C：矩阵向量积，乘以两个双倍数给出错误的符号

我试图执行一个简单的矩阵乘以向量乘法，由于某种原因，我在几个乘法的结果中得到了错误的符号。我不知道为什么会发生这种情况，任何指点都将不胜感激

这是我的全部代码，即来自mex的矩阵*向量函数和调用者函数（或任何调用函数）——我正在通过mex从Matlab运行代码

#include "mex.h"

void mxv(int m, int n, double *A, double *b, double *c) {
    double sum;
    int i, j;

    for (i = 0; i < m; i++) {
        sum = 0.0;
        for (j = 0; j < n; j++) {
            sum += A[i * n + j] * b[j];
        }
        c[i] = sum;
    }
}

void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[]) {

    double *A, *b, *c;
    int i, j, Am, An;

    A = mxGetPr(prhs[0]);
    Am = (int)mxGetM(prhs[0]);   
    An = (int)mxGetN(prhs[0]); 

    c = malloc(Am * sizeof(double));
    b = mxGetPr(prhs[1]);

    mxv(Am, An, A, b, c); 

    for (i = 0; i < Am; i++)
        printf("c[%d] = %1.4f\n", i, c[i]);
 }

正确的结果应该是：

1.3290
1.8983
1.7924
0.1799

但我明白了

c[0] = 1.3290
c[1] = 1.8983
c[2] = -0.4923
c[3] = -1.7329

所以前两个是正确的（c[0]和c[1]），但后两个不是

我在代码中添加了一组print语句，试图找出错误发生的位置：

#include "mex.h"

void mxv(int m, int n, double *A, double *b, double *c) {
    double sum;
    int i, j;

    for (i = 0; i < m; i++) {
        sum = 0.0;
        printf("********\n");
        for (j = 0; j < n; j++) {
            printf("A[%d][%d] = %1.10f\nb[%d] = %1.10f\n", i , j, A[i + n * j], j, b[j]);
            printf("A[%d][%d]*b[%d] = %1.10f\n", i, j, j, A[i * n + j] * b[j]);
            sum += A[i * n + j] * b[j];
            printf("sum = %1.10f\n", sum);
        }
        c[i] = sum;
    }
} 

void mexFunction(int nlhs, mxArray *plhs[],
                 int nrhs, const mxArray *prhs[]) {

    double *A, *b, *c;    
    int i, j, Am, An;

    A = mxGetPr(prhs[0]);
    Am = (int)mxGetM(prhs[0]);   
    An = (int)mxGetN(prhs[0]); 
    printf("size(A) = (%d,%d)\n", Am, An);
    for (i = 0; i < Am; i++) {
        for (j = 0; j < An; j++) {
            printf("A[%d][%d] = %1.4f\n", i, j, A[i + Am * j]);
        }
    }

    c = malloc(Am *sizeof(double));

    b = mxGetPr(prhs[1]);
    for (i = 0; i < Am; i++) {
        printf("b[%d] = %1.4f\n", i, b[i]);
    }
    mxv(Am, An, A, b, c);

    for (i = 0; i < Am; i++)
        printf("c[%d] = %1.4f\n", i, c[i]);    
}

但当我研究矩阵向量乘法时，我发现乘法结果在某些情况下由于某种原因具有错误的符号。这是它为i=2打印的内容：

********
A[2][0] = 0.0000000000
b[0] = -0.0015352100
A[2][0]*b[0] = -0.0000000000
sum = 0.0000000000
A[2][1] = 0.0000000000
b[1] = -0.0001116500
A[2][1]*b[1] = -0.0000000000
sum = 0.0000000000
A[2][2] = -1726.2421000000
b[2] = -0.0003765900
A[2][2]*b[2] = 0.6500855124
sum = 0.6500855124
A[2][3] = 2539.6267000000
b[3] = 0.0004498100
A[2][3]*b[3] = -1.1423494859 <- THIS SHOULD BE 1.142349... (no minus sign)
sum = -0.4922639735
********

********
A[2][0]=0.0000000000
b[0]=-0.0015352100
A[2][0]*b[0]=-0.0000000000
总和=0.0000000000
A[2][1]=0.0000000000
b[1]=-0.000116500
A[2][1]*b[1]=-0.0000000000
总和=0.0000000000
A[2][2]=-1726.2421000000
b[2]=-0.0003765900
A[2][2]*b[2]=0.6500855124
总和=0.6500855124
A[2][3]=25396267000000
b[3]=0.0004498100
A[2][3]*b[3]=-1.1423494859

第一个
printf
使用
A[i+n*j]
，而第二个使用
A[i*n+j]
。这些是数组中的转置位置。
是否使用了正确的索引？考虑到
A[2][3]
应该是
2539.6267
，而
A[3][2]
是
-2539.6267
。如果
A
的索引被切换，那么结果是正确的。是的，没错！代表我犯了愚蠢的错误。谢谢你看！啊！当然，愚蠢的错误！将所有索引更改为[i+m*j]成功了。谢谢你的帮助！
size(A) = (4,4)
A[0][0] = -865.6634
A[0][1] = 0.0000
A[0][2] = 0.0000
A[0][3] = 0.0000
A[1][0] = 0.0000
A[1][1] = -17002.6822
A[1][2] = 0.0000
A[1][3] = 0.0000
A[2][0] = 0.0000
A[2][1] = 0.0000
A[2][2] = -1726.2421
A[2][3] = 2539.6267
A[3][0] = 0.0000
A[3][1] = 0.0000
A[3][2] = -2539.6267
A[3][3] = -1726.2421
b[0] = -0.0015
b[1] = -0.0001
b[2] = -0.0004
b[3] = 0.0004

********
A[2][0] = 0.0000000000
b[0] = -0.0015352100
A[2][0]*b[0] = -0.0000000000
sum = 0.0000000000
A[2][1] = 0.0000000000
b[1] = -0.0001116500
A[2][1]*b[1] = -0.0000000000
sum = 0.0000000000
A[2][2] = -1726.2421000000
b[2] = -0.0003765900
A[2][2]*b[2] = 0.6500855124
sum = 0.6500855124
A[2][3] = 2539.6267000000
b[3] = 0.0004498100
A[2][3]*b[3] = -1.1423494859 <- THIS SHOULD BE 1.142349... (no minus sign)
sum = -0.4922639735
********

printf("A[%d][%d] = %1.10f\nb[%d] = %1.10f\n", i , j, A[i + n * j], j, b[j]);
printf("A[%d][%d]*b[%d] = %1.10f\n", i, j, j, A[i * n + j] * b[j]);