Leon Breedt wrote:
> can anyone tell me why this function works:
>
> int fopenfile(FILE **fp, char *fname)
> {
> *fp = fopen(fname, "r+");
> if (!*fp)
> *fp = fopen(fname, "w+"); /* create if not found */
>
> return 0;
> }
This will only work if `fp' points to writable memory. If it points to
something other than a variable which exists for this purpose, you may
be overwriting something important, causing your program to crash.
> but this one doesnt:
>
> int fopenfile(FILE *fp, char *fname)
> {
> fp = fopen(fname, "r+");
> if (!fp)
> fp = fopen(fname, "w+"); /* create if not found */
>
> return 0;
> }
This will work, but it is ignoring the value of `fp' which is passed
in, and you can't get at the file handle from outside the function, so
it isn't of much use.
> any operations i try to do with the file if i opened it with the
> second function, cause a segfault, but if i used the first function,
> i can read/write to the file no problems.
>
> to me it seems then that the second function is not modifying 'fp' at
> all,
Nope.
> but why not?
Because that's the way that C works. Parameters are passed by value,
not by reference.
void foo(int x)
{
x++;
}
void bar(void)
{
int n = 0;
printf("n = %d\n", n);
foo(n);
printf("n = %d\n", n);
}
> is it possible that the fp is only passed by
> value if its declared as FILE *fp?
Parameters are *always* passed by value.
> excuse me if this question is elementary,
It is.
> but i'm a convert from pascal :).
Pascal does exactly the same, unless you use the `var' keyword. E.g.
procedure foo(var x:integer);
begin
x := x + 1
end
In C, you have to pass a pointer to the variable. e.g.
void foo2(int *x)
{
(*x)++;
}
void bar(void)
{
int n = 0;
printf("n = %d\n", n);
foo2(&n); /* pass a pointer to `n' */
printf("n = %d\n", n);
}
--
Glynn Clements <[EMAIL PROTECTED]>
