递归函数
给定以下递归函数：
// Pre-condition: y is non-negative.
int mysterious(int x, int y) {
if (y == 0) return x;
return 2*mysterious(x, y-1);
}
神秘(3, 2)的返回值是多少？
这是我的调用堆栈：
return 2*mysterious(3, 2-1) => 2*3 => 6, 2*1 => mysterious(6,2)
return 2*mysterious(6, 2-1) => 6*2 => 12, 2*2 => mysterious(12, 2)
但看起来 y 永远不会达到 0。我做错了什么？
Given the following recursive function:
// Pre-condition: y is non-negative.
int mysterious(int x, int y) {
if (y == 0) return x;
return 2*mysterious(x, y-1);
}
What is the return value of mysterious(3, 2)?
Here is my call stack:
return 2*mysterious(3, 2-1) => 2*3 => 6, 2*1 => mysterious(6,2)
return 2*mysterious(6, 2-1) => 6*2 => 12, 2*2 => mysterious(12, 2)
But it seems like y will never reach 0. What am I doing wrong?
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如果你扩展该调用，实际上
Y 只会减少（每次调用减少 1），因此该函数显然是递归的，并且应该在
y>=0时终止if you expand that call you effectively have
Y only ever decreases (by 1 each call) so the function is clearly recursive and should terminate for
y>=0每次调用 Mystery 时（由您调用一次，通过递归调用两次），y 就会减 1。
因此，您得到（在 Mystery 中）
3 2
3 1
3 0
最终值为12(3*2*2)
Each time mysterious is called (once by you, twice by recursion), y is decremented by 1.
So, you get (in mysterious)
3 2
3 1
3 0
the final value is 12 (3*2*2)
x永远不会被修改，但是随着每次递归调用，y都会减一，并且当到达 ground 子句时（if y == 0）它返回 x（第一次调用时为 3）xis never modified, but with each recursive callyis reduced by one and when reaches the ground clause (if y == 0) it returns x (which from the first call is 3)它只不过是
or
，所以它可以很好地定义为
y的任何值It's nothing else than
or
so it could be very well defined for any value of
y