Neon Comparison [duplicate]

问题

As per the Neon documentation:

If the comparison is true for a lane, the result in that lane is all bits set to one. If the comparison is false for a lane, all bits are set to zero. The return type is an unsigned integer type.

I have written a small piece of code to check this and I observed the result as 0 and -1 instead of 0 and 1. Can any one tell me the reason behind this?

Code:

float c1[4] = {12.0f,12.0f,12.0f,12.0f}; 
float c2[4] = {13.0f,12.0f,9.0f,12.0f};
float32x4_t t1,t2;
uint32x4_t rq;
t1 = vld1q_f32(c1);
t2 = vld1q_f32(c2);
rq = vceqq_f32(t1,t2);
printf("start\n");
for( int i = 0;i < 4; i++){
    printf("%d\n",rq[i]);
}
printf("end\n");

Result:

start
0
-1
0
-1
end

回答1:

You're displaying the values as if they are signed, because you use %d as the printf format specifier. If you use %u you'll see the equivalent unsigned values (0 or UINT_MAX).

Note that signed-ness is not particularly important in this context - you can treat the comparison results as unsigned masks (all 0s or all 1s) or as signed integer values (0 or -1) - it's the same underlying bit pattern - just use whatever makes sense in the particular context where you are using the result.

回答2:

Here as per the neon document ,It is also mentioned as all bit will set to one .

That is Here :

Binary 1111 1111 1111 1111 1111 1111 1111 1111 = Decimal 4294967295

And use %u instead of %d .You will get result as:

Result: 0 4294967295 0 4294967295

You are getting result as -1 means all bits are set to one .That You can use for further logical operation like bitwise and and bitwise or operation .

来源：https://stackoverflow.com/questions/29984542/neon-comparison

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arm

simd

neon

cortex-a