【发布时间】：2017-03-29 18:23:36
【问题描述】：
我在编译代码时遇到问题。 我不断收到这些错误。
“C: 194 警告传递 'matrix_column_subtract' 的参数 3 使指针从整数而不进行强制转换”
"C: 12 注意：预期为 'double**' 但参数类型为 'int'
“C: 194 错误函数'matrix_column_subtract'的参数太少
我想我知道发生了什么我正在调用 matrix_column_multiply 是一个 void，我需要将它称为我认为的指针指针，我不知道如何更改它。如果有人对我如何编译它有一些想法，将不胜感激！
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define DEBUG 0
#define MAX_ITER 10000
double *eigen (int n, double **A);
void qr_decomp (int n, double **A, double **Q, double **R);
void matrix_copy_column(double **msrc, int col1, double **mdst,int col2, int rows);
void matrix_column_subtract(double **m1, int c1, double **m2, int c2, int rows);
void matrix_column_multiply(double **m, int c, double k, int rows);
int main() {
int i, n = 126;
double *eig, **Am;
FILE *BinInp, *TxtOut;
/* Input Code: Reads bv, Am in binary form from CGrad.bin */
if ( (BinInp = fopen("CGrad.bin","r")) == NULL ) {
fprintf(stderr, "Cannot open matrix binary file INPUT... exiting\n");
exit(-1);
}
Am = (double**)malloc (n*sizeof(double*));
Am[0] = (double*)malloc (n*n*sizeof(double));
for (i = 1; i < n; ++i) {
Am[i] = Am[0] + i*n;
}
for (i = 0; i < n; i++) {
if (i==0) { /* Read one extra line that is discarded (bv is still in bin file) */
if (!fread(Am[i], sizeof(double), n, BinInp)) {
fprintf(stderr, "Cannot read row Am[%03d] of matrix... exiting\n", i+1);
exit(1);
}
}
if (!fread(Am[i], sizeof(double), n, BinInp)) {
fprintf(stderr, "Cannot read row Am[%03d] of matrix... exiting\n", i+1);
exit(1);
}
}
if (fclose(BinInp) == EOF) {
fprintf(stderr, "Cannot close matrix binary file INPUT... exiting\n");
exit(-1);
}
/* COMPUTE EIGENVALUES HERE USING FUNCTIONS. RETURN EIGENVALUES (AS 1D VECTOR) TO eig. */
if (DEBUG) printf ("Calling eigen\n");
eig = eigen (n, Am);
/* Output Code: Writes eig in text form to Eigs.txt */
if ( (TxtOut = fopen("Eigs.txt", "w")) == NULL ) {
fprintf(stderr, "Cannot open matrix text file OUTPUT... exiting\n");
exit(-1);
}
for (i = 0; i < n; i++) {
fprintf (TxtOut, "%18.14e ", eig[i]);
}
if (fclose(TxtOut) == EOF) {
fprintf(stderr, "Cannot close matrix text file INPUT... exiting\n");
exit(-1);
}
return 0;
}
double* eigen (int n, double **Acur)
{
double err = 1, eps = 1e-2;
double ndenom, nnumer, temp;
double *eig, **Anex, **Qsub, **Rsub;
int i, j, k, iters = 1;
/* Malloc memory for the three matricies */
Anex = malloc (n*sizeof(double*));
Anex[0] = malloc (n*n*sizeof(double));
for (i = 1; i < n; ++i) {
Anex[i] = Anex[0] + i*n;
}
Qsub = malloc (n*sizeof(double*));
Qsub[0] = malloc (n*n*sizeof(double));
for (i = 1; i < n; ++i) {
Qsub[i] = Qsub[0] + i*n;
}
Rsub = malloc (n*sizeof(double*));
Rsub[0] = malloc (n*n*sizeof(double));
for (i = 1; i < n; ++i) {
Rsub[i] = Rsub[0] + i*n;
}
/* Malloc memory for the return eig vector */
eig = malloc (n*sizeof(double));
for (i = 0; i < n; i++) {
eig[i] = 0;
}
/* Enter main iteration loop for eigenvalues */
while (err > eps && iters < MAX_ITER) {
/* QR Decompose Acur then find next iterate value in Anex */
qr_decomp (n, Acur, Qsub, Rsub);
// FIND NEXT ITERATE VALUE, PUT IN Anex.
