not sure why this doesn't compile

**Dave Couture** · 08-17-2012

Code:

#include <stdio.h>
#include <cs50.h>
#include <stdlib.h>

int
main(int argc; char *argv[])
{
    int i = atoi(argv[1]);
    if (argc != 2 || i < 1)
    {
        printf("You must enter only 1 argument, it must be a positive integer.");
        return 1;
    }
}

I'm running Fedora and compiling with gcc. I get these errors when compiling:

gcc -ggdb -std=c99 -Wall -Werror ceasar.c -lcrypt -lcs50 -lm -o ceasar
ceasar.c:6:10: error: parameter 'argc' has just a forward declaration
ceasar.c:6:1: error: first argument of 'main' should be 'int' [-Werror=main]
ceasar.c:6:1: error: 'main' takes only zero or two arguments [-Werror=main]
ceasar.c: In function 'main':
ceasar.c:9:9: error: 'argc' undeclared (first use in this function)
ceasar.c:9:9: note: each undeclared identifier is reported only once for each function it appears in
cc1: all warnings being treated as errors

make: *** [ceasar] Error 1

**Matticus** · 08-17-2012

Code:

main(int argc; char *argv[])

Check this line again very carefully.

**anduril462** · 08-17-2012

EDIT: OP fixed formatting, related comments redacted.

Some things I see:

You need a comma to separate function parameters, not a semicolon
Not technically wrong, but you should be explicit and return 0 at the end of main for success

**Dave Couture** · 08-18-2012

Originally Posted by anduril462

EDIT: OP fixed formatting, related comments redacted.

Some things I see:

You need a comma to separate function parameters, not a semicolon
Not technically wrong, but you should be explicit and return 0 at the end of main for success

Thanks a lot, it's my first time using argc and argv so I didn't know. It seems to work fine except for one thing, if I run the program with no arguments, it gives me:

jharvard@appliance (~/Desktop/cs50 projects): ./ceasar
Segmentation fault (core dumped)

And I thought that by saying "if (argc != 2)", when the user doesn't input arguments, the program would say the line "You must enter only 1 argument, it must be a positive integer." It works when I enter more than 1 argument, but not if I don't enter any arguments. Why is that?

**smokeyangel** · 08-18-2012

The check is correct, but you're using argv[1] before you do the check.

Code:

int
main(int argc; char *argv[])
{
    int i = atoi(argv[1]);   <------- !
    if (argc != 2 || i < 1)

If you pass 2 arguments, then the atoi() call has a valid piece of data as an argument. The program will then exit anyway.
However with no arguments, there is nothing in argv[1] -- it is NULL, which will cause the crash. You should check for the correct number of arguments before you do anything with them.

**ZuK** · 08-18-2012

Originally Posted by Dave Couture

And I thought that by saying "if (argc != 2)", when the user doesn't input arguments, the program would say the line "You must enter only 1 argument, it must be a positive integer." It works when I enter more than 1 argument, but not if I don't enter any arguments. Why is that?

True, but before that you already access argv[1] and that doesn't exist in this case.
Kurt

EDIT: Typing slow. smokeyangel already told everything

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