获取gsl_vector_view的大小

时间:2019-01-05 19:46:39

标签: c++ c vector gsl

如何获得gsl_vector_view的大小?我具有以下用于采样多元高斯向量的功能

void mvnrnd(gsl_vector *x, gsl_matrix *Sigma,gsl_vector *Mu, int K, const gsl_rng *seed){

    gsl_matrix *A = gsl_matrix_alloc(K, K);
    gsl_matrix_memcpy(A, Sigma);
    gsl_linalg_cholesky_decomp (A);

    for (int k=0; k<K; k++){
        gsl_vector_set (x, k, gsl_ran_ugaussian (seed));
        }

    gsl_blas_dtrmv (CblasLower, CblasNoTrans, CblasNonUnit, A, x);
    gsl_vector_add (x, Mu);

    }

为了传递向量和矩阵,我正在做以下事情:

gsl_matrix *MuH    = gsl_matrix_calloc(Kest*Kest,1);

gsl_matrix *vecH   = gsl_matrix_calloc(Kest*Kest,1);
gsl_matrix *SigmaH = gsl_matrix_calloc(Kest*Kest,Kest*Kest);
gsl_matrix_memcpy (SigmaH, &Q_view.matrix);
gsl_vector_view vecH_view = gsl_matrix_subcolumn (vecH, 0, 0, Kest*Kest);
gsl_vector_view MuH_view =  gsl_matrix_column (MuB, 0);
mvnrnd(&vecH_view.vector, SigmaH , &MuH_view.vector, Kest*Kest, seed);

但我收到此错误:

gsl: ./oper_source.c:27: ERROR: vectors must have same length
Default GSL error handler invoked.

所以我想打印gsl_vector_view。怎么做呢? 我使用了& vecH_view.vector->size,它返回错误。

1 个答案:

答案 0 :(得分:1)

“试错”始终对我有用。我在下面提供一个最小的示例:

#include <stdio.h>
#include <stdlib.h>
#include <gsl/gsl_vector.h>
#include <gsl/gsl_matrix.h>
int main()
{
    gsl_matrix * y = gsl_matrix_calloc(6,6);
    gsl_vector_view z = gsl_matrix_column(y,0);
    printf("size = %ld\n",(&z.vector)->size);
    gsl_matrix_free(y);
}

$ gcc -Wall -pedantic-errors -O0 se.c -lgsl -lgslcblas -lm && ./a.out
size = 6