C error C2660: 'Menu' : function does not take 3 arguments - c

Question

I am new to programming and do not understand this error.
I have the same arguments in the Menu () function and when I call the same function in the menu_principal () function.
In function menu_principal(), I want execute the switch-case statement by the function Menu() with 'option' variable.
Can you help please?
int main()
{
void menu_principal();
return 0;
}
void menu_principal()
{
bool stop = true;
int option;
const char *title = "MENU PRINCIPAL";
const char *options_menu[] = { "ARTIGOS", "CLIENTES", "ORCAMENTOS", "SAIR" };
int n_options = 4;
do
{
option = Menu(title, options_menu, n_options);
switch (option)
{
case 1:
Menu_Item();
break;
case 2:
Menu_Client();
break;
case 3:
Menu_Billing();
break;
case 4:
stop = false;
break;
}
} while (stop);
}
int Menu(const char *title1, const char *options_menu1[], int n_options1)
{
int OptionSelected= 1;
int key;
bool stop = true;
do
{
system("cls");
gotoxy(5, 3 + OptionSelected); printf(">>");
gotoxy(15, 2); printf("%s", title1);
for (int i = 0; i < n_options1; i++)
{
gotoxy(10, 4 + i);
printf("%s ", options_menu1[i]);
}
do
{
key = _getch();
} while (key != KEY_UP && key != KEY_DOWN && key != KEY_ENTER );
switch (key)
{
case KEY_UP:
OptionSelected--;
if (OptionSelected < 1)
{
OptionSelected = n_options1;
}
break;
case KEY_DOWN:
OptionSelected--;
if (OptionSelected > n_options1)
{
OptionSelected = 1;
}
break;
case KEY_ENTER:
stop = false;
break;
}
} while (stop);
return OptionSelected;
}

Answer 1

The compiler reads your program top to bottom, so it sees:
option = Menu(title, options_menu, n_options);
On this line, you call a previously unknown function, Menu.
Because the function is unknown, the compiler assumes it will be int Menu(void). (takes no parameters, returns an int).
That assumption is obviously different from how Menu eventually gets declared.
To fix this, declare the function properly near the top of your file:
int Menu(const char *title1, const char *options_menu1[], int n_options1);
Then, when the compiler encounters your function-call, it will not assume a declaration, it will use the declaration you already provided (takes 3 parameters, and returns an int)