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Hello!
We've checked your code with AppChecker static analyzer and found possible bugs:
else if ((optype & 0x10) == 0x1c) {
0x10 can be represented as 0000 0000 0001 0000 0x1c can be represented as 0000 0000 0001 1100
There is no such value optype, that using operator '&' with 0x10 will rasult 0x1c, as bits number 3 and 4 will be equal to 0 anyway.
((buf[1] & 0xf0) == 0x08) ? 2 : 1, rd());
0xf0 - 0000 0000 1111 0000 0x08 - 0000 0000 0000 1000 The same as above, the fourth bit as a result of "and" operation will be 0.
Possible defects were found by Echelon Team with AppChecker static analyzer
The text was updated successfully, but these errors were encountered:
Sorry, something went wrong.
There is this code in version 1.4.0 also.
f5948a9
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Hello!
We've checked your code with AppChecker static analyzer and found possible bugs:
0x10 can be represented as 0000 0000 0001 0000
0x1c can be represented as 0000 0000 0001 1100
There is no such value optype, that using operator '&' with 0x10 will rasult 0x1c, as bits number 3 and 4 will be equal to 0 anyway.
https://github.com/radare/radare2/blob/ad54e9ddf18dd88570fec2593920f8f0223d4e8f/libr/anal/p/anal_h8300.c#L333
0xf0 - 0000 0000 1111 0000
0x08 - 0000 0000 0000 1000
The same as above, the fourth bit as a result of "and" operation will be 0.
Possible defects were found by Echelon Team with AppChecker static analyzer
The text was updated successfully, but these errors were encountered: