opening a file based on a variable.

[Levi Campbell]

I have a function that as one of its arguments is a variable for a file
name, how would I use the variable to fill in the file name when
calling fopen?

[Peter Nilsson]

Levi Campbell wrote:

I have a function that as one of its arguments is a variable for a
file name, how would I use the variable to fill in the file name
when calling fopen?

Exactly the same way you'd use a string literal for a fixed file
name.

If you have actual code that is failing, I suggest you post it
because your problem probably has nothing to do with fopen per se.

--
Peter

[Ben Pfaff]

"Levi Campbell" <le*********@gmail.com> writes:

I have a function that as one of its arguments is a variable for a file
name, how would I use the variable to fill in the file name when
calling fopen?

void
function (const char *filename)
{
FILE *file = fopen (filename, "r" /* etc. */);
if (file != NULL) {
/* ...use file... */
}
}
--
"If I've told you once, I've told you LLONG_MAX times not to
exaggerate."
--Jack Klein

[Keith Thompson]

"Levi Campbell" <le*********@gmail.com> writes:

I have a function that as one of its arguments is a variable for a file
name, how would I use the variable to fill in the file name when
calling fopen?

By calling fopen() with the argument.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Martin Ambuhl]

Levi Campbell wrote:

I have a function that as one of its arguments is a variable for a file
name, how would I use the variable to fill in the file name when
calling fopen?

As you would use a string as an argument to any other function taking a
string as an argument. For example,

#include <stdio.h>
int main(void)
{
FILE *fp;
char foobar[] = "random.filename.txt";
if (!(fp = fopen(foobar,"r")))
{
fprintf(stderr,"Oh crap! I couldn't open \"%s\" for input.\n",
foobar);
}
else fclose(fp);
return 0;
}