c – 具有较高尾数的fp如何代表较小的数字？

发布时间：2020-12-16 03:27:57 所属栏目：百科来源：网络整理

导读：我爱FP;每次我想到它,我明白了解它:) This是一个我不明白的例子.我总结8次相同的数字(0.1),我打印结果,总和和“原始”： std::cout.precision(100);int numIteration = 8;double step = 0.1;double sum = 0.0;for(int i = 0; i numIteration; i++) { sum +=

我爱FP;每次我想到它,我明白了解它:)

This是一个我不明白的例子.我总结8次相同的数字(0.1),我打印结果,总和和“原始”：

std::cout.precision(100);

int numIteration = 8;
double step = 0.1;
double sum = 0.0;

for(int i = 0; i < numIteration; i++) {
    sum += step;
}

std::cout << "orig stored as " << numIteration / 10.0 << std::endl;
std::cout << " sum stored as " << sum << std::endl;

0.1存储为0.1000000000000000055511151231257827021181583404541015625,我预计在8和之后,它将被存储大于或等于0.8,其存储为0.8000000000000000444089209850062616169452667236328125.

但结果震惊了我.事实上在8和之后,结果是0.79999999999999993338661852249060757458209991455078125,这是更小的.

另外,如果我检查两者的二进制输出,我可以看到总和比“原始”更“高”：

0.8 stored as binary 0 02222211110 1001100110011001100110011001100110011001100110011001 // smaller
sum stored as binary 0 02222211110 1001100110011001100110011001100110011001100110011010 // higher

但是0.79999999999999993338661852249060757458209991455078125< 0.8000000000000000444089209850062616169452667236328125. 你能照我吗？编辑：对不起,我复制/粘贴二进制文件时出错.他们是对的.

解决方法

每次算术运算后都会发生 IEEE floating-point rounding.四舍五入可能会上升或下降.
如果在每次迭代时打印sum的值,您应该看到：

sum is 0.1000000000000000055511151231257827021181583404541015625
sum is 0.200000000000000011102230246251565404236316680908203125
sum is 0.3000000000000000444089209850062616169452667236328125
sum is 0.40000000000000002220446049250313080847263336181640625
sum is 0.5
sum is 0.59999999999999997779553950749686919152736663818359375
sum is 0.6999999999999999555910790149937383830547332763671875
sum is 0.79999999999999993338661852249060757458209991455078125

你假设四舍五入只能上升.但是,由于“Round to nearest,ties to even”是IEEE 754中的默认舍入模式,因此在每次迭代时都会选择最接近的二进制可表示值,因此结果不必大于0.8.

另一方面

std::cout << 0.1 * 8.0 << std::endl;

会产生预期的

0.8000000000000000444089209850062616169452667236328125

更新：如注释中提到的@Evg,可以使用std::fesetround更改浮点舍入方向.

（编辑：李大同）

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