Dark Horse Programmer -08-Pointer

First, the pointer Prelude 1. The importance of pointers

Pointers are a very important type of data in C, and if you say that you have a good study in c except for pointers, you simply say you have not learned C language.

2. Small needs

l Voidchange (int n) after the function call is complete, change the value of the argument

L Analysis: Change the value of the argument, find storage space, address

Second, the pointer variable definition 1. The format of the definition

L class name identifier * pointer variable name;

L int*p;

2. Define the Post assignment first

L Simple Value

int a = 10;

int *p;

p = &a;

3. Assigning values at the same time as defined

int a = 10;

int *p = &a;

4. Attention Points

The pointer variable is used to hold the address of the variable, do not give it arbitrary assignment of a constant. The following wording is wrong.

l int *p;

l p = 200; This is wrong.

L%p The address value stored in the output pointer

L Other pointer type description, such as float *p; Char *p;

L can not use the type, such as int a = 10; Float *p =&a;

5. Clear the pointer

L p = 0;

L P =null;

Third, pointer instances

L Assign values to the variables pointed to by the pointer

1 char a = 10;

2 printf ("Before modification, value of a:%d\n", a);

3

4//pointer variable p points to variable a

5 Char *p = &a;

6

7//Modify the value of variable a indirectly by pointer variable p

8 *p = 9;

9

printf ("Modified, value of a:%d", a);

A value is the 10,P value is the address of variable a FFC3.

Notice that the 5th and 8th lines have a "*" and their meanings are different:

(1) the "*" in line 5th is used to indicate that p is a pointer variable

(2) The 8th line of "*" is a pointer operator, where the *p represents a P-value ffc3 this address to access the corresponding storage space, that is, the storage space of variable A, and then the right value 9 is written to this storage space, equivalent to a = 9;

The output is:, it can be found that we indirectly modified the value of variable A through the variable p.

l Example!!! This is the case that Miss MJ gave me a very deep impression of!!!!!!!!!!!!!!!!!!!

Iv. the research of pointers

1. A pointer variable consumes 8 bytes of storage space

V. Pointers and Arrays 1. Pointer to a one-dimensional array element

1//define an array of type int 2 int a[2]; 3 4//Define a pointer of type int 5 int *p; 6 7//Let the pointer point to the NO. 0 element of the array 8 p = &a[0]; 9 10//Modify the value of the pointed element one by one *p = 10;12 13//Print the value of the first element in printf ("a[0] =%d", a[0]);

2. Traverse one-dimensional array elements with the pointer

1//define an array of type int 2 int a[4] = {1, 2, 3, 4}; 3 4//Defines a pointer of type int and points to the No. 0 element of the array 5 int *p = A; 6 7 int i; 8for (i = 0; i < 4; i++) {9//Take the pointer operator * To remove the value of the array element int value = * (p+i); printf ("a[%d] =%d \ n", I, value); 13}

P is a pointer, a is an array

Vi. Pointers and strings

Char *s;

s = "MJ";

The point above is also correct: Define the pointer variable first, and then point to the string. If a character array is not allowed to do this, the following is an error:

1 Char s[10];

2 s = "MJ";

The compiler is sure to report the error of line 2nd, because S is a constant, representing the first address of the array, and cannot be assigned a value operation.

It is also important to note that the following practices are also incorrect:

1 char *s = "MJ";

2

3 *s = "like";

The 3rd line of code made 2 errors:

· The 3rd line of code is equivalent to the string "like" into the memory space of S point, as the 1th line of code can be seen, S points to "MJ" the first character ' M ', that is, a piece of char type S point of storage space, only 1 bytes, to "like" in 1 bytes of space, Positive memory Overflow

· As you can see from the 1th line of code, the pointer s is pointing to the string constant "MJ"! So it is no longer possible to modify the string contents by pointers! Even *s = ' A ' seems to be the wrong way of writing, because s points to a constant string that does not allow the modification of characters inside it.

Now want to change the string "LMJ" the first character ' l ' to ' l ', the solution is a variety of

1 char *p2 = "LMJ";

2 *p2 = ' L ';

3

4 printf ("%s", p2);

It seems to be possible, but this is the error code, which is wrong on line 2nd. First look at line 1th, pointer variable p2 point to a string constant, because it is a constant, so its internal characters are not allowed to modify.

Someone might have been blindfolded, here's the 1th line of code char *P2 = "LMJ"; the 2nd line of code in the first scenario char a[] = "LMJ"; not the same? It's not the same.

· Char a[] = "LMJ"; define a String variable!

· Char *p2 = "LMJ"; defines a string constant!

Dark Horse Programmer -08-Pointer