How to initialize argv array in C

c

Solution

char* dummy_args[] = { "dummyname", "arg1", "arg2 with spaces", "arg3", NULL };

int main( int argc, char** argv)
{
    argv = dummy_args;
    argc = sizeof(dummy_args)/sizeof(dummy_args[0]) - 1;

    // etc...

    return 0;
}

One thing to be aware of is that the standard `argv` strings are permitted to be modified. These replacement ones cannot be (they're literals). If you need that capability (which many option parsers might), you'll need something a bit smarter. Maybe something like:

int new_argv( char*** pargv, char** new_args) 
{
    int i = 0;
    int new_argc = 0;
    char** tmp = new_args;

    while (*tmp) {
        ++new_argc;
        ++tmp;
    }

    tmp = malloc( sizeof(char*) * (new_argc + 1));
    // if (!tmp) error_fail();

    for (i = 0; i < new_argc; ++i) {
        tmp[i] = strdup(new_args[i]);
    }
    tmp[i] = NULL;

    *pargv = tmp;

    return new_argc;
}

That gets called like so:

argc = new_argv( &argv, dummy_args);

Problem

I am trying to initialize *argv with these values : test_file model result Can anyone help me how to directly initialize the argv instead of using command line. I am doing it like this: ``` *argv[]= {"test_file","model","output",NULL}; ``` but its not working. I know its simple but i am new to programming. Can anyone help me?

Original source