Word-sized OUT to a byte IO register? Setting the Sequence Controller Register with out instructions in old VGA code

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I'm having trouble understanding how the Sequence Controller Registers are set in this example with some old VGA code:

 mov     dx,SC_INDEX
 mov     ax,0604h
 out     dx,ax   ;disable chain4 mode

The example is from Michael Abrash's VGA book: https://www.phatcode.net/res/224/files/html/ch47/47-02.html

According to the description he wants to set the Memory Mode register which is at Index 4. He wants to disable chain4 which is at Bit Index 3 (all according to the Programmer's Guide to the EGA, VGA and Super VGA Cards by Richard F. Ferraro).

I fail to understand why he puts 2 bytes 06 and 04 into ax and out's them. The out register is only one byte and he doesn't explicitly set the index of the register for addressing. Is there some magic happening that I don't understand?

ioports vga x86 assembly
1 Answers

The I/O ports are 8-bit, but 2/4 consecutive ports can also be treated as a single 16/32-bit port - that out instruction will write to both SC_INDEX and SC_INDEX+1.

You can check eg. the Intel 80386 Reference Programmer's Manual's section on I/O addressing for the port width thing, and osdev.org even seems to mention the practice specifically when discussing that specific port.

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