when I compiled the code
"VMOV.I16 q1, #2730; "
I got an error of
Error: immediate out of range
What's may be the reason of this error?I knew that S16 is within [-32768,32767],and what should I do if I what to to store constant 2730 in register Q or D?thanks!
Every ARM instruction is 32bit wide, and only a limited number bits are dedicated to the immediate values - if at all.
2730 is 0xaaa in hex, and as you can see, you need 11 bits to express the literal, and vmov only accepts 8 bits with a two bits for shift left: 8bit<<(n*8); where n can be 0 to 3
The best way of loading ANY 16bit value into a NEON register is (2730 in your case):
movw %[temp], #2730
vdup.16 q1, %[temp]
movw is an ARM instruction that accepts 16bit literals, and vdup.n is a NEON instruction that does the same as vmov.in except the source operand is an ARM integer register instead of an immediate value.
Alternate way:
You can load any 8bit value into a NEON register; You make an 8bit load with the value 0xaa, then clear the most significant four bits of 16bit values
vmov.i8 q1, #0xaa
vbic.i16 q1, q1, #0xf000
Pleas note that however, that this one works only because the bits 16~19 are the same as the bits 0~3 in your case.