带有静态变量的递归函数 [英] Recursive function with static variable

问题描述

我有一个带有静态变量count"的递归函数.该函数递归地递增计数，并且由于它具有文件作用域，当我第二次调用 foo() 时，计数仍然等于 5.是否有一种技术可以在第二次调用 foo() 之前将计数重置为 0?

基本上，我不希望 count 具有文件范围，但我希望它通过不同的迭代保留其值.

我能想到的一种方法是在 foo() 中有一个参数来初始化 foo().如 foo(int count).但是还有别的方法吗?

#include 使用命名空间标准；空 foo(){静态整数计数 = 0;如果(计数<5){计数++；cout<<计数<5"<

解决方案

更惯用的方式是将其拆分为两个功能:

void foo() {foo_recursive(0);}void foo_recursive(int count) {如果(计数＜5){计数++；cout<<计数<<结束;foo_recursive(计数)；} 别的 {cout<<计数＞5"＜＜结束;}}

这样做的好处是不需要调用者为 foo() 提供参数，而且你也不需要静态变量(我一直觉得这是个坏主意).

I have a recursive function with a static variable "count". The function increments count recursively and since it has file scope, when I call foo() a second time, count is still equal to 5. Is there a technique to reset count to 0 before the second time foo() is called?

Basically, I don't want count to have file scope but I want it to retain its value through different iterations.

One way I can think of doing it is have an argument in foo() to initialize foo(). Such as foo(int count). But is there another way?

#include <iostream>

using namespace std;

void foo()
{
    static int count = 0;

    if(count<5)
    {
        count++;
        cout<<count<<endl;
        foo();
    }
    else
    {
        cout<<"count > 5"<<endl;
    }
}

int main()
{
    foo();  //increment count from 0 to 5
    foo();  //count is already at 5

    return 0;
}

解决方案

A more idiomatic way is to split it into two functions:

void foo() {
   foo_recursive(0);
}

void foo_recursive(int count) {
    if (count < 5) {
        count++;
        cout << count << endl;
        foo_recursive(count);
    } else {
        cout << "count > 5" << endl;
    }
}

Which has the benefit of not requiring the caller to supply an argument to foo() and also you don't need a static variable (which I always feel is a bad idea).

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