I am making a timer but i have an error blocking the way. Please help me

[aravind12345]

ok so i am getting an error in this program i did not test it out because of the error ok so the error is
(expected primary-expression before ')' token) that and i am getting two warnings ([Warning] character constant too long for its type) the other warning is the same but for another line. So if any one can fix this problem i will be thank full.

#include <iostream>
#include <conio.h>
#include <windows.h>
 
using namespace std;
int main()
{
    int min;
    int sec;
    cout<<"Enter Minutes: ";
    cin>>min;
    cout<<"Enter Seconds: ";
    cin>>sec;
    cout<< "To Start Timer. Press Any Key.";
    getch();
    while(){
            cout<< min <<" : "<< sec<< endl;
 
          Sleep(750);
    switch (min)
    {
           case 'min > 0':--min;
           break;
           case 'min == 0':min = 0;
           break;
           }
           if (min - 1){
                   sec = 59;
                   --sec;
                   }
    if (min == 0 & sec == 0){
            cout<<"Timer is done. Press any key to Exit";
            getch();
            break;
            }
            }
            return 0;
            }

[weaknessforcats]

Your while loop cannot be while(). You must supply the condition for exiting the loop. If you want an infinite loop, then code while(1).

Then this:

  switch (min)
    {
           case 'min > 0':--min;
           break;
etc...

A case is not the same as an if statement. For a case you specify the value of min for the case like 5 or 10 or 0. It must be an integer.

This code:

 if (min == 0 & sec == 0){
            cout<<"Timer is done. Press any key to Exit";
            getch();
etc...

If you mean min s zero AND sec is zero, then use the && operator and not &.

[Nepomuk]

The error is in the while loop; line 16 which is

while(){

makes no sense. You have to include the condition for which the loop should be continued; if you want to continue it until you exit with break the condition would simply be true.

[chrisgates289]

Which line it reported to have warning?

[aravind12345]

The Warning is on line 22 and 24 so please fix this if you can

[Nepomuk]

You can't use the switch-case structure like that; the cases have to be concrete values, not conditions. The warnings came because anything in single quotes (' ') is interpreted to be a character and as in C characters can be mapped to integers, the code compiles. However it would probably just use the first character from the expression, so it would check whether min == 'm'. Not what you want.

If you want more flexibility, you'll have to use if-else statements; so what you're trying to do would look something like this:

if (min > 0) {
    --min;
} else if (min == 0) {
    min = 0;
}

[chrisgates289]

Sorry that I was away. I think Nepomuk has solved the warning.

[aravind12345]

Yup he did solve the thing