Do while loop with switch case in C language not working

Question

I'm trying to do this loop in C, where you would press enter and be in it and you either press 0 to exit or 3 to continue in it. But somehow the Switch commands are no being activated. Note that there is a different messages on each of them that is supposed to differentiate them from the other outcomes. Can someone help me understand the problem with this code?`Note: the code is obviously within int main().

int I = 1;
printf("Press enter to start the loop...");
getchar();
do
{
    printf("\nYou are in a LOOP. Would you like to stay in it, or leave it? \nPress 0 to leave the loop or press 3 to stay in it: ");
    scanf_s("%d", &I);
    getchar();

    switch (I)
    {
    case'3':
        printf("\nYou are STILL inside the LOOP. Press 0 to leave it or press 3 to stay in it: ");
        getchar();
        break;
    case'0':
        printf("\nExiting the LOOP...");
        break;
    default:
        printf("\nPlease, enter a valid command...: ");
        if (scanf_s("%d", &I) != 3 || scanf_s("%d", &I) != 0);
        {
            fflush(stdin);
        }
        break;
    }

while (I != 0);
printf("\nCongratulations! You are OUT of the LOOP!");

Answer 1

As others have pointed out, you are reading in an integer with scanf_s(), but comparing against a character literal inside the switch statement.
switch'3': should be written as switch 3: (using an integer literal instead). Same goes for the 0 case.

Also, your example is missing a curly brace } before while(I != 0); .

It is also worth mentioning that scanf_s() "returns the number of fields successfully converted and assigned" (from https://msdn.microsoft.com/en-us/library/w40768et.aspx ). That is, your scanf_s() call in the default switch case will only ever return 0 or 1 when reading in a single integer (or EOF on error).