CODEWITHSUNDEEP
c
user2730962
user2730962

Reputation: 19

How do can you do make an input take 3 arguments

I am trying to make it so that you would call the method with the function to perform for example -r,-u,-l and then a file name or work with standard input how would I make it so that it can take 3 inputs I've tried a couple things but I just started using c and have almost no clue what I am doing.What I can't figure out is how to make an input take three then be able to compare the strings to choose what operation to do.

#include <stdlib.h>

void upper(FILE *src, FILE *dest)
{
    int c;
    {
        fprintf(dest, "%c", toupper(c));
    }
}

void lower(FILE *src, FILE *dest)
{
    int c;
    while ((c = fgetc(src)) != EOF)
    {
        fprintf(dest, "%c", tolower(c));
    }
}

void rot13(FILE *src, FILE *dest)
{
    int c;
    while ((c = fgetc(src)) != EOF)
    {
        fprintf(dest, "%c", c+13);
    }
}

FILE * input_from_args(int argc,char choice, const char *argv[])
{
    if (argc == 1)
    {
        return stdin;
    } else
    {
        return fopen(argv[1], "r");
    }
}

FILE * input_from_args(int argc, const char *argv[])
{
    if (argc == 1)
    {
        return stdin;
    } else
    {
        return fopen(argv[1], "r");
    }
}

int main(int argc,char** choice,const char *argv[])
{
  FILE *src = input_from_args(argc, argv);
  FILE *dest = stdout;

    if (src == NULL)
    {
        fprintf(stderr, "%s: unable to open %s\n", argv[0], argv[1]);
        exit(EXIT_FAILURE);
    }
    else if(*choice == '-r')
    }
      rot13(src,dest)
    }
    else if(*choice == '-u')
    {
      upper(src,dest)
    }
    else if(*choice == '-l')
    {
      lower(src,dest)
    }

    fclose(src);

    return EXIT_SUCCESS;
}

Upvotes: 0

Views: 91

Answers (1)

Some programmer dude
Some programmer dude

Reputation: 409364

That's what the argv array is for. It will contain all arguments passed to the program.

For example take this simple test program:

#include <stdio.h>

int main(int argc, char *argv[])
{
    printf("argc = %d\n", argc);

    for (int a = 0; a < argc; ++a)
    {
        printf("argv[%d] = \"%s\"\n", a, argv[a]);
    }
}

If you build that source, and execute the program such as

$ ./a.out argument1 argument2 argument3

It will output

argc = 4
argv[0] = "./a.out"
argv[1] = "argument1"
argv[2] = "argument2"
argv[3] = "argument3"

In other words, the arguments passed to a program does not match the arguments in the source code, instead they are converted to the argv array.

Upvotes: 3

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