FINDS ALL PRIME NUMBERS LESS THAN GIVEN

   METHOD IS TO FILL ODD LOCATIONS IN % C ARRAY A WITH ONE AND THEN DELETE
   MULTIPLES OF ALL POSSIBLE % C FACTORS

   READ(GIVEN); !OBTAIN DATA

   %BEGIN %INTEGER I, SQ, TAB %BYTEINTEGERARRAY A(2:GIVEN - 1) A(2) = 1;
   !SPECIAL CASE ONLY EVEN PRIME %CYCLE I = 3, 1, GIVEN - 1 %IF I&I = 0
   %THEN A(I) = 0 %ELSE A(I) = 1 %REPEAT ! I = 3 %WHILE I < MAX
   FACTOR(GIVEN) %CYCLE %IF A(I)#0 %THEN %START SQ = I2 %WHILE SQ < GIVEN
   %CYCLE A(SQ) = 0; !THIS NUMBER NOT PRIME SQ = SQ + 2*I %REPEAT %FINISH
   I = I + 2 %REPEAT ! NOW PRINT OUT ANSWERS AT TEN % C TO A LINE !
   PRINTSTRING('PRIMES LESS THAN ') WRITE(GIVEN,4) NEWLINES(2) ! TAB = 0
   %CYCLE I = 2, 1, GIVEN; % C !ERROR. SHOULD BE 'GIVEN - 1' %IF A(I)#0
   %THEN %START WRITE(I,5) TAB = TAB + 1 %IF TAB = 10 %THEN %START TAB = 0
   NEWLINE %FINISH %FINISH %REPEAT %END; !OF INNER BLOCK %INTEGERFN MAX
   FACTOR(%INTEGER N) ! !RETURNS ANSWER SUCH THAT % C (ANSWER - 1)2 < = N
   < ANSWER2 ! %INTEGER I I = 1 %WHILE I2 < = N %THEN I = I + 1 %RESULT =
   I %END; !OF INTEGER FUNCTION MAX FACTOR %ENDOFPROGRAM

   When this program is run with a datum of 99 the following output is
   obtained.

   PRIMES LESS THAN 99 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61
   67 71 73 79 83 89 97

   MONITOR ENTERED FROM IMP ARRAY BOUND FAULT 99 ENTERED FROM LINE 41 OF
   BLOCK STARTING AT LINE 12 LOCAL VARIABLES TAB = 5 SQ = 105 I = 99
   ENTERED FROM LINE 12 OF BLOCK STARTING AT LINE 1 LOCAL VARIABLES SQ =
   NOT ASSIGNED GIVEN = 99 STOPPED AT LINE 41

   Appendix 2 The object code produced for an optimising compilation of a
   piece of IMP.

   given integer i, j, k, n integer array a(0:100), b(0:n) then the code
   produced for cycle i = 1, 1, n j = j + a(i) k = k + b(i + 1) repeat is
   as follows USING STACKFRAME, 9 USING CODEBASE, 10 *PROLOGUE TO THE
   CYCLE L 4,A80 SET GR4 TO POINT AT DYNAMIC ARRAY B L 5,K ASSIGN K TO GR5
   LA 6,1 ASSIGN INCREMENT TO GR6 L 7,N AND FINAL VALUE(N) TO GR7 L 8,J
   ASSIGN J TO GR8 LA 1,1 ASSIGN I TO GR1 AND SET TO 1 END OF
   PROLOGUE=START OF CYCLE BODY CYCO1 LR 2,1 SLL 2,2 SET GR2 = 4I A 8,A(2)
   J = J + A(I) STATIC ARRAY A IS IN STACKFRAME (UNLIKE B) A 5,4(2,4) K =
   K + B(I + 1) BXLE 1, 6, CYCO1 REPEAT *CYCLE EPILOGUE ASSUMING ALL
   REGISTERS NEED TO BE UNSET *I.E. NEXT STATEMENT IS BRANCH OR ROUTINE
   CALL. ST 5,K RETURN K TO STORE ST 8,J RETURN J TO STORE ST 7,I SET I TO
   FINAL VALUE

   This code is not the optimum possible since it is possible to rearrange
   the cycle so that it is of the form cycle
   [MATH: <semantics><mrow><mi>i</mi><mo>=</mo><mn>4</mn><mo
   separator="true">,</mo><mn>4</mn><mo>∗</mo><mi>n</mi></mrow><annotation
   encoding="application/x-tex"> i = 4, 4 * n </annotation></semantics>
   :MATH]
   i=4,4∗n this saves a LR and SLL in the innerloop at the expense of more
   in the prologue and epilogue.

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   CLARK, S. R. (1966). The main feature of Atlas Autocode, The Computer
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   70-WSK-03. MILLARD, G. E., REES, D. J., and WHITFIELD, H. (1973). The
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   computer systems, Edinburgh University Ph.D. Thesis. REES, D. J.
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