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int main (int argc, const char * argv[]) { PrintChars( 32, 47 ); PrintChars( 48, 57 ); PrintChars( 58, 64 ); PrintChars( 65, 90 ); PrintChars( 91, 96 ); PrintChars( 97, 122 ); PrintChars( 123, 126 ); return 0; } PrintChars() declares a local variable, c, to act as a counter as we step through a range of chars: void char PrintChars( char low, char high ) { c;

The advantage of a list in this case is that you can map the result directly to individual variables Script 18-6 shows how you do that Script 18-6 on split_name(the_name) set first_name to first word of the_name set last_name to last word of the_name return {first_name, last_name} end split_name set {first_name, last_name} to split_name("Paul Revere") --> first_name = "Paul" --> last_name = "Revere" As you can see in this example, since the subroutine returned a list with two items and the list of variables you assigned the result to also has two items, each item from the returned list was assigned to its corresponding item in the subroutine s returned variable This is as close as it gets to returning multiple values.

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We ll use low and high to print a label for the current line, showing the range of ASCII characters to follow. Notice that we use %d to print the integer version of these chars. %d can handle any integer types no bigger than an int.

This means this part of the scope chain is set in stone before the function ever runs However, any variables, functions, or arguments contained by the call and global objects comprising that scope chain are live Then, when you invoke a function, JavaScript adds any variables defined locally within the function, named parameters, the arguments object, and this to the scope chain So these variable objects will differ every time you invoke the function JavaScript looks in this complete set of items in the scope chain when it is looking up the value of a variable In other words, when you invoke a function that uses a variable, JavaScript can only use the variable if it is declared in the scope chain Normally, after an invoked function returns, everything added to the scope chain when you invoked the function is destroyed.

Another fact that makes this example easy to follow is that the list names of the variables you used in the subroutine s return statement inside the subroutine are the same as the list names of the variables you assigned the subroutine s result to in the calling statement..

printf( "%d to %d ---> ", low, high );

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