问题描述
这是我炸弹圈问题的汇编代码，我被困在phase2中；
炸弹实验室要求我们根据汇编代码找出正确的输入，否则它将爆炸。
This is the assembly code from my bomb lap question, I am stuck in phase2;The bomb lab require us to find out the correct input based on assembly code or it will exploded.
来自< +20>我知道％rbp -0x30（48）== 0否则它将调用< +32>并炸弹；因此％rbp = 48（DEC）
From <+20> I know that %rbp -0x30(48) == 0 or it will call <+32> and explode the bomb; so %rbp = 48(DEC)
之后（+26）％rbp-0x2c（44）必须等于1否则它将炸弹爆炸...
但由于％rbp = 48，炸弹会在任何地方爆炸，所以我现在感到困惑...
After that(+26) %rbp - 0x2c(44) must equal 1 or it will explode the bomb...But since %rbp = 48, the bomb will explode anywhere so I am confuse now...
我认为我误解了compl，je / jne或如何计算这些东西...
I think I misunderstand the compl , je/jne or how to calculate these things...
推荐答案
-0x30（％ebp）并不意味着使用值％ebp-0x30 。这是要读取的内存地址。指令（ cmpl ）的后缀为 l ，因此它处理的是4个字节的数量。因此，实际上发生的是它从地址％ebp-0x30 中读取了一个4字节的数字，并检查它是否为零。
-0x30(%ebp) doesn't mean to use the value %ebp - 0x30. It's a memory address to read from. The instruction (cmpl) has an l suffix, so it's dealing with a 4 byte quantity. So what's actually happening is that it reads a 4 byte number from the address %ebp - 0x30 and checks whether it's zero.
（前缀 $ 表示它是立即数，而不是地址。这就是为什么 0x0 照字面意义使用并且未取消引用。）
(The $ prefix means it's an immediate value, not an address. This is why 0x0 is taken literally and not dereferenced.)
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