Linux C申請內存三種基本方式

一份代碼可以知道具體方式和原理：

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;
}

輸出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都是堆棧中的棧內存申請，因int占用四個字節，故f0 -> f4。

2. cd都是靜態存儲變量申請內存，在編譯時已經申請分配好，不釋放。

3. ef都是動態申請內存，屬於堆棧的堆內存申請，此處返回一個指針。

 

情況1

        heap_e = (int *)malloc(20); heap_f = (int *)malloc(20);

  malloc (10) -> 10bytes內存申請

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

0xc04010

　　　　　　　　　　　　0xc04030

中間0x20 = 32bytes，由於字節對齊，申請時需要申請20bytes，系統對malloc管理是讓其在32bytes后再開辟新的內存空間。

情況2
        　

 heap_e = (int *)malloc(30);
 heap_f = (int *)malloc(30);

 malloc (10) -> 10bytes內存申請

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

0xc04010

　　　　　　　　　　　　          0xc04040

中間0x30 = 48bytes，由於字節對齊，申請時需要申請30bytes，系統對malloc管理是讓其在48bytes后再開辟新的內存空間。

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

　　　　前后內存地址不一致，malloc多次內存是不會連續的。