看看这段有点傻的代码。
3.times do |round|
cell = uninitialized Int32
if cell == 60
puts "cell = 60, round = #{round}"
puts " (by chance!)" if round == 0
end
cell = 60
end是每次迭代都重新创建 cell，碰巧放在上一次迭代中 cell 使用的地址上，所以第三次经过时值才是 60？
还是同一个变量一直存在于全部三次迭代中，因此我可以依赖这种行为？
Consider this rather silly snippet:
3.times do |round|
cell = uninitialized Int32
if cell == 60
puts "cell = 60, round = #{round}"
puts " (by chance!)" if round == 0
end
cell = 60
endDoes every iteration create cell again, coincidentally placing it at the address used by cell on the preceding iteration (which would explain its value of 60 on the third pass)?
Or does one variable persist across all three iterations, so that I can rely on that?
使用 uninitialized 时，变量指向的内存里原先有什么数据，就还是什么数据。这正是 uninitialized 存在的原因，它既不会清空那块内存，也不会赋上初始默认值。
这里的 cell 碰巧在每次迭代中都处于相同地址。栈的工作方式可能让这种行为一直能被观察到，但依赖它是错误的。并不是同一个变量在各次迭代之间一直存在。
编译器或代码生成器如果愿意，可以重新安排数据的位置。
With uninitialized, the variable refers to memory containing whichever data happens to be there already. That's precisely why uninitialized exists: it neither clears that memory nor assigns an initial default value.
Here, cell coincidentally resides at an identical address on every iteration. The way the stack operates probably makes this behaviour consistently observable, but depending on it would be a mistake. A single variable isn't persisting from one iteration to another.
If it chooses, the compiler / code generator may rearrange where things live.
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