Commodore PET User Manual page 224

Then. using the MID$ function at line 1101. the leftmost character (a blank repre-
senting a positive sign value) is truncated:
1101 CL$=MID$(CL$.2.LEN(CLS)-I)
CL$=MID$(jtalaI4151 4 1 4 1°12.6)
CL$= 1814151414101
At 1102. if the length of CL$ is shorter than F. zeros from ZERO$ are con-
catenated onto the front of CL$. You will need ta add an assignment statement
for a string of Os for variable ZERO$; we have done sa at line 15. In this case. the
length of CL$ is equal ta F. therefore no leading zeros are needed.
15 ZEF.:O$=" 0(1(1(10(1(10(1(1(1(10 (1(nZ1 "
1102 CL$=LEFT$(ZERO$.F-LENCCL$»+CL$
CL$=LEFT$ (ZERO$.6-6)+CL$
CL$=LEFT$ (ZERO$.O)+CL$
Then. at line 1110, CH$ is assigned the string integer value of AH-BH:
1110 CHS=STRSCINTCAH-BH»
CH$=STR$(ta123455-ta57)
CH$=STR$(ta12339a)
CH$-lta!11213!319151
Then. using the MID$ function, the leftmost blank character is truncated off:
1111 CH$=MIDSCCH$.2.LENCCHS)-1)
CH$=MID$(/taI112131319Ial.2.6)
CH$= 11121313191al
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