Intel syntax - Meaning of jmp/call instruction with $+5 operand

Question

I'm currently parsing a lot of assembly files and don't understand a specific jmp or call with $+5 as operand:

call $+5
 jmp $+5

To provide more context I grepped some of the occurrences:

mov esp, [ebp+ms_exc.old_esp]
and [ebp+ms_exc.registration.TryLevel], 0
or [ebp+ms_exc.registration.TryLevel], 0FFFFFFFFh
call $+5
jmp sub_4493CA
===== S U B R O U T I N E =======================================
push esi

[...]

mov esp, [ebp+ms_exc.old_esp]
and [ebp+ms_exc.registration.TryLevel], 0
or [ebp+ms_exc.registration.TryLevel], 0FFFFFFFFh
call $+5
jmp sub_45746A
===== S U B R O U T I N E =======================================
mov eax, dword_4778F8

[...]

mov eax, ebx
test al, 2
jnz loc_100994B8
jmp $+5
-----------------------------------------------------------------
mov eax, [ebp+var_34]
mov [ebp+var_40], eax

What is the meaning of the $+5 operand?

Answer 1

opcode for call $+5 is e8 00000000 so it calls the next instruction
opcode for jmp $+5 is e9 00000000 so it jumps to the next insturction

76E95FE0                        E8 00000000 CALL    76E95FE5         ;  <ntdll.call here>
76E95FE5 <ntdll.call here>      00          DB      00
76E95FE6                        E9 00000000 JMP     76E95FEB         ;  <ntdll.jmp_here>
76E95FEB <ntdll.jmp_here>       00          DB      00
76E95FEC                        EB 02       JMP     SHORT 76E95FF0   ;  <ntdll.jmp+4>
76E95FEE                        00          DB      00
76E95FEF                        00          DB      00
76E95FF0 <ntdll.jmp+4>          00          DB      00

Answer 2

$ = Current position (beginning of the instruction)

+5 = +5 bytes from the beginning of the instruction

jmp $+5 = jmp 3 bytes past the jmp instruction (short jmp takes 2 bytes + 3 bytes past that)