Avoid shrinking stacks to often
Shrink a stack only when the final stack size can be at most 2/3 the previous size with half of its entries empty. This commit also improves the clarity of 'luaD_growstack'.
This commit is contained in:
@@ -2,7 +2,7 @@
|
||||
-- See Copyright Notice in file all.lua
|
||||
|
||||
|
||||
print"testing C-stack overflow detection"
|
||||
print"testing stack overflow detection"
|
||||
|
||||
-- Segmentation faults in these tests probably result from a C-stack
|
||||
-- overflow. To avoid these errors, you should set a smaller limit for
|
||||
@@ -98,4 +98,52 @@ do
|
||||
print("final count: ", count)
|
||||
end
|
||||
|
||||
|
||||
if T then
|
||||
print("testing stack recovery")
|
||||
local N = 0 -- trace number of calls
|
||||
local LIM = -1 -- will store N just before stack overflow
|
||||
|
||||
-- trace stack size; after stack overflow, it should be
|
||||
-- the maximum allowed stack size.
|
||||
local stack1
|
||||
local dummy
|
||||
|
||||
local function err(msg)
|
||||
assert(string.find(msg, "stack overflow"))
|
||||
local _, stacknow = T.stacklevel()
|
||||
assert(stacknow == stack1 + 200)
|
||||
end
|
||||
|
||||
-- When LIM==-1, the 'if' is not executed, so this function only
|
||||
-- counts and stores the stack limits up to overflow. Then, LIM
|
||||
-- becomes N, and then the 'if' code is run when the stack is
|
||||
-- full. Then, there is a stack overflow inside 'xpcall', after which
|
||||
-- the stack must have been restored back to its maximum normal size.
|
||||
local function f()
|
||||
dummy, stack1 = T.stacklevel()
|
||||
if N == LIM then
|
||||
xpcall(f, err)
|
||||
local _, stacknow = T.stacklevel()
|
||||
assert(stacknow == stack1)
|
||||
return
|
||||
end
|
||||
N = N + 1
|
||||
f()
|
||||
end
|
||||
|
||||
local topB, sizeB -- top and size Before overflow
|
||||
local topA, sizeA -- top and size After overflow
|
||||
topB, sizeB = T.stacklevel()
|
||||
xpcall(f, err)
|
||||
topA, sizeA = T.stacklevel()
|
||||
-- sizes should be comparable
|
||||
assert(topA == topB and sizeA < sizeB * 2)
|
||||
print(string.format("maximum stack size: %d", stack1))
|
||||
LIM = N -- will stop recursion at maximum level
|
||||
N = 0 -- to count again
|
||||
f()
|
||||
print"+"
|
||||
end
|
||||
|
||||
print'OK'
|
||||
|
||||
Reference in New Issue
Block a user