[英]Increment/Decrement Confusion
What is happening when we decrement the code here:当我们在这里递减代码时会发生什么:
temp[--countArray[getDigit(position, input[tempIndex], radix)]]
If temp is 1 in this case: are we decrementing first so that we are assigning to 0?如果在这种情况下 temp 为 1:我们是否先递减以便分配为 0? How immediate is this decrement?
这种减少有多直接? It always seems to confuse me within array brackets.
它似乎总是让我在数组括号内感到困惑。
Try unpacking the brackets on different levels of indentation:尝试拆开不同缩进级别的括号:
temp[ // get this index in temp
-- // decrement by 1
countArray[ // get this index in countArray
getDigit(position, input[tempIndex], radix) // call getDigit()
]
]
In human-readable terms, it calls getDigit()
to index into countArray
, then decrements that value and uses it to index into temp
.用人类可读的术语来说,它调用
getDigit()
来索引countArray
,然后递减该值并使用它来索引temp
。
The decrement operator --x
is different from x--
because of what it returns.递减运算符
--x
与x--
不同,因为它返回的内容不同。 By the end of the operation, x
always ends up as one less than it was, but --x
returns the new value of x
, while x--
returns the old value of x
from before it was decremented.在操作结束时,
x
总是比原来小 1,但--x
返回x
的新值,而x--
返回x
递减之前的旧值。 The same applies for ++x
and x++
.这同样适用于
++x
和x++
。
Let me break this down some.让我把它分解一些。 Here's some code that's equivalent to above:
下面是一些与上面等效的代码:
int digit = getDigit(position, input[tempIndex], radix);
countArray[digit]--;
int count = countArray[digit];
temp[count] // Do something with the value
Incidentally, this is a classic illustration of why you shouldn't sacrifice clarity to brevity.顺便说一下,这是一个经典的例子,说明为什么不应该为了简洁而牺牲清晰度。
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