Question

I’m having an issue with scanf() getting stuck, and the only way to continue the program is to type exit.

My input is something like this:

L 1 1 3 4 C 2 3 4

where the numbers are the parameters for L and C.

int main ()
{
// Get use input and put in array
int count, x;
int array[4];
char type;

scanf("%c", &type);

for (x = 0; x < 2; x++)
{
    if (type == 'L')
    {
        for (count = 0; count < 4; count++)
        {
            scanf("%d", &array[count]);
        }

    } else if (type == 'C')
    {
        for (count = 0; count < 3; count++)
        {
            scanf("%d", &array[count]);
        }

    } 
    if (scanf("%*[ \n\t]%c", &type) == 0)
    {
        printf("Error");
        break;
    }
}

The scanf statement in question:

scanf("%*[ \n\t]%c", &type)

works fine if I’m not at the end of the line, but breaks otherwise. However, I won’t know how many L and C objects I will have so can’t rely on the value from the for loop.

Answer 1

I think your problem is here:

if (scanf("%*[ \n\t]%c", &type) == 0)

You’re trying to check if you’re out of characters to read, so you want to check if scanf returns EOF (which is -1, but just use the EOF constant). If you break when you see EOF returned you’ll drop out of the loop as soon as you end input to the process (Ctrl+D on Linux, I believe Ctrl+Z Enter on Windows)