CODEWITHSUNDEEP
ccommand-line-argumentsfopen
user5405027
user5405027

Reputation:

What does the fopen() function do when I assign it to a variable?

I'm a beginner programmer working on a project that requires my code to read characters from an input file, manipulate them without changing the original file, and then print out the modified version to an output file. To do this, I need to use the fopen() function at some point in the program, but I'm curious: what is actually happening here? Are the contents of the input file copied into the variable input1?

#include <stdio.h>

int main(int argc, char *argv[])
{
    FILE *input1;

    input1 = fopen(argv[1], "r");

    return 0;
}

Upvotes: 1

Views: 815

Answers (3)

rocky
rocky

Reputation: 25

Read file handling APIs here.

http://www.thegeekstuff.com/2012/07/c-file-handling/

Upvotes: -1

machine_1
machine_1

Reputation: 4454

The fopen function opens a stream for I/O to the file filename, and returns a pointer to the stream. so,in your statement :

input1 = fopen(argv[1], "r");

fopen() opens the file argv[1] for reading and the address returned by the function is assigned to input1.This does not involve copying of the contents of the file.

you can then use the file pointer to read data from the file.for example :

char c = getc(input1);

Upvotes: 1

Greg Hewgill
Greg Hewgill

Reputation: 992707

The return value of fopen() is a file handle, which is like a token that you can use later to interact with the file. You can pass your input1 to a function such as fgets() or fread(), depending on what you want to do with the file next.

The contents of the file are not copied anywhere by fopen().

Upvotes: 4

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