07:33 Jun 15, 2016 |
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English to Serbian translations [PRO] Medical - Medical: Instruments / automatski mikrotom | |||||||
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4 | Postavi broj reči po bloku |
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set block offset Postavi broj reči po bloku Explanation: Postavi broj reči po bloku Remark: Demo of tristate drivers in logisim (controlled registers). Improvement: Multiword Blocks The setup we have described does not take any advantage of spatial locality. The idea of having a multiword block size is to bring into the cache words near the referenced word since, by spatial locality, they are likely to be referenced in the near future. We continue to assume (for a while) that the cache is direct mapped and that all references are for one word. The terminology for “byte offset” and “block offset” is inconsistent. The byte offset gives the offset of the byte within the word so the offset of the word within the block should be called the word offset, but alas it is not in both the 2e and 3e. I don't know if this is standard (poor) terminology or a long standing typo in both editions. The figure to the right shows a 64KB direct mapped cache with 4-word blocks. What addresses in memory are in the block and where in the cache do they go? * The word address = the byte address / number of bytes per word = the byte address / 4 * for the 4-byte words we are assuming. * The memory block number = the word address / number of words per block = * the byte address / number of bytes per block. * The cache block number = the memory block number modulo the number of blocks in the cache. * The block offset (i.e., word offset) = the word address modulo the number of words per block. * The tag = the memory block number / the number of blocks in the cache = * the word address / the number of words in the cache = the byte address / the number of bytes in the cache Consider the cache shown in the diagram above and a reference to word 17003. * 17003 / 4 gives 4250 with a remainder of 3 . * So the memory block number is 4250 and the block offset is 3. * 4K=4096 and 4250 / 4096 gives 1 with a remainder of 154. * So the cache block number is 154 and the tag is 1. * Summary: Memory word 17003 resides in word 3 of cache block 154 with tag 154 set to 1. * http://cs.nyu.edu/~gottlieb/courses/2000s/2007-08-fall/arch/... Sadržaj indeksnog registra se često povećava (izvođenjem instrukcije INX) ili smanjuje (DEX) prilikom izvođenja programske petlje u cilju pristupa do podataka skladištenih u memoriji u obliku tablica. Program koji sledi je primer korištenja indeksnog načina adresiranja, uz promenu sadržaja indeksnog registra s instrukcijom INX. Program izvodi zbrajanje niza od šest brojeva. CLRA OBRIŠI AKUMULATOR A LDAB #06 U AKUMULATOR B POSTAVI BROJ REČI. LDX #$0040 U IX ADRESA PRVOG BROJA. OPET ADDA 0,X PRIBROJI SADRŽAJU AKUMULATORA A SADRŽAJ ODREĐEN SADRŽAJEM INDEKSNOG REGISTRA. INX POVEĆAJ SADRŽAJ INDEKSNOG REGISTRA. DECB SMANJI BROJILO ZA 1. BNE OPET GRANAJ AKO BROJILO NIJE=0 STAA $A2 SKLADIŠTI SUMU NA 00A2. SWI PROGRAMSKI PREKID. http://www.tf.uns.ac.rs/~omorr/radovan_omorjan_003a/VI-sem-3... |
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