副标题#e#
本来指望可以或许有div_fr1x16之类的函数来实现fract16的除法，可是很遗憾vdsp居然不直接提供这样的函数，让人颇为难过，预计是因为其CPU不直接提供fract除法的缘故。不外vdsp文档内里提供了一个做除法的例子：
fract16 saturating_fract_divide(fract16 nom, fract16 denom)
{
int partialres = (int)nom;
int divisor = (int)denom;
fract16 rtn;
int i;
int aq; /* initial value irrelevant */
if (partialres == 0) {
/* 0/anything gives 0 */
rtn = 0;
} else if (partialres >= divisor) {
/* fract16 values have the range -1.0 <= x < +1.0, */
/* so our result cannot be as high as 1.0. */
/* Therefore, for x/y, if x is larger than y, */
/* saturate the result to positive maximum. */
rtn = 0x7fff;
} else {
/* nom is a 16-bit fractional value, so move */
/* the 16 bits to the top of partialres. */
/* (promote fract16 to fract32) */
partialres <<= 16;
/* initialize sign bit and AQ, via divs(). */
partialres = divs(partialres, divisor, &aq);
/* Update each of the value bits of the partial result */
/* and reset AQ via divq(). */
for (i=0; i<15; i++) {
partialres = divq(partialres, divisor, &aq);
}
rtn = (fract16) partialres;
}
return rtn;
}
这个计较进程在不打开优化的环境下将需要500个cycle，在打开优化之后将只需要42个cycle！对较量于将fract16转换为float举办计较照旧要快得多。
#p#副标题#e#
可是很显然，例子中给出的这个函数并没有思量到标记位，好比计较0.2 / -0.4这样的式子时它将直接返回1，因而我们需要对它举办适当的修改。
fract16 saturating_fract_divide(fract16 nom, fract16 denom)
{
int partialres;
int divisor;
fract16 rtn;
int i;
int aq; /* initial value irrelevant */
int sign = (nom ^ denom) & 0x8000;
partialres = abs_fr1x16(nom);
divisor = abs_fr1x16(denom);
if (partialres == 0) {
/* 0/anything gives 0 */
rtn = 0;
} else if (partialres >= divisor) {
/* fract16 values have the range -1.0 <= x < +1.0, */
/* so our result cannot be as high as 1.0. */
/* Therefore, for x/y, if x is larger than y, */
/* saturate the result to positive maximum. */
rtn = 0x7fff;
} else {
/* nom is a 16-bit fractional value, so move */
/* the 16 bits to the top of partialres. */
/* (promote fract16 to fract32) */
partialres <<= 16;
/* initialize sign bit and AQ, via divs(). */
partialres = divs(partialres, divisor, &aq);
/* Update each of the value bits of the partial result */
/* and reset AQ via divq(). */
for (i=0; i<15; i++) {
partialres = divq(partialres, divisor, &aq);
}
rtn = (fract16) (partialres);
if(sign)
rtn = negate_fr1x16(rtn);
}
return rtn;
}
优化之前需要522个cycle，优化之后需要50个cycle。
