What does an asterisk * before an address mean in x86-64 AT&T assembly?

Question

What does the following line mean:

...
401147: ff 24 c5 80 26 40 00    jmpq   *0x402680(,%rax,8)
...

What does the asterisk in front of the memory address mean? Also, what does it mean when the memory access method is missing it's first register value?

Usually its something like ("%register", %rax, 8), but in this case it doesn't have the first register.

Any tips?

Solution 1

It's AT&T assembly syntax:

• source comes before destination
• mnemonic suffixes indicate the size of the operands (q for quad, etc.)
• registers are prefixed with % and immediate values with $
• effective addresses are in the form DISP(BASE, INDEX, SCALE) (DISP + BASE + INDEX * SCALE)
• Indirect jump/call operands indicated with * (as opposed to direct).

So, you have a jmpq for jumping to the absolute address which is stored in %rax * 8 + 0x402680, and is a quad word long.

---

AT&T syntax needed a way to distinguish RIP = foo (jmp foo) from RIP = load from some symbol address (jmp *foo). Remember that movl $1, foo is a store to the absolute address foo.

With other addressing modes, there's no ambiguity between what kind of jump / call you're doing, anything other than a bare label must be indirect. (GAS will infer that and warn about an indirect jump without * if you do jmp %rax or jmp 24(%rax) or anything other than a bare symbol name.)

(In 64-bit mode you'd normally actually use jmp *foo(%rip) to load a global variable into RIP, not use a 32-bit absolute address like jmp *foo. But the possibility exists, and before x86-64 when AT&T syntax was designed, was the normal way to do things.)

Solution 2

Actually this is computed table jmp, where the 0x402680 is address of tabele and rax is index of 8 byte (qword) pointer.

Solution 3

Getting things into Intel syntax always makes stuff clearer:

FF24C5 80264000  JMP QWORD PTR [RAX*8+402680]

Solution 4

It's a jump to an address contained in memory. The address is stored in memory at address rax*8+0x402680, where rax is the current rax value (when this instruction executes).

Solution 5

jmpq is just a un-conditional jump to a given address. The 'q' means that we're dealing with quad words (64 bits long).

*0x402680(,%rax,8) : This is a way to write an address in x-86 assembly. You are correct in saying that usually there is a register before the first comma, but you still follow the same rules if no register is specified.

The format works this way : D(reg1, reg2, scalingFactor) where D stands for displacement. Displacement is basically just an integer. reg1 is the first or base register. reg2 is the second register and scalingFactor is one of 2, 4, 8 (maybe even 1, but I'm not sure about that). Now, you can obtain your address by simply adding the values in this way: Displacement + (value at reg1) + scalingFactor*(value at reg2).

I'm not completely sure as to what the asterisk in front of the address is for, but my guess is that it means that the displacement value is stored at that address.

Hope this helps.

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