为什么将Int32转换为Float64会导致数据发生变化？

Question

我正在教自己CIL并且到目前为止一直做得很好(昨天真的开始)但我遇到了一个我无法弄清楚的问题.我正在提示用户输入一个int(int32),然后将其存储并转换为浮点并显示它.然而,无论我输入什么都是不同的浮动.这是我的代码:

.assembly variables {}
.method public static void main() cil managed
{
    .entrypoint
    .maxstack 8
    .locals init (float64)

    ldstr "Enter a digit: "
    call void [mscorlib]System.Console::WriteLine(string)
    call int32 [mscorlib]System.Console::Read()
    conv.r8
    stloc.0
    ldstr "as a float: "
    call void [mscorlib]System.Console::WriteLine(string)
    ldloc.0
    dup
    call void [mscorlib]System.Console::Write(float64)
    stloc.0
    ldstr "Stored in location 0"
    call void [mscorlib]System.Console::WriteLine(string)
    ldloc.0
    conv.i4
    call void [mscorlib]System.Console::WriteLine(int32)
    call int32 [mscorlib]System.Console::Read() // to pause before closing window
    pop
    ret
}

我只是在和CIL搞砸,但我觉得为了清晰起见,我会把整个例子都扔进去.它编译得很好但是当我输入5时它返回53作为浮点数和转换后的int32.

有人可以说明我做错了什么!

---

编辑:感谢Marc Gravell,我能够弄明白.对于那些感兴趣的人,这里是正确的代码:

.assembly variables {}
.method public static void main() cil managed
{
    .entrypoint
    .maxstack 8
    .locals init (float64)

    ldstr "Enter a digit: "
    call void [mscorlib]System.Console::WriteLine(string)
    call string [mscorlib]System.Console::ReadLine()
    call int32 [mscorlib]System.Int32::Parse(string)
    conv.r8
    stloc.0
    ldstr "as a float: "
    call void [mscorlib]System.Console::WriteLine(string)
    ldloc.0
    dup
    call void [mscorlib]System.Console::Write(float64)
    stloc.0
    ldstr "Stored in location 0"
    call void [mscorlib]System.Console::WriteLine(string)
    ldloc.0
    conv.i4
    call void [mscorlib]System.Console::WriteLine(int32)
    call int32 [mscorlib]System.Console::Read() // to pause before closing window
    pop
    ret
}

Answer 1

Console.Read返回Unicode代码点,或EOF返回-1.53是字符的代码点(不是整数)'5'.

你也许可以使用Console.ReadLine和int.Parse.