It may be a very stupid question but everywhere I read, it says a 32 bit register can represent maximum of 4GByte of memory but should it represent 4Gbit of memory? As 2^2 . 2^30 gives 4G. For addition of byte there must be another factor of 2^3. Can anyone help me if I am missing something here?
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It's 4e9 of whatever unit it's indexing. It could be bits, bytes, or megabytes, depending on context. – Oliver Charlesworth Jan 31 '15 at 13:24
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possible duplicate of [Math behind 4GB limit on 32 bit systems](http://stackoverflow.com/questions/16849702/math-behind-4gb-limit-on-32-bit-systems) – Roman R. Jan 31 '15 at 13:30
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It depends on what is your 32 bits register meaning. If this is byte address, it can handle 4G bytes address space. If it is index of 512 bytes sectors it can handle 2Tera bytes of storage space. And finaly if it is a bit index only 512M bytes (4G bits).
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32-bit register can address 2^32 bytes. 2^32 in decimal is 4,294,967,296. Hence number of addressable BYTES is 4+ billion. 4 GB addressable space.
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