Matlab Coder C++ matrix multiplication issue

Question

Ali Jooya 2015-5-31

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/221617-matlab-coder-c-matrix-multiplication-issue

编辑： Titus Edelhofer 2015-6-3

Hello, I used Matlab coder to translate some Matlab code into C++. The C++ generated code produces different result from the original Matlab code. I traced the C++ code and found out the problem is with matrix multiplication. I tried a simple code to make sure the bug is not related to my Matlab code and still getting wrong result from generated code. Here is the Matlab code:

function [Final Delta] = Test_Coder(x,y,w)
  Delta = x*y;
  Final= Delta * w;
end

and this is the script that calls the function:

clear all
clc
x = [2,4;5,6]; 
y = [7,3;5,1];
w = [3;5];
[Final Delta] = Test_Coder(x,y,w)

Here is the generated C++ code ("printf" instructions are added later for debugging purpose):

    // 'Test_Coder:3' Delta = x*y;
    for (i0 = 0; i0 < 2; i0++) {
      for (i1 = 0; i1 < 2; i1++) {
        Delta[i0 + (i1 << 1)] = 0.0;
        for (i2 = 0; i2 < 2; i2++) {
          Delta[i0 + (i1 << 1)] += x[i0 + (i2 << 1)] * y[i2 + (i1 << 1)];
          printf("\n Delta[%d] += x[%d] * y[%d] = %lf * %lf >> %lf  ",i0 + (i1 << 1),i0 + (i2 << 1) ,i2 + (i1 << 1), x[i0 + (i2 << 1)],y[i2 + (i1 << 1)] ,Delta[i0 + (i1 << 1)]);
        }
      }
      // 'Test_Coder:5' Final= Delta * w;
      Final[i0] = 0.0;
      for (i1 = 0; i1 < 2; i1++) {
        Final[i0] += Delta[i0 + (i1 << 1)] * w[i1];
        printf("\n Final[%d] +=  Delta[%d] * w[%d] = %lf * %lf >> %lf ",i0, i0 + (i1 << 1), i1, Delta[i0 + (i1 << 1)],  w[i1], Final[i0]);
      }
    }
  }

This is Matlab generated results:

Final =
   152
   300
Delta =
    34    10
    65    21

and here is C++ results:

   Delta[0] += x[0] * y[0] = 2.000000 * 7.000000 >> 14.000000  
   Delta[0] += x[2] * y[1] = 5.000000 * 3.000000 >> 29.000000  
   Delta[2] += x[0] * y[2] = 2.000000 * 5.000000 >> 10.000000  
   Delta[2] += x[2] * y[3] = 5.000000 * 1.000000 >> 15.000000  
   Final[0] +=  Delta[0] * w[0] = 29.000000 * 15.000000 >> 435.000000 
   Final[0] +=  Delta[2] * w[1] = 15.000000 * 5.000000 >> 510.000000 
   Delta[1] += x[1] * y[0] = 4.000000 * 7.000000 >> 28.000000  
   Delta[1] += x[3] * y[1] = 6.000000 * 3.000000 >> 46.000000  
   Delta[3] += x[1] * y[2] = 4.000000 * 5.000000 >> 20.000000  
   Delta[3] += x[3] * y[3] = 6.000000 * 1.000000 >> 26.000000  
   Final[1] +=  Delta[1] * w[0] = 46.000000 * 15.000000 >> 690.000000 
   Final[1] +=  Delta[3] * w[1] = 26.000000 * 26.000000 >> 1366.000000

Any idea what is wrong and how to fix it? I am using Matlab R2015a.

Thanks, Ali

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Walter Roberson 2015-5-31

在 MATLAB Online 中打开

I would change the format slightly

            printf("\n Delta[%d] += x[%d] * y[%d] = %lf * %lf >> %lf  ",i0 + (i1 << 1),i0 + (i2 << 1) ,i2 + (i1 << 1), x[i0 + (i2 << 1)],y[i2 + (i1 << 1)] ,Delta[i0 + (i1 << 1)]);

as a reminder that the values are being added to the existing and that what shows up last on the line is not the current multiplication but rather the cumulative sum for the location.

Ali Jooya 2015-6-1

Thanks Walter. I changed the format.

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Answer 1

Titus Edelhofer 2015-6-1

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/221617-matlab-coder-c-matrix-multiplication-issue#answer_181209

在 MATLAB Online 中打开

Hi,

note that MATLAB uses column format for storing arrays, so you need to initialize

double X[]={2,5,4,6};
double Y[]={7,5,3,1};

to get the same results as in MATLAB.

Titus

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Ali Jooya 2015-6-2

Thanks Titus. That solved the problem! This is a NN code I have converted into C++. It involves a lot of floating point number calculations. For low number of Epochs the results are the same, but over higher Epochs, e.g. 1000, I am getting different results. Any suggestion?

Titus Edelhofer 2015-6-3

编辑：Titus Edelhofer 2015-6-3

Hi Ali,

it depends on "different results". Note that numerical differences can always occur, so it would be necessary to see "how" different they are. With such a high number of Epochs I would be more astonished to see no differences ...

Titus

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