C——内存管理1

array limitations：

• the size of the array must be known beforehand
• the size of the array cannot be changed in the duration of your program

So there comes Dynamic memory allocation:

Dynamic memory allocation?in C is a way of circumventing these problems.

The standard C function?malloc function 定义在 stdlib.h or malloc.h 中，取决于你使用的操作系统。

Malloc.h contains only the definitions for the memory allocation functions and not the rest of the other functions defined in stdlib.h.Usually you will not need to be so specific in your program,and if both are supported,you should use <stdlib.h>,since that is ANSI C,and what we will use here.

The corresponding call to release allocated memory back to the operating system is?free. When dynamically allocated memory is no longer needed,?free?should be called to release it back to the memory pool.

Overwriting a pointer that points to dynamically allocated memory can result in that data becoming inaccessible. （对于指向动态分配的内存的指针，指向最好不要发生变化，否则会导致动态分配的内存里的数据无法被获取。）If this happens frequently,eventually the operating system will no longer be able to allocate more memory for the process. Once the process exits,the operating system is able to free all dynamically allocated memory associated with the process.

create an array:

? ? ? ? ? ? ? ? ? ? ? ? ? int array[10]; /* array?can be considered a pointer which we use as an array. */

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有时候在写程序的时候，我们不知道需要为我们的数据分配多少内存。在这种情况下，我们想要在程序已经开始运行后，为我们的程序动态分配其所需要的内存。?To do this we only need to declare a pointer,and invoke?malloc?when we wish to make space for the elements in our array,?or,we can tell?malloc?to make space when we first initialize the array. Either way is acceptable and useful.

We also need to know how much an int takes up in memory in order to make room for it; fortunately this is not difficult,we can use C‘s builtin?sizeof?operator. For example,if?sizeof(int)?yields 4,then one?int?takes up 4 bytes. Naturally,?2*sizeof(int)?is how much memory we need for 2?ints,and so on.

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So how do we?malloc?an array of ten?ints like before? If we wish to declare and make room in one hit,we can simply say

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int *array = malloc(10*sizeof(int)); 

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We only need to declare the pointer;?malloc?gives us some space to store the 10?ints,and returns the pointer to the first element,which is assigned to that pointer.

重要警示!?malloc?并不初始化 array; this means that the array may contain random or unexpected values! Like creating arrays without dynamic allocation,the programmer must initialize the array with sensible values before using it. Make sure you do so,too. (See later the function?memset?for a simple method.)

It is not necessary to immediately call?malloc?after declaring a pointer for the allocated memory. Often a number of statements exist between the declaration and the call to?malloc,as follows:

int *array = NULL; printf("Hello World!!!"); /* more statements */ array = malloc(10*sizeof(int)); /* delayed allocation */ /* use the array */

int *pt = malloc(3 * sizeof(int)); if(pt == NULL) { fprintf(stderr,"Out of memory,exitingn"); exit(1); }