how can I get the ARM MULL instruction to produce its output in a uint64_t in gcc?

问题

I would like to introduce some assembly code into a c99 codebase. I want to use the UMULL instruction from the ARM CPU to multiply 2 uint32_t and get the result immediately into a uint64_t.

Now a uint64_t needs 2 registers, so how do I specify the output and the constraints of the asm block?

回答1:

Good question!

The following code outputs what you want using GCC -O or higher without resorting to assembler:

uint32_t a, b;
uint64_t c;
...
c = (uint64_t)a * (uint64_t)b;

or if you feel you must use machine-specific asm, you can go:

uint32_t a, b;
uint64_t c;

asm ("umull %Q0, %R0, %1, %2" : "=r"(c) : "r"(a), "r"(b));

c's register name is the first of the register pair, and %Q and %R pick out the lower and upper 32-bit registers of the pair. See gcc/config/arm/arm.md -> umulsidi3 for an example.

However, if you can stay in C, that gives the optimizer a chance to do more and is kinder on readers of your program.

回答2:

The umull instruction produces its results into two 32-bit registers. I suggest explicitly reassembling the 64-bit value with something like that:

/* assuming the 64-bit result was stored in "hi" (upper
   half) and "lo" (lower half) */
uint64_t v = ((uint64_t)hi << 32) | (uint64_t)lo;

The compiler optimizer should notice that the left-shift is pure data routing, and the resulting code should be fine. To be sure, just use -S to check the compiler output.

来源：https://stackoverflow.com/questions/2193542/how-can-i-get-the-arm-mull-instruction-to-produce-its-output-in-a-uint64-t-in-gc