如何重置函数内的静态结构

Question

So let's say I have a function

struct coinTypes {
    int tenP = 0;
    int twentyP = 0;
    int fiftyP = 0;
};

coinTypes numberOfCoins(int coins)
{

    static coinTypes types;

    // incrementing structs values
}

Let's say I have used this function for some time, and values in coinTypes struct are no longer 0. Then I decided to use this function for another purpose and I need the values to be 0 again. Is there any way to reset coinTypes struct?

Answer 1

Dont use static in this case. Try to use: Class Coin with object params coinType, couinNumber for example.

Answer 2

Unless you are just misunderstanding what the keyword static does (you probably are), this is what you are asking for:

run online

#include <iostream>

struct britishCoins {
    int tenP;
    int twentyP;
    int fiftyP;
};

britishCoins& getCoins () {
    static britishCoins coins {0, 0, 0};
    return coins;
}

void resetCoins () {
    getCoins() = {0, 0, 0};
}

britishCoins numberOfCoins(int coins)
{
    britishCoins& result = getCoins();
    result.tenP += coins / 10;
    //...
    return result;
}

int main () {
    std::cout << numberOfCoins(0).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    resetCoins();
    std::cout << numberOfCoins(0).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    std::cout << numberOfCoins(10).tenP << std::endl;
    return 0;
}

Prints:

0
1
2
3
4
0
1
2
3
4

If you just want to convert int coins into britishCoins without storing it's value inside the function it's simple:

run online

#include <iostream>

struct britishCoins {
    int tenP;
    int twentyP;
    int fiftyP;
};

britishCoins numberOfCoins(int coins)
{
    britishCoins result;
    result.fiftyP = coins / 50;
    coins %= 50;
    result.twentyP = coins / 20;
    coins %= 20;
    result.tenP = coins / 10;
    return result;
}

int main () {
    for (int i = 0; i < 3; ++i) {
        britishCoins coins = numberOfCoins(130);
        std::cout << coins.fiftyP << "*50 + " << coins.twentyP << "*20 + " << coins.tenP << "*10" << std::endl;
    }
    return 0;
}

output:

2*50 + 1*20 + 1*10
2*50 + 1*20 + 1*10
2*50 + 1*20 + 1*10