从 'char*' 到 'char' 的无效转换 [-fpermissive] 如何打印？ 但在变量内部

Question

struct student{
    char name[30]...
};

struct element
{
    struct element *ant;
    struct student databse;
    struct element *prox;
};

Element *no = *pointer_to_pointer;

cout<< (*no).datbase.name << endl;

I printed the name correctly.

Element *no = *pointer_to_pointer;

char variable = (*no).datbase.name;
    cout<< variable << endl;

Error. Why? How to print? But inside the variable . Thanks.

Answer 1

You are trying to cast the char array name into a single char, this won't work, these types are incompatible.

If you want to use the content of name , you should use:

const char* variable = (*no).datbase.name;
std::cout << variable << std::endl;

If you want to index inside the name char array, use [] to index it, like this:

/* Take first character of name */
char variable = (*no).datbase.name[0];
std::cout << variable << std::endl;

Answer 2

The type of datbase.name is char[] , which means you can not assign it to a char variable directly. Use char *variable instead:

char *variable = (*no).datbase.name;
    cout<< variable << endl;

if you want to print just one char acter, write:

char variable = (*no).datbase.name[0]; //the first character of 'name'