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> Main > 32-bit value when concatenating two dwords |
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revolution 05 Mar 2017, 10:45
Assembles fine for me.
Code: use64 thread: mov r8d,'my n' mov r9d,'li m' xor r10,r10 add r10d,r8d shl r10,20h add r10d,r9d;ce 6.6 debug error. ret Perhaps you can post a minimal example that shows the error? |
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05 Mar 2017, 10:45 |
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zhong 05 Mar 2017, 10:51
add r10d,r9d;ce 6.6 debug error.
this is the key, if you use cheat engine create process to debug, the r10 will never become 'my nli m', but instead, become 'li m'. |
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05 Mar 2017, 10:51 |
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revolution 05 Mar 2017, 11:10
You are using the 32-bit register r10d. If you need a 64-bit value then you need to use the full register r10.
Code: use64 thread: mov r8d,'my n' mov r9d,'li m' xor r10,r10 add r10d,r8d shl r10,20h or r10,r9 ;<--- use a 64-bit register here ret |
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05 Mar 2017, 11:10 |
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zhong 05 Mar 2017, 11:33
thread:
mov al,'m' mov bl,'l' xor cx,cx add cl,al shl cx,8h add cl,bl ret this time i have to use dosbox and tasm's td, smile. |
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05 Mar 2017, 11:33 |
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revolution 05 Mar 2017, 11:44
In x86-64 CPUs: When you use a 32-bit register as a destination from any operation the upper 32-bit are zeroed. This is the design of the hardware. It is not a bug in the assembler. And it is not a bug in the CPU. It is by design.
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05 Mar 2017, 11:44 |
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zhong 05 Mar 2017, 11:51
thanks, i just transform my qb64 thread into fasm thread.
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05 Mar 2017, 11:51 |
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