本文通过模拟汇编里的stack机制,构建一个自己的stack,然后将上一篇blog末尾的递归函数void bst_walk(bst_node_t *root)非递归化。
o libstack.h
#ifndef _LIBSTACK_H
#define _LIBSTACK_H #ifdef __cplusplus
extern "C" {
#endif typedef void * uintptr_t; /* generic pointer to any struct */ uintptr_t *stack_init(size_t size);
void stack_fini();
int stack_isFull();
int stack_isEmpty();
void push(uintptr_t e);
void pop(uintptr_t *e); #ifdef __cplusplus
}
#endif #endif /* _LIBSTACK_H */
o libstack.c
#include <stdio.h>
#include <stdlib.h>
#include "libstack.h" /**
* Basic Stack OPs are supported, including:
*
* 1. Construct/Destruct a stack
* 2. Tell stack is full or empty
* 3. push() and pop()
*
* == DESIGN NOTES ==
*
* There are 3 static variables reserved,
*
* ss: stack segment
* sp: stack pointer
* sz: stack size
*
* And the stack looks like:
*
* | RED | ss[-1] ; SHOULD NEVER BE ACCESSED
* low-addr +-----+ <------------TopOfStack-----------
* ^ | | ss[0]
* | | | ss[1]
* | | ... |
* | | |
* | | | ss[sz-1]
* | +-----+ <------------BottomOfStack--------
* high-addr | RED | ss[sz] ; SHOULD NEVER BE ACCESSED
*
* (1) If (sp - ss) == 0, stack is full
* (2) If (sp - ss) == sz, stack is empty
* (3) Push(E): { sp -= 1; *sp = E; }
* (4) Pop(&E): { *E = *sp; sp += 1; }
*/ static uintptr_t *ss = NULL; /* stack segment */
static uintptr_t *sp = NULL; /* stack pointer */
static size_t sz = ; /* stack size */ int stack_isFull() { return (sp == ss); }
int stack_isEmpty() { return (sp == ss + sz); } uintptr_t *
stack_init(size_t size)
{
ss = (uintptr_t *)malloc(sizeof (uintptr_t) * size);
if (ss == NULL) {
fprintf(stderr, "failed to malloc\n");
return NULL;
} sz = size;
sp = ss + size;
return ss;
} void
stack_fini()
{
free(ss);
} void
push(uintptr_t e)
{
sp -= ;
*sp = e;
} void
pop(uintptr_t *e)
{
*e = *sp;
sp += ;
}
1. 一旦栈被初始化后,栈指针sp一定是指向栈底,*sp不可访问(尤其是写操作),因为不在分配的内存有效范围内;
2. 对于入栈操作(push), 第一步是将sp-=1, 第二步是写入要入栈的元素 (*sp = E); (因为初始化后*sp的内存不可写,所以push操作一定率先改写sp)
3. 对于出栈操作(pop), 顺序与push相反,第一步取出sp指向的内存地址里的内容(E = *sp), 第二步才是将sp+=1;
o foo.c (简单测试)
/**
* A simple test against stack OPs, including:
* o stack_init(), stack_fini()
* o stack_isFull(), stack_isEmpty()
* o push(), pop()
*/ #include <stdio.h>
#include "libstack.h" static void
dump_stack(uintptr_t *ss, size_t size)
{
(void) printf("%p: ", ss);
for (int i = ; i < size; i++) {
if (ss[i] != NULL)
(void) printf("%-10p ", *(ss+i));
else
(void) printf("0x%-8x ", 0x0);
}
printf("\n");
} int
main(int argc, char *argv[])
{
size_t size = ; uintptr_t *ss = stack_init(size);
dump_stack(ss, size); for (int i = ; !stack_isFull(); i++) {
push((uintptr_t)(ss+i));
dump_stack(ss, size);
} (void) printf("\n"); uintptr_t e = NULL;
for (; !stack_isEmpty();) {
pop(&e);
(void) printf(" (pop) got %-10p\n", e);
} stack_fini(); return ;
}
o Makefile
CC = gcc
CFLAGS = -g -Wall -std=gnu99 -m32
INCS = TARGET = foo all: ${TARGET} foo: foo.o libstack.o
${CC} ${CFLAGS} -o $@ $^ foo.o: foo.c
${CC} ${CFLAGS} -c $< ${INCS} libstack.o: libstack.c libstack.h
${CC} ${CFLAGS} -c $< clean:
rm -f *.o
clobber: clean
rm -f ${TARGET}
o 编译和运行测试
$ make
gcc -g -Wall -std=gnu99 -m32 -c foo.c
gcc -g -Wall -std=gnu99 -m32 -c libstack.c
gcc -g -Wall -std=gnu99 -m32 -o foo foo.o libstack.o $ ./foo
0x8ecc008: 0x0 0x0 0x0 0x0
0x8ecc008: 0x0 0x0 0x0 0x8ecc008
0x8ecc008: 0x0 0x0 0x8ecc00c 0x8ecc008
0x8ecc008: 0x0 0x8ecc010 0x8ecc00c 0x8ecc008
0x8ecc008: 0x8ecc014 0x8ecc010 0x8ecc00c 0x8ecc008 (pop) got 0x8ecc014
(pop) got 0x8ecc010
(pop) got 0x8ecc00c
(pop) got 0x8ecc008
$
测试简单且直接,不解释。如果还不确信,可以用gdb调试。
现在对上一篇blog末尾的递归函数使用上面实现的stack进行去递归化改写,改写后的代码如下:
void
bst_walk(bst_node_t *root)
{
if (root == NULL)
return; (void) stack_init(STACK_SIZE); while (root != NULL || !stack_isEmpty()) {
if (root != NULL) {
push((uintptr_t)root);
root = root->left;
continue;
} pop((uintptr_t *)(&root));
printf("%d\n", root->key); root = root->right;
} stack_fini();
}
为方便阅读,下面给出使用meld进行diff后的截图,
- L7: 构建一个stack, 其中STACK_SIZE是一个宏
- L22: 将stack销毁
- L9-14: 首先遍历左子树,不断将结点压入栈中,直到到达最左的叶子结点,那么则执行L16-17 (最左的叶子结点也会被压入栈中)
- L16-17: 出栈并打印结点的key
- L19: 将新的根结点设置为刚刚出栈的结点的右儿子, 重新执行L9-17, 直到所有结点都被遍历到(当然, stack为空)
注: 左图中的函数使用了两次递归,所以将其转化成非递归函数的难度相对较大。