Passing FIle pointer as argt

I want to pass input and output file as command line arguments...

I want output file pointer have to be passed as argument to a function..

How can i do it? Is this Ok??

But I am getting Error :cannot convert `_iobuf' to `FILE**' for argument `2' to `void encode(char*, FILE**)'

Code:

---


int main(int argc,char *argv[])
{
    FILE *fp,*fp2;
    char str[80],*str1;
    fp=fopen(argv[1],"r");
    fp2=fopen(argv[2],"w");  // file opened here

...............

 encode(str1,*fp2);




}



       
void encode(char *str,FILE **fp2)
{
    ................
        // Here i want to write it to file
          fprintf(*fp2,"%d",no);
       

}

---

Help me how to do please?

The address of operator is "&" not "*".

But from the looks of it, you shouldn't be using **fp2, and thus not use & either - unless you intend to CHANGE file-pointer in the called function, all you need to pass is the file-pointer itself, as "FILE *fp", and just give it's name as the argument to the function.

--
Mats