const int x = 0;

BYTE *foo(int address)
 { return (BYTE *) address;
 }

int a[] = {1, 2, 3, 4, 5};

assignment-operator: one of
    = *= /= %= += -= >>= <<= &= ˆ= |=

42 = 40+2;

char **arr;
(arr[x]=argv[1])[0] = '\0';

a[10] = 2;

*(a + 10) = 2;

       substr EXPR,OFFSET,LENGTH,REPLACEMENT
       substr EXPR,OFFSET,LENGTH
       substr EXPR,OFFSET
               Extracts a substring out of EXPR and returns it.  First character is at offset 0, or whatever you've set $[ to (but don't do that).  If OFFSET
               is negative (or more precisely, less than $[), starts that far from the end of the string.  If LENGTH is omitted, returns everything to the end
               of the string.  If LENGTH is negative, leaves that many characters off the end of the string.

                   my $s = "The black cat climbed the green tree";
                   my $color  = substr $s, 4, 5;       # black
                   my $middle = substr $s, 4, -11;     # black cat climbed the
                   my $end    = substr $s, 14;         # climbed the green tree
                   my $tail   = substr $s, -4;         # tree
                   my $z      = substr $s, -4, 2;      # tr

               You can use the substr() function as an lvalue, in which case EXPR must itself be an lvalue.  If you assign something shorter than LENGTH, the
               string will shrink, and if you assign something longer than LENGTH, the string will grow to accommodate it.  To keep the string the same length
               you may need to pad or chop your value using "sprintf".

               If OFFSET and LENGTH specify a substring that is partly outside the string, only the part within the string is returned.  If the substring is
               beyond either end of the string, substr() returns the undefined value and produces a warning.  When used as an lvalue, specifying a substring
               that is entirely outside the string is a fatal error.  Here's an example showing the behavior for boundary cases:

                   my $name = 'fred';
                   substr($name, 4) = 'dy';            # $name is now 'freddy'
                   my $null = substr $name, 6, 2;      # returns '' (no warning)
                   my $oops = substr $name, 7;         # returns undef, with warning
                   substr($name, 7) = 'gap';           # fatal error

               An alternative to using substr() as an lvalue is to specify the replacement string as the 4th argument.  This allows you to replace parts of the
               EXPR and return what was there before in one operation, just as you can with splice().

                   my $s = "The black cat climbed the green tree";
                   my $z = substr $s, 14, 7, "jumped from";    # climbed
                   # $s is now "The black cat jumped from the green tree"

               Note that the lvalue returned by the 3-arg version of substr() acts as a 'magic bullet'; each time it is assigned to, it remembers which part of
               the original string is being modified; for example:

                   $x = '1234';
                   for (substr($x,1,2)) {
                       $_ = 'a';   print $x,"\n";      # prints 1a4
                       $_ = 'xyz'; print $x,"\n";      # prints 1xyz4
                       $x = '56789';
                       $_ = 'pq';  print $x,"\n";      # prints 5pq9
                   }