如何在MATLAB中更改浮点运算的舍入模式？

Question

我想更改MATLAB中浮点运算的舍入模式。根据IEEE 754-2008，有五种舍入策略：

• 四舍五入到最接近的关系
• 四舍五入到最接近，领带从零开始
• 向零舍入
• 向上舍入（向正无穷大）
• 向下舍入（向负无穷大）

MATLAB是否支持这5种策略？如何在MATLAB中更改浮点运算的舍入模式？

Answer 1

回答

的种类。有一个未公开的feature('setround')函数调用，您可以使用它来获取或设置 Matlab 使用的舍入模式。

所以，它可以做到，但你不应该这样做。:)

警告：这是一个未记录的、不受支持的功能！使用后果自负！

这feature('setround')支持 5 种 IEEE-754 舍入模式中的 4 种：只有一种“最近”模式，我不知道它是“与偶数相关”还是“与零相关”。

支持的模式：

• feature('setround') – 获取当前舍入模式
• feature('setround', 0.5) – 向最近的舍入（不知道它是否与偶数或远离零有关）
• feature('setround', Inf) – 向上取整（朝向 +Inf）
• feature('setround', 0) – 向零舍入
• feature('setround', -Inf) – 向下舍入（朝向 -Inf）

测试注意事项：IEEE-754 舍入模式不影响round()及其相关性。相反，它控制着算术运算在浮点精度限制下的行为方式。

示范

%ROUNDINGEXAMPLE Demonstrates IEEE-754 Rounding Mode control
%
% This uses a completely undocumented and unsupported feature!
% Not for production use!

%% Setup
clear; clc

n = 2000;
X = ones(n)*1E-30; % matrix with n^2 elements
defaultRoundingMode = feature('setround'); % store default rounding mode

%%
feature('setround',0.5);
r1 = prettyPrint('Nearest', sum(X(:)));
%{
  sign   exponent                       mantissa
     0 01110110001 0011010101111100001010011001101001110101010000011110
     | \_________/ \__________________________________________________/
     |      |             ______________________|___________________________
     |      |            /                                                  \
(-1)^0 2^( 945 - 1023) 1.0011010101111100001010011001101001110101010000011110 = 4e-24
%}

%%
feature('setround',-Inf);
r2 = prettyPrint('To -Infinity', sum(X(:)));
%{
  sign   exponent                       mantissa
     0 01110110001 0011010101111100001010011001101001011100000111000110
     | \_________/ \__________________________________________________/
     |      |             ______________________|___________________________
     |      |            /                                                  \
(-1)^0 2^( 945 - 1023) 1.0011010101111100001010011001101001011100000111000110 = 4e-24
%}

%%
feature('setround',Inf);
r3 = prettyPrint('To Infinity', sum(X(:)));
%{
  sign   exponent                       mantissa
     0 01110110001 0011010101111100001010011001101010100011101100100001
     | \_________/ \__________________________________________________/
     |      |             ______________________|___________________________
     |      |            /                                                  \
(-1)^0 2^( 945 - 1023) 1.0011010101111100001010011001101010100011101100100001 = 4e-24
%}

%%
feature('setround',0);
r4 = prettyPrint('To zero', sum(X(:)));
%{
  sign   exponent                       mantissa
     0 01110110001 0011010101111100001010011001101001011100000111000110
     | \_________/ \__________________________________________________/
     |      |             ______________________|___________________________
     |      |            /                                                  \
(-1)^0 2^( 945 - 1023) 1.0011010101111100001010011001101001011100000111000110 = 4e-24
%}

%%
feature('setround',defaultRoundingMode);
r5 = prettyPrint('No accumulated roundoff error', 4e-24);
%{
  sign   exponent                       mantissa
     0 01110110001 0011010101111100001010011001101010001000111010100111
     | \_________/ \__________________________________________________/
     |      |             ______________________|___________________________
     |      |            /                                                  \
(-1)^0 2^( 945 - 1023) 1.0011010101111100001010011001101010001000111010100111 = 4e-24
%}

%% Helper function
function r = prettyPrint(s, r)
    fprintf('%s:\n%65.60f\n\n', s, r); 
end

我得到：

Nearest:
   0.000000000000000000000003999999999966490758963870373537264729

To -Infinity:
   0.000000000000000000000003999999999789077070014108839608005726

To Infinity:
   0.000000000000000000000004000000000118618095059505975310731249

To zero:
   0.000000000000000000000003999999999789077070014108839608005726

No accumulated roundoff error:
   0.000000000000000000000003999999999999999694801998206811298525

致谢

感谢 MathWorks 技术支持的 Ryan Klots 让我直截了当并提供了很好的演示代码！