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
|--------------------|.....|--------------------|0xc040100xc04030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
|------------------------------|.....|------------------------------|0xc040100xc04040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。