Example app in C that takes 2 positional parameters

Question

I am looking for an example of a simple app that will use printf to express two different strings based on a positional parameter.

In bash I would use:

case $1 in
    -h | --help )           showHelp
                            exit
                            ;;
    * )                     manPipe
                            exit 1
esac

And prior to this I would list that a function called showHelp would be called if the operater types either $ foo -h or $ foo -help into the Terminal. Anything else like $ foo -bar would request that the function manPipe would get called.

I have this code so far:

#include <stdio.h>
#include <cstring>

int secretFunction() {
    printf("Success! You found the secret message!");
}

int main() {

str posParam;
posParam = X;

printf("Enter a number:");
scanf("%s",&posParam);

if ( posParam == "X" ){
    printf("Welcome to app!\nType: " + $0 + " t\nto show a message");
}else{
    if (posParam == "t" ){
        secretFunction();
        }
return 0;
}
return 0;

I know this code is really crappy, I was trying to make an example of the above code in bash. I am not trying to convert a bash script into a C app, I'm trying to play around with it. I drew the idea of something I want to work on from the Wikipedia article on the MD5 checksum C app that takes a string and calculates the MD5 checksum for it. I cannot seem to work out what part they get the positional parameter to pass to the application.

This is a little different, I do understand, because it has prompted the user to provide an answer and then assign it to a value. I would rather use it as a positional parameter in the first instance.

So, I'm assuming from the language tags that `` isn't the C++ standard library header for encompassing the C-equivalent ``. Because in neither C nor C++ is `str` even a valid type provided by the standards. — WhozCraig, Oct 15 '13 at 08:22

score 2 · Accepted Answer · answered Oct 15 '13 at 08:22

2

What is $1 in Bash (et al) is argv[1] in a C program:

#include <stdio.h>

int main(int argc, char *argv[])
{
    if (argc > 1)
    {
        printf("You provided at least one argument (or parameter)\n");
        printf("The first argument is \"%s\"\n", argv[1]);
    }

    return 0;
}

The argument argc is the number of valid entries in the argv array. argv[0] is the executable name, and you can access up to argv[argc - 1]. (Actually you can access argv[argv] as well, it is always a NULL pointer).

answered Oct 15 '13 at 08:22

Some programmer dude

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How do I specify `argv[x]`, where `x` is predetermined arguments, like `-k` and `-v` but anything else is considered garbage? Do you need an array to be predetermined with each `argv[]`, can I do `argv[-k,-v]` and then use `if (argv[1] == "-k") { do something;}` or will that not work? – Danijel-James W Oct 15 '13 at 08:49
1

@DanijelJ You don't. Just like in Bash you actually have to search for it. And you can't use `argv[1] == "-k"`, you have to use e.g. `strcmp(argv[1], "-k") == 0`. There are many example of argument parsing in C, and if you're on a POSIX platform (like Linux, BSD or OSX) then you can use the [`getopt`](http://pubs.opengroup.org/onlinepubs/9699919799/functions/getopt.html) function to help you. – Some programmer dude Oct 15 '13 at 08:54

David Ranieri · Answer 2 · 2013-10-15T09:05:47.317

1

As Joachim says, replace $0 with argv[0], also (assuming that str is a char*):

scanf("%s",&posParam);
           ^ there is no need to use &, posParam is already a pointer.

if ( posParam == "X" ){

strings can not be compared with ==, instead use:

if (strcmp(posParam, "X") == 0){

edited Oct 15 '13 at 09:05

answered Oct 15 '13 at 08:51

David Ranieri

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Example app in C that takes 2 positional parameters

2 Answers2