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。