Explanation of the program that reads files and displays them

[suvojit168]

Hi all
I was trying to understand file handling in Cpp. I came accross the followig code:

#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
FILE *fp;
char ch;
if(argc!=2) {
printf("You forgot to enter the filename.\n");
exit(1);
}
if((fp=fopen(argv[1], "r"))==NULL) {
printf("Cannot open file.\n");
exit(1);
}
ch = getc(fp); /* read one character */
while (ch!=EOF) {
putchar(ch); /* print on screen */
ch = getc(fp);
}
fclose(fp);
return 0;
}

Overall prog is clear to me but i dont understand following:

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

I have never seen main in this form. can anyone explain me the parameters and their purpose?

if(argc!=2) {
printf("You forgot to enter the filename.\n");
exit(1);
}

What is happening in the above code snippet?

thanks in advance for replies.....
Regards

[yzaykov]

That's the another standard way of declaring main. This way you can use the arguments, passed to your program. argc is the number of the arguments (the first one, e.g. [0] is always the name of your program) and the following items in the array argv are the arguments. For example, if your program is called fileReader and you execute it this way:
fileReader my_file.txt
then argv[0] woulb be fileReader, argv[1] would be my_file.txt and argc would be 2. That's the way your program wants to be called.

[Peter_APIIT]

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

I have never seen main in this form. can anyone explain me the parameters and their purpose?

This was called command line arguments where user supplied argument to the program upon startup and mostly use in command line environment.

[suvojit168]

thanks