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
211
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