[Bpf] 64-bit immediate instructions clarification

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The spec defines:
> As discussed below in `64-bit immediate instructions`_, a 64-bit immediate
> instruction uses a 64-bit immediate value that is constructed as follows.
> The 64 bits following the basic instruction contain a pseudo instruction
> using the same format but with opcode, dst_reg, src_reg, and offset all set to zero,
> and imm containing the high 32 bits of the immediate value.
[...]
> imm64 = (next_imm << 32) | imm

The 64-bit immediate instructions section then says:
> Instructions with the ``BPF_IMM`` 'mode' modifier use the wide instruction
> encoding defined in `Instruction encoding`_, and use the 'src' field of the
> basic instruction to hold an opcode subtype.

Some instructions then nicely state how to use the full 64 bit immediate value, such as
> BPF_IMM | BPF_DW | BPF_LD  0x18    0x0  dst = imm64                                integer      integer
> BPF_IMM | BPF_DW | BPF_LD  0x18    0x2  dst = map_val(map_by_fd(imm)) + next_imm   map fd       data pointer
> BPF_IMM | BPF_DW | BPF_LD  0x18    0x6  dst = map_val(map_by_idx(imm)) + next_imm  map index    data pointer

Others don't:
> BPF_IMM | BPF_DW | BPF_LD  0x18    0x1  dst = map_by_fd(imm)                       map fd       map
> BPF_IMM | BPF_DW | BPF_LD  0x18    0x3  dst = var_addr(imm)                        variable id  data pointer
> BPF_IMM | BPF_DW | BPF_LD  0x18    0x4  dst = code_addr(imm)                       integer      code pointer
> BPF_IMM | BPF_DW | BPF_LD  0x18    0x5  dst = map_by_idx(imm)                      map index    map

How is next_imm used in those four?  Must it be 0?  Or can it be anything and it's ignored?
Or is it used for something?

Dave