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The IADD instruction in IJVM adds two 1-word numbers. When I add EEEEEEEE to itself I get DDDDDDDC. What happens to the carry 1? How can I get it? Is it saved in a register?

Boann
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user2925939
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  • The MIC1 that interprets IJVM only has two condition codes, N and Z. The carry out from the ALU is not stored. The microarchitecture could be modified to store the carry out, like it stores the N and Z bits. – downeyt Nov 11 '15 at 04:46
  • @downeyt Would you mind if I quote your comment in my answer? – Jonathon Reinhart Nov 11 '15 at 12:22
  • @Jonathon Reinhart - You may quote my comment in your answer. – downeyt Nov 11 '15 at 13:09
  • If you create a table for adding 2 two-bit numbers and showing the results, without the carry bit, you might get an idea of how to detect carry without needing a carry flag. – downeyt Nov 29 '15 at 22:51

1 Answers1

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It appears that the carry-out bit is lost.

No version of the IJVM Assembly Language Specification that I've come across says anything about a carry-out bit, or carry flag.

IADD    Pop two words from stack; push their sum

downeyt adds:

The MIC1 that interprets IJVM only has two condition codes, N and Z. The carry out from the ALU is not stored. The microarchitecture could be modified to store the carry out, like it stores the N and Z bits.

Community
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Jonathon Reinhart
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