I am currently reading Windows Assembly Language and System Programming book. The book laid out the differences between the processors 8086 and 8088.

8086 : 16-bit wide data bus
8088 : 8-bit wide data bus

The book also says that 8086 is better because the data bus is 8-bit. But I thought that the wider the data bus the better it is since more or larger information can be carried by the bus at the time; which is the reason i am confused.

Can anybody clarify this for me please?
Posted on 2010-12-30 11:13:43 by banzemanga
The 8088 is indeed an 8086 processor on an 8-bit bus.
The 8086 wasn't used on a wide scale though. Nearly all IBM PCs and compatibles in the early days used the 8088 processor (IBM PC and XT).
From a software point-of-view, they are indistinguishable from eachother.
The 8088 was 'better' because the 8-bit bus meant that a much cheaper 8-bit motherboard and chipset could be used. This was the main reason why IBM chose the Intel processor. Otherwise they would probably have gone for the Motorola 68000 processor. All other 16-bit machines used it at the time, since it was a more powerful CPU than the 8086, and it delivered good price/performance.

The first widespread 16-bit x86 processor was the 80286 in the IBM AT. There was an 80186, but like the 8086, it never found widespread use.
Posted on 2010-12-30 11:54:47 by Scali