K＆安培; - [R的malloc code没有意义？ [英] K & R malloc code doesn't make sense?
问题描述
这code是从K＆放大器; - [R书 - 第8章第7节：举例 - 存储分配器。这code，至少对我来说，没有任何意义。 头是一个结构和最严格的对齐类型，这是一个长型的联合。然后的malloc会发现头部大小的倍数的大小足够大的自由空间。
This code is from the K & R book - Chapter 8 Section 7: Example - Storage Allocator. This code, at least for me, doesn't make sense. "Header" is a union of a struct and a "most restrictive alignment type", which is a long type. Malloc will then find a big enough free space with size of a multiple of the header size.
static Header base; /* empty list to get started */
static Header *freep = NULL; /* start of free list */
/* malloc: general-purpose storage allocator */
void *malloc(unsigned nbytes)
{
Header *p, *prevp;
Header *morecore(unsigned);
unsigned nunits;
nunits = (nbytes+sizeof(Header)-1)/sizeof(Header) + 1;
if ((prevp = freep) == NULL) { /* no free list yet */
base.s.ptr = freeptr = prevptr = &base;
base.s.size = 0;
}
for (p = prevp->s.ptr; ; prevp = p, p = p->s.ptr) {
if (p->s.size >= nunits) { /* big enough */
if (p->s.size == nunits) /* exactly */
prevp->s.ptr = p->s.ptr;
else { /* allocate tail end */
p->s.size -= nunits;
p += p->s.size;
p->s.size = nunits;
}
freep = prevp;
return (void *)(p+1);
}
if (p == freep) /* wrapped around free list */
if ((p = morecore(nunits)) == NULL)
return NULL; /* none left */
}
}
这code的奇数部分是语句 nunits =（为nbytes +的sizeof（头）-1）/的sizeof（头）+ 1; 这是那么在比较中使用如果（对GT; s.size＆GT; = nunits）来找到一台足够大的空间，在头的规模而言。应该不是前者是 nunits =（为nbytes +的sizeof（头））/的sizeof（标题）只？原来的code将评估为低于它应该是一个值。什么是与+ -1s？为什么分配空间小于期望。
The odd part of this code is the statement nunits = (nbytes+sizeof(Header)-1)/sizeof(Header) + 1; which is then used in the comparison if (p->s.size >= nunits) to find a big enough space with units in terms of the size of Header. Shouldn't the former be nunits = (nbytes+sizeof(Header)) / sizeof(Header) only? The original code would evaluate to a value less than it ought to be. What is with the +-1s? Why allocate space less than the desired.
推荐答案
的 1 和 +1 是考虑到不属于该块大小的相乘值。
The -1 and +1 are to account for values that aren't multiplies of the block size.
例如，假设块的大小是10，如果你尝试使用公式 N / 10 来获得所需的块数，那么你将获得1块N = 15，这是不对的，你需要2。
For example, suppose the block size is 10. If you try to use the formula n / 10 to get the number of required blocks then you would get 1 block for n = 15. This is wrong, you need 2.
如果您更改公式为 N / 10 + 1 那么它也将是错误的。当n = 20你只需要2个街区，但公式将返回3。
If you change the formula to be n / 10 + 1 then it will also be wrong. When n = 20 you only need 2 blocks, but that formula will return 3.
正确的公式是（N - 1）/ 10 + 1 。这就是你怎么收尾整数除法。与此唯一的区别，你问公式是额外的的sizeof（标题），这是刚需的头本身。
The correct formula is (n - 1) / 10 + 1. That's how you round up with integer division. The only difference with this and the formula you asked about is the extra sizeof(Header), which is just the extra space needed for the header itself.
这篇关于K＆安培; - [R的malloc code没有意义？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
