递归函数可以通过内联？ [英] Can a recursive function be inline?

问题描述

inline int factorial(int n)
{
    if(!n) return 1;
    else return n*factorial(n-1);
}

当我在读这，发现上面code会导致无限汇编如果不通过编译器正确处理。

As I was reading this, found that the above code would lead to "infinite compilation" if not handled by compiler correctly.

编译器如何决定是否内联函数或不？

How does the compiler decide whether to inline a function or not ?

推荐答案

首先，在一个函数在线规范只是一个提示。编译器可以（而且经常）完全忽略了在线预选赛的presence与否。随着中说，一个编译器的可以的内联递归函数，就像它可以展开一个无限循环。它只是放置一个上限水平，它会展开功能。

First, the inline specification on a function is just a hint. The compiler can (and often does) completely ignore the presence or absence of an inline qualifier. With that said, a compiler can inline a recursive function, much as it can unroll an infinite loop. It simply has to place a limit on the level to which it will "unroll" the function.

这是优化编译器可能会将这个code：

An optimizing compiler might turn this code:

inline int factorial(int n)
{
    if (n <= 1)
    {
        return 1;
    }
    else
    {
        return n * factorial(n - 1);
    }
}

int f(int x)
{
    return factorial(x);
}

这个code：

int factorial(int n)
{
    if (n <= 1)
    {
        return 1;
    }
    else
    {
        return n * factorial(n - 1);
    }
}

int f(int x)
{
    if (x <= 1)
    {
        return 1;
    }
    else
    {
        int x2 = x - 1;
        if (x2 <= 1)
        {
            return x * 1;
        }
        else
        {
            int x3 = x2 - 1;
            if (x3 <= 1)
            {
                return x * x2 * 1;
            }
            else
            {
                return x * x2 * x3 * factorial(x3 - 1);
            }
        }
    }
}

在这种情况下，我们基本上是内联函数的3倍。一些编译器的不的执行该优化。我记得MSVC ++有一个设置调整，将在递归函数（最多20个，我相信）进行内联的水平。

In this case, we've basically inlined the function 3 times. Some compilers do perform this optimization. I recall MSVC++ having a setting to tune the level of inlining that would be performed on recursive functions (up to 20, I believe).

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