代码是针对一个简单的蛇克隆而且我不想让蛇我能够离开它,如果它已经正确的话。
如果我只是在向右方向按下LEFT但是如果我按下UP然后在时间范围内左转它开始向左移动它是有效的。
function self.update(dt)
if love.keyboard.isDown(self.left) and self.prevvelocity.x ~= 1 then
self.velocity.x = -1
self.velocity.y = 0
end
if love.keyboard.isDown(self.right) and self.prevvelocity.x ~= -1 then
self.velocity.x = 1
self.velocity.y = 0
end
if love.keyboard.isDown(self.up) and self.prevvelocity.y ~= 1 then
self.velocity.x = 0
self.velocity.y = -1
end
if love.keyboard.isDown(self.down) and self.prevvelocity.y ~= -1 then
self.velocity.x = 0
self.velocity.y = 1
end
if self.timeSinceLastMove < self.speedinverted then
self.timeSinceLastMove = self.timeSinceLastMove + dt
else
table.remove(self.tail, 1)
tail = { x = self.position.x, y = self.position.y }
table.insert(self.tail, tail)
self.position.x = self.position.x + self.velocity.x * tileSize
self.position.y = self.position.y + self.velocity.y * tileSize
self.prevvelocity = self.velocity
self.timeSinceLastMove = 0;
end
end
答案 0 :(得分:2)
function self.update(dt)
if love.keyboard.isDown(self.left) and self.prevvelocity.x ~= 1 then
self.velocity.x = -1
self.velocity.y = 0
end
if love.keyboard.isDown(self.right) and self.prevvelocity.x ~= -1 then
self.velocity.x = 1
self.velocity.y = 0
end
if love.keyboard.isDown(self.up) and self.prevvelocity.y ~= 1 then
self.velocity.x = 0
self.velocity.y = -1
end
if love.keyboard.isDown(self.down) and self.prevvelocity.y ~= -1 then
self.velocity.x = 0
self.velocity.y = 1
end
self.timeSinceLastMove = self.timeSinceLastMove + dt
if self.timeSinceLastMove >= self.speedinverted then
self.timeSinceLastMove = self.timeSinceLastMove - self.speedinverted
self.position.x = self.position.x + self.velocity.x * tileSize
self.position.y = self.position.y + self.velocity.y * tileSize
table.remove(self.tail, 1)
local head = { x = self.position.x, y = self.position.y }
table.insert(self.tail, head)
self.prevvelocity = { x = self.velocity.x, y = self.velocity.y }
end
end