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Why example in Rule A5-0-2 is non-complient

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Why example in Rule A5-0-2 is non-complient
kafka Offline
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#1
27-07-2022, 07:13 PM
Could someone please explain why the "if (u && (boolean1 <= boolean2));" in Rule A5-0-2 is non-compliant.
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Why example in Rule A5-0-2 is non-complient - by kafka - 27-07-2022, 07:13 PM

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