Need help understanding E8 asm call instruction x86

Question

I need a helping hand in order to understand the following assembly instruction. It seems to me that I am calling a address at someUnknownValue += 20994A?

E8 32F6FFFF - call std::_Init_locks::operator=+20994A

Solution

Whatever you're using to obtain the disassembly is trying to be helpful, by giving the target of the call as an offset from some symbol that it knows about -- but given that the offset is so large, it's probably confused.

The actual target of the call can be calculated as follows:

• E8 is a call with a relative offset.
• In a 32-bit code segment, the offset is specified as a signed 32-bit value.
• This value is in little-endian byte order.
• The offset is measured from the address of the following instruction.

e.g.

<some address>       E8 32 F6 FF FF         call <somewhere>
<some address>+5     (next instruction)

• The offset is 0xFFFFF632.
• Interpreted as a signed 32-bit value, this is -0x9CE.
• The call instruction is at <some address> and is 5 bytes long; the next instruction is at <some address> + 5.
• So the target address of the call is <some address> + 5 - 0x9CE.

Comments

[Michael]

Tank you so much. Your example is spot on!

[Rafa]

@Matthew could the call instruction be more than 5 bytes long? (In a x86 archi. can the next instric. be at <some address> + 6)? In what case?

[Vladislav Ivanishin]

@Rafa, call relative offset instruction is 5 bytes, because max relative offset must fit in 4 bytes. If the target is farther than 2**31 bytes away, mov reg, imm64; call reg is used.