a == b，指数表示法。八度怪异？
如果我将其输入八度音阶：
a = 8*10^-5;
b = 8.0e-005;
a==b
我会得到答案：
ans = 1
但是如果我将其输入八度音阶：
a = 3*10^-5;
b = 3.0e-005;
a==b
我会得到这个答案
ans = 0
我是否遗漏了什么？这不是测试平等的正确方法吗？
If I type this into octave:
a = 8*10^-5;
b = 8.0e-005;
a==b
I get the answer:
ans = 1
But if I type this into octave:
a = 3*10^-5;
b = 3.0e-005;
a==b
I get this answer
ans = 0
Am I missing something? Is this not the right way to test for equality?
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这是涉及浮点数的计算期间精度损失的结果，例如在 Java 中
3 * Math.pow(10, -5)会给你“2.9999999999999997E-5”，它不再匹配b 在你的第二个例子中。这与浮点值在表达式求值过程中的表示和演变方式有关。对于更详细的观点，您可以在谷歌上搜索“精度损失”后缀以及您用来评估这些表达式的上下文/环境。希望这有帮助。This is a result of loss of precision during computations involving floating point numbers, for example in Java
3 * Math.pow(10, -5)would give you '2.9999999999999997E-5' which no longer matches b in your second example. This is linked to how floating point values are represented and evolve during the evaluation of an expression. For more elaborate perspectives, you can google 'loss of precision' suffixing that with the context / environment you are using to evaluate these expressions. Hope this helps.这是计算机系统中浮点数表示方式的常见问题。
许多十进制“四舍五入”的数字在计算机使用的二进制表示形式中变成了重复分数，并且它们必须在某个点被截断，从而造成非常轻微的精度损失。如果两个浮点数的得出方式不同，则必须将它们与一点点斜率进行比较，例如
abs(ab)abs(ab)abs(ab)abs(ab)abs(ab) < EPSILON其中EPSILON是一个适当的常数，如 1e-12。This is a common problem with the way floating point numbers are represented in computer systems.
Many numbers which are 'round' in decimal become repeating fractions in the binary representation used by the computer, and they have to be truncated at some point, for a very slight loss of precision. If two floating point numbers are arrived at differently, you have to compare them with a little bit of slop, e.g.
abs(a-b) < EPSILONwhereEPSILONis an appropriate constant like 1e-12.