CODEWITHSUNDEEP
assemblyx86disassembly
Aman
Aman

Reputation: 783

How is this instruction calculating jump destination?

I am trying to understand what this line is doing

   0x0000000000400f7e <+59>:    jmpq   *0x401be0(,%rax,8)

Let us assume $rax = 2. So we have *0x401be0+(8*2).

*0x401be0 points to 4198277. So we have 4198277+16 = 4198293 or 0x400F95.

So I am expecting the code to jump to instruction at 0x400F95. But it jumps to 0x400fc9. What is actually happening?

Upvotes: 1

Views: 1350

Answers (1)

R.. GitHub STOP HELPING ICE
R.. GitHub STOP HELPING ICE

Reputation: 215221

The code reads a 64-bit value (a code address) from the data at address 0x401be0+8*rax and jumps to the address read.

Upvotes: 2

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