如何为增量序列号制作例程

我假设它以十六进制递增，因此可以使用以下BT-Basic代码。
dim SN $（0:15）+ *！Initializae序列号字符串数组* + S $ =“302QA125TT06”+ *！
为I = 0到15 + *定义序列号前缀* +！
启动for循环* + SN $（I）= S $＆amp;
ith $（I）+ *！
连接S和I * +的整数到十六进制值打印SN $（I）+ *！
打印阵列* +下一个I + *！
for循环结束* +结束+ *！
程序结束（如果一部分的subrutine应当subend）* +该程序的输出是：302QA125TT060 302QA125TT061 302QA125TT062 302QA125TT063 302QA125TT064 302QA125TT065 302QA125TT066 302QA125TT067 302QA125TT068 302QA125TT069 302QA125TT06A 302QA125TT06B 302QA125TT06C 302QA125TT06D 302QA125TT06E 302QA125TT06F对不起由评论上的每一行生成的扰频
码。
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     以下为原文

  I assume that it increments in hexadecimal so the following BT-Basic code can be used.

dim SN$(0:15)                       +*!Initializae Serial Number String Array*+
S$="302QA125TT06"           +*! Define Serial Number prefix*+
for I = 0 to 15                        +*! Initiate a for loop*+
SN$(I) = S$ & ith$(I)               +*! Concatenate S and the Integer-to-Hex value of I*+
print SN$(I)                           +*! Print the array*+
next I                                     +*! end of for loop*+
end                                        +*! end of program (if part a subrutine should be subend)*+

The output of this program is:


302QA125TT060
302QA125TT061
302QA125TT062
302QA125TT063
302QA125TT064
302QA125TT065
302QA125TT066
302QA125TT067
302QA125TT068
302QA125TT069
302QA125TT06A
302QA125TT06B
302QA125TT06C
302QA125TT06D
302QA125TT06E
302QA125TT06F

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