Passing file as an argument into my program and then into a function?

Question

I am using CodeBlocks to create my program in C and I'm having trouble with the following issue.

I am trying to open and read from a .txt file and it works fine if I put it into the main function like this:

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

FILE *input;

input = fopen("file.txt", "r");

char singleLine[50];

while(!feof(input)){

fgets(singleLine, 50, input);
puts(singleLine);
}

However, if I set the name of the file "file.txt" as an argument in CodeBlock using the "Set Program's Arguments option, and then want to pass it into a function that would read from it like this:

void read(char *name){

File *input;

....
....

}

And call it like this:

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


read(&argv[1]);

}

It does not work and the program crashes.

Please see Why is “while ( !feof (file) )” always wrong? — Sourav Ghosh
– Sourav Ghosh, Commented Nov 27, 2015 at 16:09

Sourav Ghosh · Accepted Answer · 2015-11-27 16:12:01Z

3

If your function prototyope is

void read(char *name)

then you need to call it like

read(argv[1]);

because , argv[1] itself gives you a char *, not &argv[1].

FWIW, always check for the validity of argv[n] before using it directly.

edited Nov 27, 2015 at 16:12

answered Nov 27, 2015 at 16:11

Sourav Ghosh

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6 Comments

Daeto Over a year ago

Hey there and thanks for your reply!I definitely made a mistake there but it still does not work. Maybe something to do with the location of the file?

Weather Vane Over a year ago

@Daeto after checking argc > 1 try printing the file name in main() and again in read(). You must also test input == NULL after attempting to open the file. Finally please don't use read() as a function name - it's already used for a library function.

Daeto Over a year ago

The printf function prints the name of the file both in main and in the read function so I guess passing the name of the file as an argument is not being an issue here. However, I added the condition to check whether opening the file was successful, it it turns out that it was not. Any idea why? The commands in the function look just like I described them above in main.

Weather Vane Over a year ago

@Daeto you are opening the file for reading, so it must it already exist, on a path that the program will look in.

Weather Vane Over a year ago

@Daeto you have shown code for what worked, but you didn't show the code for what did not work.

|

theWalker · Accepted Answer · 2015-11-27 16:18:45Z

0

Test

read(&argv[0]);

and see what is in argv in console

printf(&argv[0]);

answered Nov 27, 2015 at 16:18

theWalker

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2 Comments

Daeto Over a year ago

I'm confused. Sourav said I should not be passing arguments into the read function like this. Also, argv[0] is the location of my program isn' t it? Why would I need that? If I print argv[1], it prints file.txt which is the name of the file located in the same directory as my program. So I guess it is really there.

Sourav Ghosh Over a year ago

printf(&argv[0]); is really a bad example, even ignoring the logic, whatsoever.

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