On x86-64, you need an extra prefix to do 64-bit operations (while 64-bit addressing is the default), so it's a question of "are you sure you need the 64 bits and 32 isn't enough?"

> Shouldn't they be 64 bits on most modern systems then?

Arguably so, but then one would lose the ability to natively name 16-bit integer types because "short" would be 32 bits.

An earlier comment addresses x86-64. AArch64 (pedantically, the A64 instruction set used for AArch64's 64-bit execution mode) is similar, in that addresses are 64 bits wide but ALU instructions typically encode a width bit, called "sf", that selects either 32- or 64-bit data registers and arithmetic. See, for example, https://arm.jonpalmisc.com/latest_aarch64/add_addsub_ext .

It should indeed, and in hindsight it would have been better to move int to 64 bits (especially for C's integer promotion, which only really makes sense when the promotion happens to the register width.

But porting 32-bit code to 64-bit was a big deal back then, and C99 with its new fixed-width integer types overlapped with the first AMD64 CPUs (and Microsoft's MSVC didn't start to support C99 until around 2015 anyway), I guess keeping int on 32-bits in the popular compilers was deemed 'safer' for porting existing code. I guess we can already be lucky that all the big compilers agreed on the same int width.

Memory addresses being 64 bits due to needing to address more than 4 GiB memory doesn't mean most integer instructions operate most efficiently with 64 bits. Instructions for 32 bit integers are still more efficient than 64 bit, whereas 16 bit operands require a prefix byte meaning they're less compact and cache efficient (on AMD64 anyway).