
{
//*(p+i) is equivalent to getting the i-th row of the two-dimensional array and the array name of the i-th row
//The array name represents the address of the first element, which is also the address of the first element in line i
printf("%d", *(*(p+i)+j));
// p[i][j]
//p is the address of the first line
//p+i is the address of line i
//*(p+i) is the address of the first element in line i
//*(* (p+i)+j) is the value of the first element in line i
}
printf("\n");
}
}
int main()
{
int arr[3][5] = { {1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7} };
print(arr, 3, 5);
return 0;
}practice
int arr[5]; //arr is an integer array with five elements, each of which is of type int int* parr1[10]; //Parr1 is an array with 10 elements. Each element is of type int *, so parr1 is an array of pointers int (*parr2)[10]; //The combination of parr2 and * is a pointer of type int(*)[10], which points to an array. The array has 10 elements, and each element is of type int, so parr2 is an array pointer int (*parr3[10])[5]; //The combination of parr3 and [] shows that parr3 is an array with 10 elements, and each element type of the array is int(*)[5] (an array pointer, which points to an array with 5 elements of type int)
After text transfer: ↓
int arr[5];
//arr is an integer array with five elements, each of which is of type int
int* parr1[10];
//Parr1 is an array with 10 elements. Each element is of type int *, so parr1 is an array of pointers
int (*parr2)[10];
//The combination of parr2 and * is a pointer of type int(*)[10], which points to an array. The array has 10 elements, and each element is of type int, so parr2 is an array pointer
int (*parr3[10])[5];
//The combination of parr3 and [] shows that parr3 is an array with 10 elements, and each element type of the array is int(*)[5] (an array pointer, which points to an array with 5 elements of type int)
