write a loop to a file, getopt()

I decided today to re-write a simple calculator I wrote a while back in Java, to C. This time I'm trying to make the calculator a bit more advanced with a few command line options. Among many a -v option for verbose. When the -v option is set I want the output from the calculations (the -c option start the loop) to be written to a file. But there's a few problems. I'm using the getopt() function to retrieve the options set. If I use ./calc -vc (or -v -c) the verbose mode is on, and the calculation loop start. But if I do ./calc -cv (or -c -v), only the calculation loop starts. Why? The real problem here is I don't know how to write the input/output from the loop to a file. Should I create a buffer? How do I tell (check) the loop that the -v option is set?
Here's some of the code (I have cut most of it off, if more is needed let me know):

Code:

---

    while((argument = getopt(argc, argv, "hvcV")) != -1) {

        switch(argument) {

            case 'h':
                help();
                break;

            case 'v':
                printf("verbose mode on\n");
                verbose();
                break;

            case 'c':
                printf("options:\t+ - / *\n");
                calculate();
                break;

---

A part of the calculate(), which loops until either Q or q is pressed:

Code:

---

        switch(option) {

            case '+':
            case '1':
                printf("first value:\t");
                first = userInput();
                printf("second value:\t");
                second = userInput();
                total = first + second;
                printf("total:\t\t%.3f\n", total);
                break;

---

Any suggestions? Thanks in advance. :)

>> But if I do ./calc -cv (or -c -v), only the calculation loop starts. Why?
Because getopt() hits the c option first which calls calculate() directly. getopt() is then called again and hits the v option, but by then it's too late.

Do something like:

Code:

---

int optCalculate = 0;
int optVerbose = 0;

while((argument = getopt(argc, argv, "hvcV")) != -1) {

        switch(argument) {

            case 'h':
                help();
                break;

            case 'v':
                printf("verbose mode on\n");
                optVerbose = 1;
                break;

            case 'c':
                optCalculate = 1;
                break;
}

if (optVerbose) printf ("I'll tell you what I'm doing...\n");

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Sorry for the delay in replying.

I don't think I completely understood what you ment. I tried the following (note that the verbose() calls the calculate() again):

Code:

---

int setVerbose = 0;
[...]

while((argument = getopt(argc, argv, "hvcdV")) != -1) {

 switch(argument) {

  case 'h':
    help();
    break;

  case 'v':
    setVerbose = 1;
    break;

  case 'c':
    printf("options:\t+ - / *\n");

    if(setVerbose == 1) {
      printf("verbose mode:\ton\n");
      verbose();                   
    } else {
      printf("verbose mode:\toff\n");
      calculate();
    }
    break;

---

The problem is I can see this isn't a good sollutuion. It doesn't work if you use -c -v, only with -v -c. In theory the ideal thing would be to place the if sentence after all the options, but is that a good idea?

>>place the if sentence after all the options, but is that a good idea?<<
In essence, yes.

Normally, I'd finish parsing the args before invoking any functions that do any real work. It would be acceptable to put code like puts("verbose mode is now on"); in the loop you've shown though.

So, maybe it'd look like this:

Code:

---

while((argument = getopt(argc, argv, "hvcdV")) != -1) {

 switch(argument) {

  case 'h':
    help();
    return 0;  /* End of program */
    break;

  case 'v':
    setVerbose = 1;
    puts ("Verbose mode is on");
    break;

  case 'c':
    DoCalculate = 1;
    break;
}

if (DoCalculate)
{
    if (setVerbose) puts ("Calling Calculate()");
    Calculate();
}

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