SMMUL instruction in ARM Assembly

Refer to: http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.dui0553a/CHDHGFEF.html

Writes the most significant signed 32 bits of the result in Rd.

SMMUL is meant to multiply to vairable together but I do not understand how it calculates the "most significant signed 32 bits of the result"

Thanks in advance

When you multiply 2 32-bit value you get a 64-bit value and you have to save the result in two registers since a register is 32-bit. However may be you are not interested in lowest 32-bit and only highest 32-bit.

SMMUL instruction provides you that. First calculates 64-bit result then it may simply discard/truncate lower 32-bits or it can round them into higher 32-bits (SMMULR).

SMULL Rdlo, Rdhi, src1, src2 produces both halves in two output registers. (docs). Most Cortex-M microcontrollers either have both smmul and smull, or neither depending on whether they have a 64-bit multiplier. Cortex-M3 has smull (Thumb-2 group) but not smmul (DSP group).

If you only need the high half, SMMUL avoids overwriting a second output register, and has the rounding version.

$ cat smmul.c 
int smull(int a, int b, int c, int d) {
    asm volatile("smmul r0, r2, r3");
  // only works in debug builds.  Use Extended Asm for real uses
}

int main() {
    int a, b, c;
    a = 0x00000001;
    b = 0x00000001;
    c = smull(0, 0, a, b);
    printf("%x(%d) %x(%d) %x(%d)\n", a, a, b, b, c, c);
    a = 0x40000000;
    b = 0x00000004;
    c = smull(0, 0, a, b);
    printf("%x(%d) %x(%d) %x(%d)\n", a, a, b, b, c, c); 
    return 0;
}

$ smmul
1(1) 1(1) 0(0)
40000000(1073741824) 4(4) 1(1)

A less hacky version of the inline asm that actually tells the compiler what's going on:

// untested, otherwise I'd edit the original test case.
int smull(int a, int b, int c, int d)
{
   // b is unused, kept only to match the original
    int retval;
    asm("smmul %0, %1, %2"
      : "=r"(retval)
      : "r"(c), "r"(d)
      : // no clobbers
    );
    return retval;
}