如何用Lua的and/or运算符简化赋值代码？现有代码问题求解

Let's start by aligning on your original logic to make sure we're working from the same baseline:

• When left is true and right is false: xPos = -1
• When right is true and left is false: xPos = 1
• When both are true or both are false: xPos = 0

Why Your Simplified Code Fails #

Your attempt:

local xPos = right and not left and 1 or -1

Relies on Lua's short-circuit evaluation rules for and/or, which work like this:

• a and b: Returns a if a is falsy; otherwise returns b
• a or b: Returns a if a is truthy; otherwise returns b

Let's walk through every scenario to see where it breaks:

1. left=true, right=false: right is falsy, so right and not left returns false. Then false or -1 returns -1 (this matches your desired logic).
2. left=false, right=true: right and not left evaluates to true, so true and 1 returns 1. Then 1 or -1 returns 1 (this also works).
3. left=true, right=true: right and not left becomes true and false = false. Then false or -1 returns -1—but your original logic expects 0 here.
4. left=false, right=false: right is falsy, so right and not left returns false. Then false or -1 returns -1—again, original logic expects 0.

The core issue is your expression only handles two of the four possible state combinations. There's no branch that returns 0 for the "both true" or "both false" cases, so it defaults to -1 incorrectly.

Correct Simplifications #

Here are a few clean, idiomatic ways to implement your logic with and/or (and other Lua tricks):

1. Explicit and/or Branching

This directly translates your original if-else chain into a single line, covering all cases:

local xPos = (left and not right) and -1 or (right and not left) and 1 or 0

How it works:

• First checks if left is true and right is false: returns -1 if so
• If not, checks if right is true and left is false: returns 1 if so
• If neither condition is met, returns 0 (covers both true or both false)

2. Arithmetic Shortcut (Most Clean!)

Lua treats true as 1 and false as 0 in arithmetic contexts, so you can calculate xPos directly with simple math:

local xPos = (right and 1 or 0) - (left and 1 or 0)

Or equivalently:

local xPos = (right and 1 or 0) + (left and -1 or 0)

Testing this against all cases:

• left=true, right=false: 0 - 1 = -1 (correct)
• left=false, right=true: 1 - 0 = 1 (correct)
• left=true, right=true: 1 - 1 = 0 (correct)
• left=false, right=false: 0 - 0 = 0 (correct)

3. Readable Ternary-Style Check

If you prioritize clarity over brevity, you can first check if left and right differ, then return the appropriate value:

local xPos = left ~= right and (left and -1 or 1) or 0

This reads like: "If left and right are different, return -1 if left is true, else 1; if they're the same, return 0."

内容的提问来源于stack exchange，提问作者arsonist