Command line arguments, reading a file

Question

If i entered into the command line C: myprogram myfile.txt

How can I use myfile in my program. Do I have to scanf it in or is there an arbitrary way of accessing it.

My question is how can I use the myfile.txt in my program.

int
main(){
    /* So in this area how do I access the myfile.txt 
    to then be able to read from it./*

Answer 1

You can use int main(int argc, char **argv) as your main function.

argc - will be the count of input arguments to your program.
argv - will be a pointer to all the input arguments.

So, if you entered C:\myprogram myfile.txt to run your program:

• argc will be 2
• argv[0] will be myprogram.
• argv[1] will be myfile.txt.

More details can be found here

To read the file:
FILE *f = fopen(argv[1], "r"); // "r" for read

For opening the file in other modes, read this.

Answer 2

1. Declare your main like this

   int main(int argc, char* argv [])

   • argc specified the number of arguments (if no arguments are passed it's equal to 1 for the name of the program)

   • argv is a pointer to an array of strings (containing at least one member - the name of the program)

   • you would read the file from the command line like so: C:\my_program input_file.txt

2. Set up a file handle:

   File* file_handle;

3. Open the file_handle for reading:

   file_handle = fopen(argv[1], "r");

   • fopen returns a pointer to a file or NULL if the file doesn't exist. argv1, contains the file you want to read as an argument

   • "r" means that you open the file for reading (more on the other modes here)

4. Read the contents using for example fgets:

   fgets (buffer_to_store_data_in , 50 , file_handle);

   • you need a char * buffer to store the data in (such as an array of chars), the second argument specifies how much to read and the third is a pointer to a file

5. Finally close the handle

   fclose(file_handle);

All done :)