Re: gcc not compiling

From: Derek Martin (code_at_pizzashack.org)
Date: 10/31/05

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    Date: Mon, 31 Oct 2005 16:01:03 -0500
To: fedora-list@redhat.com

    On Mon, Oct 31, 2005 at 01:30:44PM -0600, STYMA, ROBERT E (ROBERT) wrote:
    > The syntax:
    > int main(int argc, char **argv)
    > works, but most of the C books I have seen recommend
    > the *argv[] version.

    I meant to comment that these two different notations are functionally
    identical; an array name is nothing more than a pointer. For example,
    if we have the following code:

            #include <stdio.h>
            void main (void){
            
                    char foo[] = "This is a string\n";
                    char *bar = foo;
            
                    /* these expressions all print "T\n" */
                    printf("%c\n", foo[0]);
                    printf("%c\n", *bar);
                    printf("%c\n", bar[0]);
                    printf("%c\n", *foo);
            
                    /* these expressions all print "i\n" */
                    printf("%c\n", foo[2]);
                    printf("%c\n", *(bar + 2));
                    printf("%c\n", bar[2]);
                    printf("%c\n", *(foo + 2));
            }

    The results may surprise programming students:

            $ gcc -o ptr ptr.c
            ptr.c: In function `main':
            ptr.c:2: warning: return type of 'main' is not `int'
            [ddm@archonis ~]
            $ ./ptr
            T
            T
            T
            T
            i
            i
            i
            i
            

    In reality, foo[x] is just "syntactic sugar" for *(foo + x). This
    is called pointer math, and works properly regardless of the size and
    type of foo, so long as it was declared properly before being used.

    [I'm very bored today...] 

    -- 
Derek D. Martin
http://www.pizzashack.org/
GPG Key ID: 0x81CFE75D



    


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