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1. Assembly to Pseudo-Code
Rewrite the following assembly routine into pseudo-code:
Solutions:
Option 1: D0 = 10; D0 = D0 - 2; while (D0 >= 0) { D1 = D1 + 1; D0 = D0 - 2; }
Option 2: D0 = 10; while (D0 - 2 >= 0) { D1 = D1 + 1; D0 = D0 - 2; }
2. Pseudo-Code to Assembly
Rewrite the following pseudo-code into assembly code:
for ( i = 0; i < 7; i++ ) {
x = x + 2;
CEG 320/520: Computer Organization and Assembly Language Programming
Fall, 2004
In Class : Flow Control - General updated: 3-oct-
1 ORG $
2 MOVE.L #10, D
3 CLR.L D
4 LOOP SUB.L #2, D
5 BLT REST
6 ADD.W #1, D
7 BRA LOOP
8 REST END
3. The Condition Code Register and Branch Instructions
Consider these three instructions when answering the following three questions.
3a. What are the contents of D0 after the above instructions have been executed (decimal or hex)?
3b. The CMP instruction sets the N, Z, V, and C bits in the CCR. What are the values of those bits after
the following instruction (1/2 point per bit)?
Solutions:
Option 1 (total of 10 words in the loop): MOVE.W #0, D CLR.L D LOOP CMP #7, D BGE ENDLOOP ADD #2, D ADD #1, D BRA LOOP ENDLOOP NOP
Option 2 (total of 6 words in the loop): MOVE.W #7, D CLR.L D LOOP BLE ENDLOOP ADDQ #2, D SUBQ #1, D BRA LOOP ENDLOOP NOP
Option 3 (total of 4 words in the loop): MOVE.W #7, D CLR.L D LOOP ADDQ #2, D SUBQ #1, D BGT LOOP ENDLOOP NOP
1 CLR.L D
2 TARGET ADDQ.L #6, D
3 CMP.L #8, D
D0: $0000 0006
N Z V C
