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我已经为填充 oracle CTF 编写了这个小函数，但在此过程中迷路了，有人可以帮忙吗？

我正在尝试从后面逐字节更新字节串。但是，当我运行它时，它是从前面的 newbyte 执行的：看起来像 b"x00/x00..." iv[0:x] 不会以某种方式变小

def test_new_bytes(stringX, stringRest, c0, x):

    print("iv 0:x", iv[0 : x - 2])

    if x > 1:
        for i in range(0, 256):

            if i < 16:
                HEX_I = hex(i)[2:]
                #   print(HEX_I)
                newbyte = bytes.fromhex("0" + HEX_I) + stringRest
                print(i, newbyte)

                iv_new = stringX + newbyte
                print("check", stringX, newbyte, stringRest)
                print(iv_new)
                if checkPadding(xor(iv_new, c0) + c0):
                    print("Correckt padding woohoo", xor(iv_new, c0))
                    test_new_bytes(iv[0 : x - 1], newbyte, c0, x=x - 1)
                    print("your i is ", i)

            #   return iv_new

            else:
                HEX_I = hex(i)[2:]
                #   print(HEX_I)
                newbyte = bytes.fromhex(HEX_I) + stringRest
                print(i, newbyte)
                iv_new = stringX + newbyte
                print(iv_new)
                if checkPadding(xor(iv_new, c0) + c0):
                    print(
                        "Correckt padding woohoo at i, x", i, x, xor(iv_new, c0)
                    )
                    print("check", stringX, newbyte)
                    test_new_bytes(iv[0 : x - 1], newbyte, c0, x=x - 1)
                    print("your i is ", i)  #   return iv_new


print(iv[0:16], b"", c0)
test_new_bytes(iv[0:15], b"", c0, 15)
test_new_bytes(c0[0:15], b"", c1, 15)
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