Originally posted by 波不动 at 2009-12-18 19:40:


其实我最早用积分算法测试过了，也只有加减乘除，但结果没有记录。可能肯定的是gcc的劣势依然明显无疑，谢谢你的建议。我会做一下这个测试的。

#include
#include
#include
#include

// Function to be integrated
// Define and prototype it here
// | sin(x) |
#define INTEG_FUNC(x) fabs(sin(x))

// Prototype timing function
double dclock(void);

int main(void)
{
// Loop counters and number of interior points
unsigned int i, j, N;
// Stepsize, independent variable x, and accumulated sum
double step, x_i, sum;
// Timing variables for evaluation
double start, finish, duration, clock_t;
// Start integral from
double interval_begin = 0.0;
// Complete integral at
double interval_end = 2.0 * 3.141592653589793238;

// Start timing for the entire application
start = clock();

printf(" \n");
printf(" Number of | Computed Integral | \n");
printf(" Interior Points | | \n");
for (j=2;j<27;j++)
{
printf("------------------------------------- \n");

// Compute the number of (internal rectangles + 1)
N = 1 << j;

// Compute stepsize for N-1 internal rectangles
step = (interval_end - interval_begin) / N;

// Approx. 1/2 area in first rectangle: f(x0) * [step/2]
sum = INTEG_FUNC(interval_begin) * step / 2.0;

// Apply midpoint rule:
// Given length = f(x), compute the area of the
// rectangle of width step
// Sum areas of internal rectangle: f(xi + step) * step

for (i=1;i {
x_i = i * step;
sum += INTEG_FUNC(x_i) * step;
}

// Approx. 1/2 area in last rectangle: f(xN) * [step/2]
sum += INTEG_FUNC(interval_end) * step / 2.0;

printf(" %10d | %14e | \n", N, sum);
}
finish = clock();
duration = (finish - start);
printf(" \n");
printf(" Application Clocks = %10e \n", duration);
printf(" \n");

return 0;
}

Originally posted by yalefield at 2009-12-6 23:34:
（1）编译时间是如何获得的？

（2）是否考虑了“库”的问题？是静态还是动态（共享）？
这不仅与可执行程序的大小有关，而且和连接(Link)时间有关。
俺猜测，MinGW和BCB得到的文件大小一样，说明它们都静态连 ...