了解C语言中的双精度运算 [英] Understanding double precision operations in C

问题描述

我想了解为什么这段代码:

I would like to understand why this code:

double r,d,rc;
scanf("%lf %lf", &r, &d);
rc = (r * r) - (d/2) * (d/2);
printf("%.2f\n", M_PI * rc);

返回比此结果更精确的结果(没有分配rc变量):

returns more precise result than this one (without rc variable assignment):

double r,d,rc;
scanf("%lf %lf", &r, &d);
printf("%.2f\n", M_PI * (r * r) - (d/2) * (d/2));

另一个相关的问题:为什么n * n比pow(n,2)好?

Another, related, question: why is n * n better than pow(n,2)?

推荐答案

第一个代码示例计算:

M_PI * ((r * r) - (d/2) * (d/2));

第二个计算:

(M_PI * (r * r)) - (d/2) * (d/2);

在大多数编译器中，对pow(n, 2)的调用与n * n相同.将会发出完全相同程序集.这是由于称为强度降低"的优化而引起的-大多数pow()实现将检查指数是否为2，并将该情况简化为一个乘法.未优化的版本稍微贵一点，因为它需要一个函数调用和一些分支.

A call to pow(n, 2) is the same as n * n, on most compilers. The exact same assembly will be emitted. This is due to an optimization called "strength reduction" -- most pow() implementations will check to see if the exponent is 2, and reduce that case to a single multiplication. The unoptimized version is slightly more expensive since it requires a function call and some branching.

请注意，M_PI不是C标准的一部分，因此您可以使用等效的代码，该代码可以编译为完全相同的代码:

Note that M_PI is not part of the C standard, so you can use the equivalent, which compiles to the exact same code:

double M_PI = 4.0 * atan(1.0);

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