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如果push和pop同时发生，这个实现是不安全的。 #1

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fanjing-whu opened this issue Mar 9, 2021 · 1 comment
Open

如果push和pop同时发生，这个实现是不安全的。 #1

fanjing-whu opened this issue Mar 9, 2021 · 1 comment

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@fanjing-whu
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push1start
这里top = 0
pop1start
这里top = -1
push2start
这里top = 0
push1modify 这里修改的是top = 0
push1end
push2modify 这里修改的也是top = 0
push2end
pop1read 这里读取到的数据是push2写入的错误数据（也可能是其他情况，总之是不安全的数据）
@fanjing-whu
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栈的实现

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