x86 - Why is there not a register that contains the higher bytes of EAX?

Question

%AX = (%AH + %AL)

So why not %EAX = (%SOME_REGISTER + %AX) for some register %SOME_REGISTER?

Answer 1

Just for some clarification. In the early microprocessor days of the 1970's, CPUs had only a small number of registers and a very limited instruction set. Typically, the arithmetic unit could only operate on a single CPU register, often referred to as the "accumulator". The accumulator on the 8 bit 8080 & Z80 processors was called "A". There were 6 other general purpose 8 bit registers: B, C, D, E, H & L. These six registers could be paired up to form 3 16 bit registers: BC, DE & HL. Internally, the accumulator was combined with the Flags register to form the AF 16 bit register.

When Intel developed the 16 bit 8086 family they wanted to be able to port 8080 code, so they kept the same basic register structure:

8080/Z80  8086
A         AX
BC        BX
DE        CX
HL        DX
IX        SI    
IY        DI

Because of the need to port 8 bit code they needed to be able to refer to the individual 8 bit parts of AX, BX, CX & DX. These are called AL, AH for the low & high bytes of AX and so on for BL/BH, CL/CH & DL/DH. IX & IY on the Z80 were only ever used as 16 bit pointer registers so there was no need to access the two halves of SI & DI.

When the 80386 was released in the mid 1980s they created "extended" versions of all the registers. So, AX became EAX, BX became EBX etc. There was no need to access to top 16 bits of these new extended registers, so they didn't create an EAXH pseudo register.

AMD applied the same trick when they produced the first 64 bit processors. The 64 bit version of the AX register is called RAX. So, now you have something that looks like this:

|63..32|31..16|15-8|7-0|
               |AH.|AL.|
               |AX.....|
       |EAX............|
|RAX...................|