OCaml字节码的可移植性(Portability of OCaml bytecode)

 根据规范 ，在"12345" >> 0 ，会发生这种情况： 
 
 
 
 "12345"通过规范的抽象ToInt32函数 ： 
  
 
   
 ToInt32通过ToNumber运行它以将字符串强制转换为标准JavaScript号（IEEE-754双精度二进制浮点值）; 它最终成为12345 
 
   
 ToInt32然后将其转换为32位整数值; 它最终成为...... 12345 
 
  
 
 
 这个32位整数被移位......完全没有，因为你使用了>> 0 ，所以我们仍然有12345 
 
 
 
 
 并将返回一个int 
 
 
 有点。 如果将结果存储在变量中，它最终将再次成为标准JavaScript编号（双精度）。 如果将这个操作与其他操作相结合，那么规范似乎允许将int值原样使用，这是有道理的。 
 
 如果您的目标只是使用默认的强制规则强制"12345"为数字，则>> 0是一种相当模糊且有些低效的方法（不是后者可能很重要）。 一元+更直接并适用相同的规则： +"12345" 

According to the specification, in "12345" >> 0, this happens:
 
 
 
"12345" is put through the specification's abstract ToInt32 function: 
  
 
   
ToInt32 runs it through ToNumber to coerce the string to a standard JavaScript number (an IEEE-754 double-precision binary floating-point value); it ends up being 12345
 
   
ToInt32 then converts it to a 32-bit integer value; it ends up being...12345
 
  
 
 
That 32-bit integer is shifted...not at all because you used >> 0, so we still have 12345
 
 
 
 
and will return a int
 
 
Sort of. If you stored the result in a variable, it would end up being a standard JavaScript number again (a double). The spec seems to allow for the possibility the int value would be used as-is if you combined this operation with others, though, which makes sense.
 
If your goal is just to force "12345" to be a number using the default coercion rules, >> 0 is a fairly obscure and somewhat inefficient way to do it (not that the latter is likely to matter). The unary + is rather more direct and applies the same rules: +"12345"