COMMENT ⊗   VALID 00002 PAGES
C REC  PAGE   DESCRIPTION
C00001 00001
C00002 00002	begin "balance"
C00005 ENDMK
C⊗;
begin "balance"

comment This will scan the address file, counting the number of customers who
	owe us amounts of money.  The classes will be 0-10, 10-20, 20-40, and
	over-40. At the end, it will print out the number in each class;

comment The program operates by retaining 4 variables, and incrementing them
	each time a customer is examined.  Later, we might actually print out
	new files with the customers in them, or create a sorted version
	of the file (sorted by amount owed);

define crlf = "('15&'12)";

comment the count variables;
integer xm, x0, x1, x2, x3, x4, total;
integer balance;

boolean eof;
integer flag, ichan, daddr, brchar, blf;
string addr;

comment Set breaktable, to read one address, up to the dollar sign;
setbreak(daddr←getbreak,"$",null,"IS"); 

total ← xm ← x0 ← x1 ← x2 ← x3 ← x4 ← 0;
eof←0;

comment Open channels for input and output;
open(ichan←getchan,"DSK",0,19,0,450,brchar,eof);

lookup(ichan,"ADDFIL.DSK",flag);
if flag then print("Can't find address file.",crlf);

comment The main loop.  Proceed through the address file one at a time;
balance←input(ichan, daddr);

do begin "next address"
	balance←intin(ichan);
	if balance < 0 then xm ← xm+1
	else if balance = 0 then x0 ← x0+1
	else if balance ≤ 10 then x1 ← x1+1
	else if balance ≤ 20 then x2 ← x2+1
	else if balance ≤ 40 then x3 ← x3+1
	else x4 ← x4+1;
	total ← total+1;
	balance←input(ichan,daddr);
    end "next address"
until eof;

comment Now print out the totals;
print(crlf, "Totals:",crlf, crlf, "Number of customers: ",total, crlf,
	crlf,"Below zero: ", xm, crlf, "Zero: ", x0, crlf,
	"0-10: ", x1, crlf, "10-20: ", x2, crlf, "20-40: ", x3, crlf, "Over 40: ",
	x4, crlf, crlf);

close(ichan);

end "balance";