MSVC 下面有：
1
2 void * _aligned_malloc(
3 size_t size,
4 size_t alignment
5 );
Unix/Linux下面有：
1 int posix_memalign(void **memptr, size_t alignment, size_t size);
自己實(shí)現(xiàn)：
1 #include <iostream>
2 #include <stdlib.h>
3 using namespace std;
4 5 //by snape 2012-3-25
6 7 void *aligned_malloc(size_t required_bytes, size_t alignment)
8 {
9 void *p1;
//orignal block
10 void **p2;
//aligned block
11 12 int offset=alignment-1+
sizeof(
void*);
13 if( (p1=(
void*)malloc(required_bytes+offset))==NULL )
14 return NULL;
15 //cast void* to void** to let p1 store into heap in the form of pointer
16 p2=(
void**)( ((size_t)(p1)+offset) & ~(alignment-1) );
//add offset to p1 and then back to the aligned address
17 p2[-1]=p1;
18 return p2;
19 }
20 21 void aigned_free(
void *p2)
22 {
23 void* p1=((
void**)p2)[-1];
//get the p1
24 free(p1);
25 }
26 27 28 29 int main()
30 {
31 int size=10, align=16;
32 int *p=(
int *)aligned_malloc(size, align);
33 34 cout<<"The
start address="<<p<<endl;
35 36 aigned_free(p);
37 }
上面實(shí)現(xiàn)很考驗(yàn)對(duì)指針的掌握。原理不是很難。
Line12：分配內(nèi)存的時(shí)候，加上alignment-1個(gè)空間，就能保證能夠找到和alignment對(duì)齊的開(kāi)始地址。多出來(lái)的sizeof（void*）是為了保存malloc最初返回的指針p1而分配的一個(gè)空間
Line 17: 保存原始的malloc返回的地址p1到對(duì)齊地址的前一個(gè)空間