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这是我对这个令人难以置信的社区的第一个问题。
这些天来,我正在为自己编写一个 Lua 模块。这是有问题的代码的最小部分
mt = {}
bf = {}
------------
-- bf.new --
------------
function bf.new(A)
local out = A
setmetatable(out,mt)
return out
end
-----------------
-- bf.tostring --
-----------------
function bf.tostring(A)
local typeA = type(A)
local out = ""
if typeA=="number" or typeA=="boolean" then
print(tostring(A))
elseif typeA=="table" then
local col = ""
local m = #A
local typeA1 = type(A[1])
for i=1,m do
if typeA1=="table" then
n = #A[1]
for j=1,n do
out = out.." "..tostring(A[i][j])
if j==n then
out = out.."\n"
end
end
elseif (typeA1=="number" or typeA1=="boolean") then
row = row.." "..tostring(A[i][j])
end
end
end
return out
end
-----------------------------
-- bf.compare(A,logical,B) --
-----------------------------
function bf.compare(A,logical,B)
if (logical~="<" and
logical~=">" and
logical~="<=" and
logical~=">=" and
logical~="==" and
logical~="~=") then
error("second input input must be a logical operator written into a string")
end
local out = {}
local ind = {}
local count = 0
if type(B)=="number" then
if type(A[1])=="table" then
for i=1,#A do
out[i] = {}
for j=1,#A[1] do
loadstring("cond ="..A[i][j]..logical..B)()
if cond then
out[i][j] = true
count = count+1
ind[count] = (i-1)*#A[1]+j
else
out[i][j] = false
end
end
end
elseif type(A[1])=="number" then
for j=1,#A do
loadstring("cond ="..A[j]..logical..B)()
if cond then
out[j] = true
count = count+1
ind[count] = j
else
out[j] = false
end
end
end
else
if (type(A[1])=="table" and type(B[1])=="table") then
if (#A==#B and #A[1]==#B[1]) then
for i=1,#A do
out[i] = {}
for j=1,#A[1] do
loadstring("cond ="..A[i][j]..logical..B[i][j])()
if cond then
out[i][j] = true
count = count+1
ind[count] = (i-1)*#A[1]+j
else
out[i][j] = false
end
end
end
else
error("The comparison can be done between "..
"two matrix with same dimension "..
"or between a matrix with a scalar value")
end
elseif (type(A[1])=="number" and type(B[1])=="number") then
if (#A==#B) then
for j=1,#A do
loadstring("cond ="..A[j]..logical..B[j])()
if cond then
out[j] = true
count = count+1
ind[count] = j
else
out[j] = false
end
end
else
error("Comparison between "..
"two vector with different dimension")
end
else
error("The comparison can be done between "..
"two matrix with same dimension "..
"or between a matrix with a scalar value")
end
end
return setmetatable(out,mt)--,ind
end
------------------------
-- metamethod setting --
------------------------
mt.__tostring = bf.tostring
mt.__lt = function(A,B) return bf.compare(A,"<",B) end
--------------------------
-- use of my metamethod --
--------------------------
A = bf.new{{1,2,3,4},{5,6,7,8},{9,10,11,12}}
B = bf.new{{3,6,1,8},{1,3,87,20},{11,2,5,7}}
C1 = bf.compare(A,"<",B)
C2 = A<B
print("What I want")
print(C1)
print("What I get")
print(C2)
What I want
true true false true
false false true true
true false false false
What I get
true
>Exit code: 0
最佳答案
Lua manual声明 __lt metamethod
的伪代码:
function lt_event (op1, op2)
if type(op1) == "number" and type(op2) == "number" then
return op1 < op2 -- numeric comparison
elseif type(op1) == "string" and type(op2) == "string" then
return op1 < op2 -- lexicographic comparison
else
local h = getbinhandler(op1, op2, "__lt")
if h then
return not not h(op1, op2)
else
error(···)
end
end
end
return not not h(op1, op2)
仅返回处理程序返回的单个(第一个)值
h
,如
not
是一元运算符。作为第二个效果,它将处理程序输出转换为标量:
not {} == false
, 和
not false == true
.
function myFun(A)
local out=A
out[1]='bar'
return out
end
A={'foo',1,2,3}
B=myFun(A)
print(table.concat(B,', ')) -- OK
print(table.concat(A,', ')) -- A also changed, because:
print(A,B) -- they are the same table!
关于Lua:__lt 元方法设置无法正常工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8605265/
这是我对这个令人难以置信的社区的第一个问题。 这些天来,我正在为自己编写一个 Lua 模块。这是有问题的代码的最小部分 mt = {} bf = {} ------------ -- bf.new -
在元表中重载 __eq、__lt 和 __le 总是将返回值转换为 bool 值。 有没有办法访问实际的返回值? 这将在下面的小 lua 脚本中用于为参数创建表达式树 用法: print(_.a +
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