错误C2664：show_info：无法从 '的char [20]' 转换参数2至“字符

Question

我有一个小的结构：

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

甲功能不起作用：

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; 
} 

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。

如上所述，错误是c2664。

Answer 1

重写功能通过以下方式

#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"; 
    } 
} 

Answer 2

void show_list(price *&head, char cur) 

应该

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

为你传递inf即char [20]在show_info(price_list,inf)

PS：有可能是其他问题太