jmp $+5, why the jump

Question

I am looking at some code that has jmp $+5 instruction at the beginning of the entry code in a library, right after pushing registers to the stack. I understand that jmp $+5 puts IP at the next instruction but why is the jump there at all? Without it, the next instruction would have been executed anyway. The only thing that comes to mind is that it is PIE related and jump with displacement of 0 that could potentially be a non-zero value, but this code is a part of a big "blob" and I don't think its individual parts would be relocatable, at least not to a point where the jump displacement would be non-zero.

The compiler is GCC 3.7?, linux 32 bit

Version needs section '.gnu.version_r' contains 2 entries:
 Addr: 0x0000000000017e0c  Offset: 0x017e0c  Link: 3 (.dynstr)
  000000: Version: 1  File: libgcc_s.so.1  Cnt: 1
  0x0010:   Name: GCC_3.0  Flags: none  Version: 7

Thanks

Answer 1

Are you sure you encounter JMP $+5? A common x86 function prologue pattern has CALL $+5 in the position you describe. In addition to continuing at the next instruction, it also pushes the address of the next instruction to the stack. It is usually followed by POP EBX to put that address into ebx and then ADD EBX, $const to modify EBX so it contains a specific reference address, IIRC the beginning of the function slots in the GOT.

The important point about this instruction sequence is that it works independently of the base address of the code, as JMP NEAR (opcode E9) and CALL NEAR contain relative addresses, so $+5 is always encoded the same way, no matter where the instruction is located. The add instruction then adds/subtracts the fixed distance between that instruction and the GOT.