2.14 Question on Direct Mapping | Cache Memory | GATE Questions in Computer Architecture in Hindi

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[@KNOWLEDGEGATE_kg]

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[@chanchalsharma1802]

Thank you so much sir , very nice explanation and first time i understand this concept by you otherwise i was scared by this concept but now every single step is clear in my mind

[@KNOWLEDGEGATE_kg]

Glad to hear that❤ Keep learning & supporting! Do visit our website www.knowledgegate.in for more amazing courses & contents 👍😊

[@mysticeye5341]

Very nice explanation sir..

[@Jaspreetsingh-ux6lo]

Very nice Sir

[@birpalsingh3888]

Thank you sir

[@somaseal1846]

I can't understand the representation of E201F in the form of tag, cache line and block offset.....
E in hexadecimal mean 15 but here tag size 4..kindly explain me..

[@vrandachag8413]

the binary conversion of the number will be 1110 0010 0000 1111. Since the tag will be the first 4 digit which will be 'E' and similarly for Cache line it will be of 12 bits which will be '201'. I hope it helps

[@rimasignouclassmate2595]

you just saved my bca exam

[@KNOWLEDGEGATE_kg]

Thanks a lot.. All the best for your exam.. Keep learning and supporting !!