Thread: Read/Write long integer to file!

You forgot the & in the fscanf function.

You mean to say: "You forgot to pay attention to your compiler."


Quzah.

One of your problems is this:

Code:

fscanf(fd2, "%li\n", long_int);

Think back to how you use scanf().

You might want to check the return value of fopen() to see if it failed. Other than that, try using 100 instead of 245749955.

Originally Posted by Ancient Dragon

The compiler probably would not have caught that error because it doesn't parse the format flags to see what parameters should be passed.

It doesn't have to. All arguments to fscanf are pointers. An integer is not a pointer, long or no.

Code:

#include<stdio.h>
int main( void )
{
    FILE *fp = NULL;
    int x;
    fscanf(fp,"%d",x);
    
    return 0;
}

fscanfargs.c: In function `main':
fscanfargs.c:6: warning: format argument is not a pointer (arg 3)

Some of us actually do know what we're talking about here.

[edit]
Yes, that's intentionally NULL, no it doesn't have anything to do with the warning. It simply serves as an illustration to get the correct warning text without me having to make some file.
[/edit]


Quzah.

You could have used stdin.

You don't realise that 245749955 is OUTSIDE the range of your long integers - try unsigned

245749955 would easily fit into a signed int.

Not on some piece of crap 16 bit compiler it won't. Unsigned won't help either. (Who'll bet me it's TC 3.0? ... Yeah, it's probably not, but you never know with people around here. So we have to be careful on our generalizations.)


Quzah.

I never though of that. But at the very least, it would fit into a long, which is what he has.

Who'll bet me it's TC 3.0?

Do you mean TC 2? That's the one that's freely downloadable.

Originally Posted by quzah

It doesn't have to. All arguments to fscanf are pointers. An integer is not a pointer, long or no.

fscanf takes a variable number of parameters and is prototyped like below, which prevents the compiler from doing type checking of the parameters beyond the first two:

Code:

int __cdecl fscanf(FILE *, const char *, ...);

Yeah, but some compilers know better. They know what scanf() is supposed to take for arguments.

It doesn't matter if it has an unspecified number of arguments, they're still all pointers. How the hell is scanf supposed to updated the contents of a variable passed to it, so it takes effect in the calling function, without using a pointer?

Yeah, so anyway, as I said, it always takes pointers for its arguments.


Quzah.