Thread: scanf in while problem

Hello everybody,
I'm new at programming in C and because of that when I was trying to make a loop in C a problem came up and I have no idea of what is going on, hope you can help me with this :

Code:

void le_posicao(int *coor_i, int *coor_j, int dim_tab)
{
    int nconv=0;
    do
    {
    printf("Insira as coordenadas de uma luz na forma (a,b): ");
    nconv=scanf("(%d,%d)",coor_i,coor_j);
    }while(nconv!=2);
}

I created a function to read data from the user but the condition goes to infinity i don't know why...

Well that's the point, what I want to do is force the user to type the '(', the ',' and the ')' but in spite of asking new values to the user it just goes to infinity. Is there a way to make it work in a way that the user must type (a,b) in this precise format?

Opportunistic scanning is your friend. If you read the input (line) into a buffer, you can use sscanf() with different patterns. For example,

Code:

    char line[200], dummy;
    double x, y;

    if (fgets(line, sizeof line, stdin) == NULL) {

        /* No more input: user pressed CTRL+D,
         * or no more piped or redirected input.
         * You should exit here, as there will
         * not be any more input.
        */

    } else
    if (sscanf(line, " ( %lf , %lf ) %c", &x, &y, &dummy) == 2 ||
        sscanf(line, " [ %lf , %lf ] %c", &x, &y, &dummy) == 2 ||
        sscanf(line, " < %lf , %lf > %c", &x, &y, &dummy) == 2 ||
        sscanf(line, " ( %lf %lf ) %c", &x, &y, &dummy) == 2 ||
        sscanf(line, " [ %lf %lf ] %c", &x, &y, &dummy) == 2 ||
        sscanf(line, " < %lf %lf > %c", &x, &y, &dummy) == 2 ||
        sscanf(line, " %lf %*1[:;,/xX] %lf %c", &x, &y, &dummy) == 2) {
        sscanf(line, " %lf %lf %c", &x, &y, &dummy) == 2) {

        /* It was a 2D vector.
         * The line did not contain anything except the vector
         * (we verified that by scanning an additional "dummy" char).
         * The components are in the x and y variables.
        */

    } else {

        /* Tell the user you couldn't understand the line, and exit.
         * Or, tell the user about the expected formats, and retry.
        */
    }

The %*1[:;,/xX] pattern is an interesting one there. It uses the [ conversion character (with the list of allowed characters immediately following it), and ends with ]. Due to the *, it does not store anything, and even if it matches, it is not counted as a conversion in the result value. Due to the 1, it matches one character (technically, always at least one, up to the specified number, of characters). That character must be one of : ; , / x X .

Because you read the line, you don't need to exit if the line could not be parsed, but can for example tell the user about the formats supported.

(The reason for allowing : / x in addition to a semicolon or a comma is that I like to specify vectors as single command-line parameters using format widthxheight or xxx:yyy or xxx/yyy/zzz -- these do not interfere with shell special characters like ; and space do. Some locales use , as the decimal point or thousands separator, so its not a good list separator. (If you just include <locale.h> and call setlocale(LC_ALL, ""); before trying to scan anything, both scanning and printing will automagically use the decimal point specified in your locale. Which is nice.)

Alright I think I understood your piece of code, thank you so much! So basically with that any type of input will be accepted, is it?

Originally Posted by khanny

So basically with that any type of input will be accepted, is it?

Why don't you test it, and find out?

Just add a printf("x=%f, y=%f\n", x, y); into the if case body, and then throw different vector formats at it to see what works. That's how I decided on those patterns, anyway.