Section 10.2

Examples of the Use of the 101 Instruction

This section contains a number of detailed examples illustrating how various types of radix conversion operations can be performed by the 101 instruction. For a specification of the 101 instruction and the notation used see section 3.10.

If a number, when converted, can be represented by-eight characters or fewer, it can usually be converted by a single 101-instruction, provided that the radices are all integers.

Example 1

If the number in A1 is a binary integer in the range -9999999 (A1)I 99999999 (-107+1 (A1)I 108-1) it can be converted to characters by

101     CHARS     RAD     A1

with  the   program   constants

RAD)         +100000000
10,10,10,10,10,10,10,10+16

the +16 in the last radix-character ensuring that zero is converted as 0.
 

Example 2

If (A1)I represents pence of a sterling quantity in the range -99.19.11 to 999.19.11 inclusive, then it may be converted to characters with minus-sign if negative, and with full stops between pounds, shillings and pence by

101     CHARS     STERL     A1

with

STERL) +240000
10,10,10+16,1+32,2,10,1+32,12

This will convert a zero in the 10/- column as SP; to convert it as 0 the 5th radix should be 2+16.

Note that in these two examples, y is the product of the radices (i.e. y = ∏ ri); this is true generally when converting an integer to eight or fewer characters.  Note that it is not possible to use the 101-function unless all radices are integral, e.g. to convert ounces to quarters, pounds and ounces would require radices such as 2.8 and 1.6 which cannot be produced.

A case in which y is not the product of the radices is that of converting a binary fraction to characters.
 

Example 3

If, in A1, is a non-negative fraction, it may be converted for printing, to 6 decimal places preceded by 0.  Thus:-

 51     A1     1
 00     A1     RD
101     NUM   CON    A1

with

RD)   +0.00000025
CON)   31,63,63,63,63,63,63,63
       2+16,1+32,10,10,10,10,10,10

The word CON) is 01147 = 1.0-2-47, i.e. it is the best single-length approximation to +1.0 attainable.  (A1) is shifted down 1 place to prevent the possibility of overflow (a) on adding the rounding constant which is in RD) and (b) on dividing by 1.0-2-47.  This shift is compensated by giving the first radix r0 the value 2.  This first radix forces a zero which is converted as 0 (because +16).  The second radix (l) also forces a zero which is converted as full-stop (or decimal point).  Thereafter, the digits proper are produced.  Note that since the 0 produced by the radix -character 2+16 is treated as significant, β being set, therefore any left-hand zeros in the fraction will be converted as numeral 0 and not as SP.

If the converted form has more than 8 characters, a single 101-instruction is not sufficient.
 

Example 4

Assuming that A1 contains any single-length integer, convert it for output.

FIRST)

 14

NUM

 0

 

| Clear NUM

 

101

NUM+1

CON

A1

| Convert l.s. 8 bits

 

 66

FIN

 4

 

| Test OVR, if clear, FIN

 

 40

A1

CON

 

| if (A1) 10

 

 65

SEC

A1

 

| Jump if quotient 0

 

 60

SEC

A2

 

| Jump if remainder = 0

 

 81

SEC

A1

 

| if <0, add 1 & test if

 

 14

NUM

30

 

| if =0, put minus sign in NUM

 

 75

...

 0

 

| and exit

CON)

+100000000

 

 

10,10,10,10,10,10,10,10+16

 

SEC)

101

NUM

CON

A1

| convert m.s. 7 digits

FIN)

 75

...

 0

 

| and exit


 

The first 101-instruction produces the l.s. 8 digits in NUM+1.  If the number occupies more than eight characters OVR is set, β is set and θ left clear.  OVR is tested and left clear by the 66-instruction; if the number occupies 8 characters or fewer, OVR is clear, the conversion is completed by the one 101-instruction and so the 66-instruction causes a jump to the exit point.  Otherwise the number is divided by 108.  The quotient in A1 corresponds to the m.s. digits and the remainder in A2 to the l.s. 8 digits.

If this quotient is non-negative, it can be converted directly by a second 101-instruction: the 65-instructicn tests for this case.  If the quotient is negative, it may be that all the l.s. 8 characters are zeros (i.e. the number is an integral multiple of 10).   If so, the m.s. characters can be directly converted by a second 101-instruction.  This case will be indicated by a zero remainder on division by 10, which is detected by the 60-instruction.

The 81-instruction is obeyed if the integer is negative and converts to more than 8 characters, i.e. N -108.   Suppose in fact, N=-(108a+b) where a and b are integers.  Then when N is divided by 108 (in the 40-instruction) the quotient will be -(a+1) and the remainder 108-b.  The digits of b have been correctly converted by the first 101-instruction and thus it is necessary to add 1 to the quotient and convert the result.  In fact if, on adding 1 to the quotient, the result is zero, it follows that OVR was set only because b has eight digits and the minus sign could not be inserted into NUM+1.  In such a case the 81-instruction does not cause a jump, the minus sign character is put in the l.s. end of NUM and the sequence left.  Otherwise a second 101-instruction is obeyed, converting the m.s. digits and inserting the minus sign automatically.
 

Example 5
 

To convert a sterling quantity in the range -9,999,999,999.19.11 to 99,999,999,999.19.11 inclusive, held as pence in A1.
 

 

 14

NUM

  0

 

 

101

NUM+1

CON

A1

 

 66

FIN

  4

 

 

 40

 A1

CON

 

 

 65

SEC

 A1

 

 

 60

SEC

 A2

 

 

 10

 A1

  1

 

 

 61

SEC

 A1

 

 

 14

NUM

 30

 

 

 75

...

  0

 

CON)

+240000

 

 

 

 

10,10,10+16,1+32,2,10,1+32,12

 

+100000000

 

 

 

10,10,10,10,10,10,10,10

SEC)

101

NUM

CON+2

A1

FIN)

 75

...

  0

 

The reasoning behind this sequence of instructions is similar to that for Example 4.  Note that the 81-instruction of Example 4 is'replaced here by a 10-instruction and a 61-instruction.  This is because, if a sterling quantity in the given range is divided by 240000 the quotient may exceed 24 bits and thus cannot be tested completely by an 81-instruction.
 

Example 6

Given a signed fraction P in A1 , convert it to 13 decimal places, preceded by SP0. if positive or by -0. if negative, allowing also for the case of -1.0000000000000

 

 14 

A3 

0 

  

| Clear A3

 

 65

EB

A1

 

| Test if F 0

 

 66

BB

10

A1

| If < 0, test if F=-1.0

 

 12

A1

0

 

| If -1.0, negate fraction

 

115

A3

30

 

| and put minus-sign in A3

BB)

 32

A1

CON+4

 

| Multiply F by 105

 

 31

A2

CON+2

 

| Multiply l.s. half by 108, rounded

 

 61

CC

CON+2

A2

| Test if latter product = 108

 

 10

A1

1

 

| If so, add 1 to m.s.digits

 

 14

A2

0

 

| and clear l.s.digits

CC)

101

NUM

CON

A1

| Convert m.s.digits

 

101

NUM

CON+2

A2

| Convert l.s.digits

 

 00

NUM

A3

 

| Add sign if F negative

 

 75

....

O

 

| Exit

CON)

+200000

 

 

1,2+16,1+32,10,10,10,10,10

 

+100000000

 

 

10+16,10,10,10,10,10,10,10

 

+100000

In this method of solution the two multiplication instructions are used to transform the fraction into two integers, that in A1 representing the five m.s. decimal places and that in A2 representing the l.s. eight decimal places.  The first radix, 1, in CON+1 serves to force a zero, converted as SP in which the minus sign may be set when appropriate.