LUA 删除元素的问题

table在删除元素时要注意,例
t = { "hello", "world", "!"}
t[1] = nil
此时print(#t) --输出3,就是说把表的元素置为nil并没有移除该表项。

但,若是:

t = {
[1] = nil,
[2] = 223,
[3] = nil
}

function count(t)
    local c = 0
    for k, v in pairs(t) do
        c = c + 1
    end
    return c
end

print(count(t))--1

 再看更诡异的

 1 t = {
 2 [1] = 12,
 3 [2] = nil,
 4 [3] = 4
 5 }
 6 
 7 function count(t)
 8     local c = 0
 9     for k, v in pairs(t) do
10         c = c + 1
11     end
12     return c
13 end
14 
15 print("------------------", #t)
16 t2 = { "hello", "world", "!"}
17 t2[1] = nil
18 t2[22] = nil
19 t2[9] = nil
20 print("--------t2---------", #t2)
21 for k, v in pairs(t2) do
22     print(k, v)
23 end
24 
25 print("--------------->")
26 for i=1, #t2 do
27     print(i, t2[i])
28 end
原文地址:https://www.cnblogs.com/timeObjserver/p/6404361.html