[英]Why doesn't the following heapsort function produce an error
I took the following code from GeeksforGeeks to try and understand heap sort 我从GeeksforGeeks中获取了以下代码,以尝试了解堆排序
def heapify(arr, n, i):
largest = i
l = 2*i + 1
r = 2*i + 2
if l < n and arr[i] < arr[l]:
largest = l
if r < n and arr[largest] < arr[r]:
largest = r
if largest != i:
arr[i],arr[largest] = arr[largest],arr[i]
heapify(arr, n, largest)
def heapSort(arr):
n = len(arr)
for i in range(n, -1, -1):
heapify(arr, n, i)
for i in range(n-1, 0, -1):
arr[i], arr[0] = arr[0], arr[i]
heapify(arr, i, 0)
arr = [7, 11, 13, 6, 5, 12]
heapSort(arr)
print ("Sorted array is", arr)
On the very first iteration, n = 6 and l = 13 Then for the following line of code 在第一次迭代中,n = 6且l = 13然后是下面的代码行
if l < n and arr[i] < arr[l]
arr[l] points to an index that doesn't exist. arr [l]指向不存在的索引。
I don't understand why this doesn't flag an error like "out of index" or something. 我不明白为什么这不标记“索引不足”之类的错误。 Even though its an "if" statement, it is still surely checking the value in arr[l]. 即使其为“ if”语句,它仍肯定会检查arr [l]中的值。 As this doesn't exist, it should "break" and flag an error? 由于不存在,因此应该“中断”并标记错误?
Thanks 谢谢
if-statement conditions are evaluated in the order that they are defined. if语句条件按照定义的顺序进行评估。 they are also optimized. 他们也进行了优化。
if l < n and arr[i] < arr[l]
The l < n
will be evaluated first. l < n
将首先被评估。 It's False
. 是False
。 Since and
ing anything with False
will be false anyway, the arr[i] < arr[l]
is never evaluated. 由于and
以1ng任何False
将是假的反正arr[i] < arr[l]
是从来没有进行评价。 Hence you never get the IndexError
因此,您永远不会得到IndexError
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