这个问题是关于Apple libm C数学库中sin函数的实现，可用here. 主罪功能从line 736开始. 简而言之:当x足够小，以至于距离sin(x)最近的浮点数只有x时，为什么该函数要执行其结果从未被使用的浮点相乘，然后返回x，而不是立即返回x？
该功能根据输入x的幅度xk使用不同的方法.我感兴趣的情况是，当xk足够小时，与sin(x)最近的浮点数只是x，lines 809-829. 我猜在这种情况下，该函数只会执行return x. 它的作用如下:
/* Make t0 volatile to force compiler to fetch it at run-time
rather than optimizing away the multiplication.
*/
static volatile const double t0 = 1/3.;
/* Get t0 once. If we wrote "t0*t0", the compiler would load it
twice, since it is volatile.
*/
const double t1 = t0;
/* Perform a multiplication and pass it into an assembly construct to
prevent the compiler from knowing we do not use the result and
optimizing away the multiplication.
*/
__asm__("" : : "X" (t1*t1));
// Return the floating-point number nearest sine(x), which is x.
return x;
因此，它创建了一个新的静态volatile const double t0等于1/3.，另一个具有相同值的const double t1，以一种声称编译器无法优化的方式将t1乘以本身..然后返回x.Why does it do this, instead of just returning 104 at the start?