递归函数中的返回类型推导 [英] Return type deduction in recursive function

问题描述

以下代码编译：

auto foo(int i) {
  if( i == 1 )
    return i;
  else 
    return foo(i-1)+i ; 
}

跟随没有， _{c ++ 1y}

While following doesn't, _c++1y

auto foo(int i) {
  return (i == 1) ? i : foo(i-1)+i ;  
}

---

为什么不能编译推导第二种情况下的返回类型？我在这想念什么吗？

---

Why can't compiler deduce the return type in second case ? Am I missing something over here ?

在第二种情况下，我知道（i == 1）之后有一个序列点，但这不应该会影响编译，对吗？

I know there's a sequence point after (i == 1) in second scenario, but that shouldn't be affecting compilation, right ?

推荐答案

由于此规则，第一个有效，即最新草案

The first works because of this rule, 7.1.6.4/11 of the latest draft

在函数中可以看到c $ c> return 语句，但是从该语句推导的返回类型
可以在函数的其余部分中使用，包括在其他<$ c $中c> return 语句。

Once a return statement has been seen in a function, however, the return type deduced from that statement can be used in the rest of the function, including in other return statements.

因此，将返回类型推导为 int 从第一个 return 语句开始；只需检查第二个，以确保它也给出 int ，并假定递归调用确实给出了。

So the return type is deduced as int from the first return statement; the second is just checked to make sure that it also gives int, assuming that the recursive call does.

第二个不编译，因为表达式的类型取决于返回类型；因此无法推断出类型。

The second doesn't compile because the type of the expression depends on the return type; so the type can't be deduced.

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