Is this a valid assembly instruction ?
MOV CX,0F9
I mean, as we see there is "0F9" in it, although the source operand looks like "hexadecimal" but there are no "H" symbols used at the end of the instruction (or the end of the source operand).
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How can you be sure the value is hex? Starting literal values with 0 usually indicates octal – cowboydan Nov 17 '12 at 06:42
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Not valid in Nasm. `0x0F9`, `0F9h`, or `$0F9` all indicate hex. – Frank Kotler Nov 17 '12 at 06:58
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@Frank Kotler So as it is not one of the three examples you have wrote, the Nasm would notice it as an invalid instruction ? – mohammad mahed Nov 17 '12 at 08:01
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Right. Nasm would say `expression syntax error`. Don't be afraid to try these things! – Frank Kotler Nov 17 '12 at 16:25
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yes. it is indeed a valid intel x86 syntax, moving 0xF9 to CX register. By convention, we assume that any immediate value is in its hexadecimal representation. The reason why there's a 0 in front of F9 is to ensure that the assembler treats the value as an immediate value (hex), rather than a label. It is not there to differentiate its base-n representation.
JosephH
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