while(step > 0.02){
for(;;){
if(x+step <= 100.0 &&
(ttlen = ttl(x+step, y)) < minlength){
minlength = ttlen;
x += step;
} else if(x-step >=0.0 &&
(ttlen = ttl(x-step, y)) < minlength){
minlength = ttlen;
x -= step;
} else if(y+step <= 100.0 &&
(ttlen = ttl(x, y+step)) < minlength){
minlength = ttlen;
y += step;
} else if(y-step >= 0.0 &&
(ttlen = ttl(x, y-step)) < minlength){
minlength = ttlen;
y -= step;
} else break;
}
step /= 2.0;
}
在 x-y 平面上寻找一个对 ttl 函数的极值点。先用 step 为步长寻找，然后步长减半再找，直到满足一定精度为止。
逻辑上过程上都很简单才对，可惜以上代码在 VC2008 运行正确，用 Gcc 却陷入死循环。
调试下发现居然出现了这种情况：在 A 点时发现 B 比 A 小，然后在 B点 发现 C 比 B 小，而在 C点 又发现 A 更小……于是就在这几个点上跳不出来了。
其实以前也看过不少强调处理浮点数比较时要特别注意的地方，但每个地方都特别写一下又很麻烦，而且觉得 double 的精度够高应该不用担心，没想到还是出了问题。VC 里大概对浮点数的比较运算做了特殊处理才没出错，感觉在这种很细节的地方 VC 非常强调安全性，但可能有时候也会觉得多此一举或是担心性能上的影响吧。
总之发现问题所在就好解决了，(ttlen = ttl(x, y+step)) < minlength 改为 (ttlen = ttl(x, y+step)) < minlength - 1e-4 即可。
今后再处理浮点数的问题时一定要相当小心咯
2022年8月18日 13:21
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2023年1月06日 22:41
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