sml - 读取 SML 中的命令行参数

Question

我正在尝试读取我的输入文件的名称 argv[1] 。这是我到目前为止所做的:

val args = CommandLine.arguments() ;
val (x::y) =  args ;
val _ = agora x 

但我不断收到此错误消息:

uncaught exception Bind [nonexhaustive binding failure] .

有人可以帮忙吗？提前谢谢你!

Answer 1

这是编译器警告您不能确定绑定(bind)模式始终有效。

例如，给定以下程序:

val args = CommandLine.arguments ()
val (x::y) = args
val _ = print (x ^ "\n")

编译运行得到:

$ mosmlc args.sml
$ ./a.out Hello
Hello
$ ./a.out
Uncaught exception:
Bind

要安全地处理可变数量的命令行参数，您可以使用case-of:

fun main () =
    case CommandLine.arguments () of
         []     => print ("Too few arguments!\n")
       | [arg1] => print ("That's right! " ^ arg1 ^ "\n")
       | args   => print ("Too many arguments!\n")

val _ = main ()

编译运行得到:

$ mosmlc args2.sml
$ ./a.out 
Too few arguments!
$ ./a.out hello
That's right! hello
$ ./a.out hello world
Too many arguments!

旁注:C 的 argv[0] 等效于 CommandLine.name ()。