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aldelaro5 Newbie cheater Reputation: 0
Joined: 16 Oct 2015 Posts: 10
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Posted: Mon Jul 24, 2017 10:54 am Post subject: Problem with signed value scan |
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Hi, I am currently developping an external ram search tool for the Dolphin emulator in the goal of making it easy to use for dolphin's purposes.
Since CE is currently the only good ram search that supports big endian type via custom types extensions (required as the GC and Wii are big endian memory), i use CE to compare my scan result count as first test.
However, I noticed something strange about the handling of signed values in scanning. Basically, CE doesn't seem to care about the sign and automatically assumes I meant unsigned, which I found weird to see no option to have a signed value scan. For example, any scan with type bigger than -1 returns no results, it likely converted -1 to unsigned and nothing is bigger than the max range. Same if I scan for smaller than -1, it returns everything.
My question: am I missing something? Does CE has any support to scan for signed values? If not, why?
Thanks.
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Dark Byte Site Admin Reputation: 457
Joined: 09 May 2003 Posts: 25262 Location: The netherlands
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Posted: Mon Jul 24, 2017 12:37 pm Post subject: |
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Practically no one scans for a decimal range that overlaps negative and positive.
If they do, it's usually the float type, which does support it
Also, I really discourage scanning for the value 0, which is another reason
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aldelaro5 Newbie cheater Reputation: 0
Joined: 16 Oct 2015 Posts: 10
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Posted: Mon Jul 24, 2017 12:57 pm Post subject: |
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Dark Byte wrote: | Practically no one scans for a decimal range that overlaps negative and positive.
If they do, it's usually the float type, which does support it
Also, I really discourage scanning for the value 0, which is another reason |
Isn't this weird considering the watcher has the option to interpret as signed though? Because it allows to even parse user input as signed.
Idk, but I expected that signed or unsigned was part of the memory type, it's just interpreting the ram differently in both cases.
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