問題描述

我正在將一些代碼從 Matlab 移植到 C++.

I am porting some code from Matlab to C++.

在 Matlab 中

format long
D = 0.689655172413793 (this is 1.0 / 1.45)
E = 2600 / D
// I get E = 3.770000000000e+03

在 C++ 中

double D = 0.68965517241379315; //(this is 1.0 / 1.45)
double E = 2600 / D;
//I get E = 3769.9999999999995

這對我來說是個(gè)問題，因?yàn)樵趦煞N情況下我都必須四舍五入到 0(Matlab 的修復(fù))，在第一種情況下(Matlab)變成 3770，而在第二種情況下(C++)變成 3769.

It is a problem for me because in both cases I have to do rounding down to 0 (Matlab's fix), and in the first case (Matlab) is becomes 3770, whereas in the second case (C++) it becomes 3769.

我意識(shí)到這是因?yàn)?C++ 案例中另外兩個(gè)最低有效數(shù)字15".鑒于 Matlab 似乎只能以雙精度存儲(chǔ)多達(dá) 15 個(gè)有效數(shù)字的精度(如上所示 - 0.689655172413793)，我如何有效地告訴 C++ 忽略后面的15"?

I realise that it is because of the two additional least significant digits "15" in the C++ case. Given that Matlab seems to only store up to 15 significant digits of precision in double precision (as shown above - 0.689655172413793), how can I effectively tell C++ to ignore the "15" at the back?

所有計(jì)算均以雙精度完成.

All calculations are done in double precision.

推薦答案

您對 C++ 和 MATLAB 打印雙精度值的不同方式感到困惑.MATLAB 的 format long 只打印15 有效數(shù)字，而 C++ 打印 17 有效數(shù)字.在內(nèi)部兩者使用相同的數(shù)字:IEEE 754 64 位浮點(diǎn)數(shù).為了在 MATLAB 中重現(xiàn) C++ 行為，我定義了一個(gè) 匿名函數(shù) disp17 打印 17 位有效數(shù)字的數(shù)字:

You got confused by the different ways C++ and MATLAB are printing double values. MATLAB's format long only prints 15 significant digits while C++ prints 17 significant digits. Internally both use the same numbers: IEEE 754 64 bit floating point numbers. To reproduce the C++-behaviour in MATLAB, I defined a anonymous function disp17 which prints numbers with 17 significant digits:

>> disp17=@(x)(disp(num2str(x,17)))

disp17 = 

    @(x)(disp(num2str(x,17)))

>> 1.0 / 1.45

ans =

   0.689655172413793

>> disp17(1.0 / 1.45)
0.68965517241379315

您在 MATLAB 和 C++ 中看到的結(jié)果是相同的，它們只是打印不同數(shù)量的數(shù)字.如果你現(xiàn)在繼續(xù)使用相同的常量在兩種編程語言中，你會(huì)得到相同的結(jié)果.

You see the result in MATLAB and C++ is the same, they just print a different number of digits. If you now continue in both programming languages with the same constant, you get the same result.

>> D = 0.68965517241379315 %17 digits, enough to represent a double.

D =

   0.689655172413793

>> ans = 2600 / D %Result looks wrong

ans =

     3.770000000000000e+03

>> disp17(2600 / D) %But displaying 17 digits it is the same.
3769.9999999999995

打印 17 或 15 位數(shù)字的背景:

The background for printing 17 or 15 digits:

• Adouble 需要在不丟失精度的情況下存儲(chǔ) 17 個(gè)有效數(shù)字，這是 C 打印的內(nèi)容.
• 對于最多 15 位數(shù)字，任何數(shù)字都可以從字符串轉(zhuǎn)換為雙精度字符串并返回原始數(shù)字，這正是 MATLAB 所做的.

• A double requires 17 significant digits to be stored without precision loss, which is what C prints.
• For up to 15 digits any number can be converted from string to double to string and results back in the original number, which is what MATLAB does.

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