VB不尊重"和"1"之间的区别

Question

我们有数据,其中键是可以包含引号的字符串.问题是Visual Basic不尊重普通双引号和斜引号之间的区别.例如声明:

MessageBox.Show("""1""" = "””1””") 

返回True.请注意,我们从python程序生成非常大的脚本文件,并在VB脚本环境中运行它们.我们不是从文件中读取数据.我怎样才能让VB尊重两种报价之间的区别？

Answer 1

您正在使用VB.Net语言规范,该规范在代码语句中使用时将三个不同的双引号字符视为相同的字符.

来自字符串文字:

字符串文字是零个或多个Unicode字符的序列,以ASCII双引号字符,Unicode左双引号字符或Unicode右双引号字符开头和结尾.在字符串中,两个双引号字符的序列是表示字符串中双引号的转义序列.

StringLiteral
    : DoubleQuoteCharacter StringCharacter* DoubleQuoteCharacter
    ;

DoubleQuoteCharacter
    : '"'
    | '<unicode left double-quote 0x201c>'
    | '<unicode right double-quote 0x201D>'
    ;

StringCharacter
    : '<Any character except DoubleQuoteCharacter>'
    | DoubleQuoteCharacter DoubleQuoteCharacter
    ;

在上面引用的规范中,"ASCII双引号字符"的用法表示英寸字符或Chrw(34).

在VS2015之前,您甚至无法"""1""" = "””1””"在不自动转换为代码编辑器的情况下粘贴到代码编辑器中"""1""" = """1""".

如果需要构造包含Unicode双引号的代码语句,则需要使用各自的字符表示来构造它们.

Const ucDoubleLeftQuote As Char = ChrW(&H201C) ' "“"c
Const ucDoubleRightQuote As Char = ChrW(&H201D) ' "”"c
Const asciiDoubleQuote As Char = ChrW(34) ' """"c

Dim asciiQuoted As String = """1"""
Dim asciiQuotedv2 As String = asciiDoubleQuote & "1" & asciiDoubleQuote

Dim unicodeQuoted As String = ucDoubleLeftQuote & "1" & ucDoubleLeftQuote

MessageBox.Show((asciiQuoted = asciiQuotedv2).ToString()) ' yields true
MessageBox.Show((asciiQuoted = unicodeQuoted).ToString()) ' yields false

编辑:要演示VB编译器替换字符串文字中任何Unicode双引号的ASCII双引号,请考虑以下代码.

Module Module1
    Sub Main()
        T1("““ 1 ””") ' unicode quotation marks left and right
        T2(""" 1 """) ' ascii quotation mark
    End Sub
    Sub T1(s As String) ' dummy method to highlight unicode quotation mark
    End Sub
    Sub T2(s As String) ' dummy method to highlight asci quotation mark
    End Sub
End Module

在ILDASM中查看时,此代码将生成以下IL.

.method public static void  Main() cil managed
{
  .entrypoint
  .custom instance void [mscorlib]System.STAThreadAttribute::.ctor() = ( 01 00 00 00 ) 
  // Code size       24 (0x18)
  .maxstack  8
  IL_0000:  nop
  IL_0001:  ldstr      "\" 1 \""
  IL_0006:  call       void ConsoleApp1.Module1::T1(string)
  IL_000b:  nop
  IL_000c:  ldstr      "\" 1 \""
  IL_0011:  call       void ConsoleApp1.Module1::T2(string)
  IL_0016:  nop
  IL_0017:  ret
} // end of method Module1::Main

IL_0001: ldstr "\" 1 \""对应于调用语句的字符串加载:T1("““ 1 ””").

你可以看到,这与IL_000c: ldstr "\" 1 \""对应于调用语句的字符串加载相对应:T2(""" 1 """).