passing a file name to main()

i am trying to pass a filename to my program from a promp but i don't know how to pass a file handling argument to main so that main can open the file and so on. any advices? thanks.

dj.

Please post your code!

Show what you are doing...

Code:

---


int main(int argc,char* argv[])
{
    char buf[20];
    FILE* fp = fopen(argv[1],"r");
    if(fp == NULL)
        return -1;
    fread(buf,1,20,fp);
    printf("%s\n",buf);
    fclose(fp);
    return 0;
}

---

explaination

int main(int argc,char* argv)

this declaration of main tells it to take command line arguments

where argc is the number of arguements;
and argv contains the actual arguements

Note: argv[0] always contains the programs name sothat would be the name of the exe

no one's answer contained a typo... main should look like this:

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

or

int main (int argc, char** argv)

mxr

thanks mix0,
that would have caused some trouble i'll edit and fix it.

fyi: no-one's code is actually right. in this case char **argv and char *argv[] accomplish the same thing. char* argv[] just means an array of character pointers, each pointer being the base address of a parameter string. either way should work with the compiler.

actually i edited it to * argv[] so it would not confuse him. it was when mix0 posted my code was * argv so his correction was quite right.

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

you can use the single star as previously stated all this is doing is point to an unbound array that stores command line inputs, I've always used the single star and not had any problems

just remember that unlike most arrays argv[] will start at 1 to what ever, this is that the actual program itsself is the argv[0] part.

so if you entered something like

compare txt1 txt2 txt3

argv[0] = compare
argv[1] = txt1
argv[2] = txt2
argv[3] = txt3

if you are uncertain write a small program that displays the content of the command line input

Code:

---

int main(int argc, char *argv[])
{
  printf("You enetered the following command prompt %s %s %s %s", argv[0], argv[1], argv[2], argv[3]);
}

---

hope this helps

-ali