/* Determine relative error change, reset "old" iterate value. */
ndenom = 0;
nnumer = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
temp = Anex[i][j]-Acur[i][j];
ndenom += temp*temp;
nnumer += Anex[i][j]*Anex[i][j];
Acur[i][j] = Anex[i][j];
}
}
err = sqrt(ndenom)/sqrt(nnumer);
/* Increment the iteration count and report error */
if (iters % 25 == 0) {
printf ("Error at end of iteration %05d = %14.10f %%\n", iters, 100*err);
}
++iters;
}
printf ("Error at end of iteration %05d = %14.10f\nCONVERGED.\n", iters-1, err);
if (iters == MAX_ITER) {
printf ("WARNING: MAX_ITER iterations reached!...\n");
}
/* Copy diagonal entries of Acur into eig for return to main */
for (i=0; i<n; i++) {
eig[i] = Acur[i][i];
}
return eig;
}
void qr_decomp (int n, double **Adec, double **myQ, double **myR)
{
int i, j, k; /* Loop Variables: this is all you should need! */
double **T, **S;
double r;
T = malloc (n*sizeof(double*));
T[0] = malloc (n*n*sizeof(double));
for (i = 1; i < n; ++i) {
T[i] = T[0] + i*n;
}
S = malloc (n*sizeof(double*));
S[0] = malloc (n*n*sizeof(double));
for (i = 1; i < n; ++i) {
S[i] = S[0] + i*n;
}
/* Main loop for decomposition */
for (i=0; i<n; i++) {
/* Column i of Q is initially column i of A */
matrix_copy_column(Adec,i,myQ,i,n);
/* For j < i-1, Perform the dot product between the j row of Q and the
i row of A to determine the R(j,i) value, then insure the Q i column
is orthogonal to all other Q columns by subtracting existing Q columns
from it. */
for (j = 0; j < i; j++) {
//r[j,i] = Qj^T * Ui
matrix_copy_column(myQ,j,T,0,n);
matrix_copy_column(Adec,i,S,0,n);
for (k=0; k<n; k++) {
r += T[k][0] * S[k][0];
}
/* Determine the R diagonal as the magnitude of the Q column, then
normalize the Q column (make it a unit vector). */
//Qi = Ui
myR[j][i] = r;
// Something wrong here.
// There is one parameter missing, as matrix_column_subtract needs 5 parameters and
// only 4 are given.
// Also, matrix_column_multiply is defined as returning a void, whereas the 3rd parameter
// of matrix_column_subtract should be a double **
matrix_column_subtract(myQ,i,matrix_column_multiply(T,0,r,n),j);
}
}
}
/* Copies a matrix column from msrc at column col1 to mdst at column col2 */
void matrix_copy_column(double **msrc, int col1, double **mdst,int col2, int rows) {
int i = 0;
for (i=0; i<rows; i++) {
mdst[i][col2] = msrc[i][col1];
}
}
/* Subtracts m2's column c2 from m1's column c1 */
void matrix_column_subtract(double **m1, int c1, double **m2, int c2, int rows) {
int i = 0;
for (i=0; i<rows; i++) {
m1[i][c1] -= m2[i][c2];
}
/*return m1;*/
}
void matrix_column_multiply(double **m, int c, double k, int rows) {
int i = 0;
for (i=0; i<rows; i++) {
m[i][c] *= k;
}
/*return m;*/
}
【问题讨论】：
-
注意：像
double **这样的东西不是矩阵（又名二维数组），也不能指向一个。指针不是数组。而且我们没有调试服务。阅读How to Ask 并提供minimal reproducible example。通过一些诊断向我们扔一堵代码墙并不是这里的工作方式。 -
我试图弄清楚那些 for 循环在
qr_decomp开头的作用。T有空间存放ndouble *的列表，而T[0]有空间存放n*n数字，但T[i] = T[0] + i*n;在做什么？这是在T[0]中分配一个二维数组，然后T的其余部分指向T[0]中的行吗？为什么？这是一种允许T[x][y]工作而只需分配两个内存块的技术吗？
标签： c