Simplified
Crazy walk trough;
void reverse(int a[], int s)
{
int temp; /* temporary value */
if (s <= 2) return; /* trigger done */
t = a[0]; /* temp = first index of a */
a[0] = a[s - 1]; /* a[0] = a[end - 1] (end including \0) */
a[s - 1] = t; /* a[end - 1] = temp */
r(&a[1], s - 2); /* pass address of a[1] and end - 2 */
}
由於字符數組"ABCDEFG"
簡化存儲表可能是:
Address Value
7 A
8 B
9 C
a D
b E
c F
d G
/* Or as used here: */
789abcd <- Simplified memory address
ABCDEFG
我們得到; main()
電話reverse(ABCDEFG, 7)
列表1
- 地址裁判。到
A
被推到
- 7被推到堆棧
- 返回地址呼叫者被壓入堆棧
- 等堆棧(A {BCDEFG})
- 函數稱爲
而且像
#::::::::::::::::::::::::::::::::::::::::::::::::::::
reverse(ABCDEFG, 7); # Push to STACK 0xB (As List 1)
#====================================================
789abcd <- Memory address.
ABCDEFG <- Values.
<- Indexes for a in recursion 1.
if (7 <= 2) return;
temp = A
+ .
a[0] = a[6] => ABCDEFG = GBCDEFG
+
a[6] = temp => GBCDEFG = GBCDEFA
reverse(BCDEFA, 5); # Push to STACK 0xC (As in List 1)
#====================================================
7 89abcd <- Memory addresses.
[G]BCDEFA <- Values
<- Indexes for a in recursion 2.
if (5 <= 2) return;
temp = B
+ .
a[0] = a[4] => BCDEFA = FCDEFA
+
a[4] = temp => FCDEFA = FCDEBA
reverse(CDEBA, 3); # Push to STACK 0xD (As in List 1)
#====================================================
78 9abcd <- Memory addresses.
[GF]CDEBA <- Values.
<- indexes for a in recursion 3.
if (3 <= 2) return;
temp = C
+ .
a[0] = a[2] => CDEBA = EDEBA
+
a[2] = temp => EDEBA = EDCBA
reverse(DCBA, 1); # Push to STACK 0xE (As in List 1)
#====================================================
789 abcd <- Memory addresses.
[GFE]DCBA <- Values.
<- Indexes for a in recursion 4.
if (1 <= 2) return; YES!
#:::: roll back stack ::::
Pop STACK 0xE
Pop STACK 0xD
Pop STACK 0xC
Pop STACK 0xB
We are back in main() and memory region 789abcd has
been altered from ABCDEFG to GFEDCBA.
IMHO行'如果(一個或多個<= 2)回報;'應該是'如果(一個或多個<2)回報;' – wildplasser 2012-04-15 14:44:42
@wildplasser你是絕對正確的;請參閱下面的答案中的分析。 – 2012-04-15 21:35:27