Question

使用原始代码，我一直在Error: Parse error: [expr level ;] expected after "in" (in [expr])

上获得let numDigits = numDigits - 1 in

原文：

let rec rev_int num =
  if num / 10 == 0 then
     num
  else
    let temp = num mod 10 in

    let numDigits = String.length(string_of_int num) - 1 in

    if num < 0 then
      let numDigits = numDigits - 1 in
    else
      let numDigits = numDigits + 0 in

    let num = (num - temp) / 10 in
    temp * int_of_float(10.0 ** float_of_int numDigits) + rev_int num

有以下各种变体：

if num < 0 then
   let numDigits = numDigits - 1 in;
else
   let numDigits = numDigits + 0 in;

if num < 0 then
   let numDigits = numDigits - 1 in
else begin
   let numDigits = numDigits + 0 in end

我修改了代码，现在它可以工作，但我想知道是否有办法用嵌套的if和更少的冗余来做。

修订：

let rec rev_int num =
  if num / 10 == 0 then
    num
  else
    let temp = num mod 10 in

    let numDigits = String.length(string_of_int num) - 1 in

    if num < 0 then
      let numDigits = numDigits - 1 in
      let num = (num - temp) / 10 in
      temp * int_of_float(10.0 ** float_of_int numDigits) + rev_int num
    else
      let numDigits = numDigits + 0 in
      let num = (num - temp) / 10 in
      temp * int_of_float(10.0 ** float_of_int numDigits) + rev_int num

Answer 1

来自Hunan Rostomyan

let rec rev_int num =
  if num / 10 == 0 then
    num
  else
    let temp = num mod 10 in

    let numDigits = String.length(string_of_int num) - 1 in

    let numDigits = numDigits - (if num < 0 then 1 else 0) in
    let num = (num - temp) / 10 in
    temp * int_of_float(10.0 ** float_of_int numDigits) + rev_int num

OCaml的问题嵌套if

1 个答案: