I think what the author was trying to show was that with C your pointers are real pointers to memory and once a variable is set up in a function its memory is fixed. If you fiddle with a pointer to it you will change it. Here is an example: 
int *mysub ();
main() {
    int* x;
    x = mysub();
    printf("in main x = %d\n", *x);
    *x = 5;
    x = mysub();
}
int *mysub () {
    int z;
    printf("in sub z = %d\n", z);
    z = 3;
    return &z;
}

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Compiling gives a warning but it still compiles. 
$ make temp
gcc     temp.c   -o temp
temp.c: In function `mysub':
temp.c:13: warning: function returns address of local variable

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running it gives you: 
$ temp
in sub z = 0
in main x = 3
in sub z = 5

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as you can see main "reached into" mysub and changed z. There is no equivalent way to "reach into" a perl function and fiddle with its my variables 
use strict;
my $x;
$x = &mysub;
printf "in main \$x = %d\n", $$x;
$$x = 5;
$x = &mysub;

sub mysub () {
    my $z;
    printf "in sub \$z = %d\n", $z;
    $z = 3;
    return \$z;
}

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outputs: 
$ perl temp.pl
in sub $z = 0
in main $x = 3
in sub $z = 0

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as you can see the main program could not "reach in" and change the value of $z in mysub even though you dereferenced the reference it returned. The my $z in mysub creates a new variable every time and does not just reference a fixed memory location as in C.

Before an expert points it out:

I know, I know, the memory locations may change in C depending on hardware, OS, etc. but I am trying just to give and example of things that are possible in C, not necessarily good programming practice.

Update -- I noticed that if you make z static in mysub 

int *mysub () {
    static int z;
...

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the compiler (gcc in my case) will not emit a warning but the output of the program is the same.

--

flounder

In reply to Re: Reference to a function by flounder99
in thread Reference to a function by Anonymous Monk

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