how will this short c++ code look in pure c?

how will this short c++ code look in pure c?

int main(void)
{
int i;
do
{
cout << "[1] Func1 << "\n";
cout << "[0] End << "\n" << endl;

cin >> i;

switch(i)
{
case 1:
Func1();
break;
case 0:
break;
default: cout << "Error" << endl;
}
}
while (i != 0);
return 0;
}

Code:

---

int main(void)
{
    int i;
    do
    {
        printf( "[1] Func1\n" );
        printf( "[0] End\n\n" );

        i = getchar();

        switch(i)
        {
            case 1:
                Func1();
                break;
            case 0:
                break;
            default:
                printf( "Error" );
                break;
        }
    }
    while (i != 0);
    return 0;
}

---

Close:
> cout << "Error" << endl;
> printf( "Error" );

Forgot the newline.

puts("Error");

Quzah.

Don't know much of C++, but it looks like the i is being used as an int, so instead of

Code:

---

i = getchar ();

---

you may want

Code:

---

scanf ("%d", &i);

---

Actually, all of the *getc*() functions return integers.

Quzah.

Yea, but if the user types in 1, then the value stored in i is not going to be 1.. it's gonna be '1'.

This is true. However, if their case statement was designed right it won't matter. But yeah, you're right there.

Quzah.

Yes...your code will work just fine. but how much important the default statement is for our purpose is my question....

The default statement is simple error checking. If the user enters 3 for instance, what happens? The result is undefined, so it's a good idea to check and see if the value input was what you want, and if not it does nothing but restart the menu.