Linux C申请内存三种基本方式

一份代码可以知道具体方式和原理：

int main()

{

        int stack_a;

        int stack_b;

        static int static_c;

        static int static_d;

        int *heap_e;

        int *heap_f;

        heap_e = (int *)malloc(10);

        heap_f = (int *)malloc(10);

        printf("The a address is %p\n",&stack_a);

        printf("The b address is %p\n",&stack_b);

        printf("The c address is %p\n",&static_c);

        printf("The d address is %p\n",&static_d);

        printf("The e address is %p\n",heap_e);

        printf("The f address is %p\n",heap_f);

        return 0;

}

输出log

root@ubuntu:/home/watson/test# ./a.out

The a address is 0x7ffd2d5894f0

The b address is 0x7ffd2d5894f4

The c address is 0x60104c

The d address is 0x601050

The e address is 0x23db010

The f address is 0x23db030

分析：

1. ab都是堆栈中的栈内存申请，因int占用四个字节，故f0 -> f4。

2. cd都是静态存储变量申请内存，在编译时已经申请分配好，不释放。

3. ef都是动态申请内存，属于堆栈的堆内存申请，此处返回一个指针。

情况1

        heap_e = (int *)malloc(20);

        heap_f = (int *)malloc(20);

  malloc (10) -> 10bytes内存申请

　The e address is 0xc04010
　The f address is 0xc04030
|--------------------|.....|--------------------|

0xc04010

　　　　　　　　　　　　0xc04030

中间0x20 = 32bytes，由于字节对齐，申请时需要申请20bytes，系统对malloc管理是让其在32bytes后再开辟新的内存空间。

情况2
        　

 heap_e = (int *)malloc(30);

 heap_f = (int *)malloc(30);

 malloc (10) -> 10bytes内存申请

　The e address is 0xc04010
　The f address is 0xc04040
|------------------------------|.....|------------------------------|

0xc04010

　　　　　　　　　　　　          0xc04040

中间0x30 = 48bytes，由于字节对齐，申请时需要申请30bytes，系统对malloc管理是让其在48bytes后再开辟新的内存空间。

修改如下的程序：

        printf("The e address is %p\n",heap_e);

        printf("The e+1 address is %p\n",heap_e + 1);

        printf("The f address is %p\n",heap_f);

        printf("The f-1 address is %p\n",heap_f - 1);

　　　　The e address is 0x12fa010
　　　　The e+1 address is 0x12fa014
　　　　The f address is 0x12fa030
　　　　The f-1 address is 0x12fa02c

　　　　0x12fa014

　　　　0x12fa02c

　　　　前后内存地址不一致，malloc多次内存是不会连续的。

Linux C申请内存三种基本方式的相关教程结束。