c++ - 错误 C2664：show_info：无法将参数 2 从 'char [20]' 转换为 'char

Question

我有一个小结构：

struct price
{
    char name[20];
    char shop[20];
    int pr;
    price *next;
};

一个不起作用的功能：

void show_info(price *&head, char cur)
{
    bool found = 0;
    price *temp = new price;
    temp->name = cur;
    for (price *i=head; i!=NULL; i=i->next)
        if (temp == i)
        {
            cout<< i->shop << i->pr;
            found = 1;
        }
        if (!found)
            cout << "The the good with such name is not found";
        delete temp;
 }

一个主文件：

int main()
{
    price *price_list=NULL;
    char inf[20];
    list_fill(price_list);
    cout << "Info about goods: ";
    show_list(price_list); //there is no problem
    cout <<"Input goods name you want to know about: ";
    cin >> inf;
    cout << "The info about good " << inf << show_info(price_list,inf)<<endl;
    system("pause");
    return 0;
}

我需要修复我的功能，以便它可以正常工作。

如前所述，错误是c2664。

Answer 1

void show_list(price *&head, char cur)

应该

void show_list(price *&head, char cur[] )

当你inf经过char [20]时show_info(price_list,inf)

PS：可能还有其他问题

Answer 2

改写函数如下

#include <cstring>

//...

void show_info( const price *head, const char *cur )
{
    bool found = false;
    const price *i = head;

    for ( ; i != NULL && !found; i = i->next )
    {
        found = strcmp( i->name, cur ) == 0;
    }

    if ( found )
    {
        cout<< i->shop << i->pr;
    }
    else
    {
        cout << "The the good with such name is not found";
    }
}