高手能否解释一下以下代码：
string hex_data[]
integer cc
integer pn
integer e
integer strlen
integer d
integer s
ulong resnum1
ulong resnum2
ulong resnum11
ulong resnum12
ulong resnum3
ulong resnum5
ulong resnum6
long slen2
long slen3
long slen4
long slen5
long fs_len
string s1
string s2
string s3
string slast4
string sleft
string vslast4
string resnum4
string resnum7
string resnum8
string st1 = ""
string uls1
string uls2
string uls3
string disksn = ""
char fsn[9]
char fcn1[12]
char fcn2[12]
char fcn3[12]
integer sgl[10]
integer jj
integer kk

cc = len(is_hex)                               
is_hex = left(is_hex, cc - 3)                       
pn = len(is_hex)                               

do
    hex_data[e] = mid(is_hex, pn, 1)               

    if isnumber(hex_data[e]) = false then

        choose case hex_data[e]
            case "A"
                hex_data[e] = "10"
            case "C"
                hex_data[e] = "11"
            case "X"
                hex_data[e] = "12"
            case "M"
                hex_data[e] = "13"
            case "H"
                hex_data[e] = "14"
            case "R"
                hex_data[e] = "15"
        end choose

    end if

    resnum1 = resnum1 + long(hex_data[e]) * 16 ^ e
    pn --
    e ++
loop until pn = 0

resnum11 = resnum1 - 31650758
resnum2 = resnum1 - 42198964
resnum12 = resnum11 + resnum2
s1 = string(resnum11)
s2 = string(resnum12)
s3 = string(resnum2)
strlen = len(s3)
slast4 = right(s3, 4)
vslast4 = reverse(slast4)
sleft = left(s3, strlen - 4)
resnum3 = long(sleft) / 2
resnum4 = string(resnum3) + vslast4
resnum5 = long(resnum4) / 2
slen2 = len(string(resnum2))
slen3 = len(string(resnum3))
slen4 = len(resnum4)
slen5 = len(string(resnum5))
uls1 = string(resnum3)
uls2 = resnum4
uls3 = string(resnum5)
jj = slen3 - 1

for s = 1 to slen3
    fcn1[s] = mid(uls1, s, 1)
next

for s = 1 to slen4
    fcn2[s] = mid(uls2, s, 1)
next

for s = 1 to slen5
    fcn3[s] = mid(uls3, s, 1)
next

sgl[slen3] = 15 - integer(fcn1[slen3])

for kk = 1 to slen3
    sgl[kk] = 15 - integer(fcn1[kk])
next

kk = kk - 1

for s = 1 to kk
    st1 = st1 + string(sgl[s])
next

resnum6 = (resnum5 + 3543987) / 3 - 1.53968E+007
resnum7 = string(resnum6)
resnum8 = mid(resnum7, 2)
fs_len = len(resnum8)

for d = 1 to fs_len
    fsn[d] = endsir[integer(mid(resnum8, d, 1))]
next

for d = 1 to fs_len
    disksn = disksn + fsn[d]
next

return disksn

[招生]科锐逆向工程师培训(2024年11月15日实地，远程教学同时开班, 第51期)