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Top Forums Programming does any one know how to solve? Post 302227569 by shamrock on Thursday 21st of August 2008 01:11:08 PM
Old 08-21-2008
Quote:
Originally Posted by mlhazan
Do you know why for loop does not work?
Don't need a for loop to go through all the record entries...when EOF is reached fread() returns zero for the number of items read and that can be used as a terminal condition.
You don't need to initialize a 1000 element employee[] array when you will be using only a few elements of it. The size of the employee[] array is known completely at compile time using the sizeof operator after the employee[] array has been initialized. See source changes highlighted in red.
Code:
#include <stdio.h>
#include <stdlib.h>

struct date
{
    int month;
    int day;
    int year;
};
struct empRec
{
    char name[25];
    char room[10];
    int jobLevel;
    struct date startDate;
};

int main()
{
    int entries;        /* number of employee records */
    FILE *data;
    static struct empRec employee[] =
    {
        {"Peter North","4B-208",35400,{10,11,1983}},
        {"John Musa","2B-118",25400,{07,10,1993}},
        {"Paula Jeminova","1A-506",18700,{02,1,1990}},
        {"Patricia Silver","6C-123",52100,{14,21,2000}},
        {"Robert mill","4D-318",42100,{01,19,2008}}
    };

    struct empRec employee2;
    entries = (sizeof employee / sizeof employee[0]);

    if ((data = fopen("database","w+")) == NULL) {
        fprintf(stderr,"Can't Create database\n");
        exit(1);
    }
    if (fwrite(&employee,sizeof(struct empRec), entries, data) != entries) {
        fprintf(stderr,"WRITE ERROR\n");
        exit(1);
    }
    rewind(data);

    /* Replace for loop with the while loop below */
    while(fread(&employee2,sizeof(struct empRec), 1, data) == 1)
        printf("%s %s %d %d %d %d\n", employee2.name, employee2.room, employee2.jobLevel,
            employee2.startDate.month, employee2.startDate.day, employee2.startDate.year);

    fclose(data);
    printf("Created database\n");
    exit(0);
}


Last edited by shamrock; 08-21-2008 at 02:23 PM..
 

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ZGELS(l)								 )								  ZGELS(l)

NAME
ZGELS - solve overdetermined or underdetermined complex linear systems involving an M-by-N matrix A, or its conjugate-transpose, using a QR or LQ factorization of A SYNOPSIS
SUBROUTINE ZGELS( TRANS, M, N, NRHS, A, LDA, B, LDB, WORK, LWORK, INFO ) CHARACTER TRANS INTEGER INFO, LDA, LDB, LWORK, M, N, NRHS COMPLEX*16 A( LDA, * ), B( LDB, * ), WORK( * ) PURPOSE
ZGELS solves overdetermined or underdetermined complex linear systems involving an M-by-N matrix A, or its conjugate-transpose, using a QR or LQ factorization of A. It is assumed that A has full rank. The following options are provided: 1. If TRANS = 'N' and m >= n: find the least squares solution of an overdetermined system, i.e., solve the least squares problem minimize || B - A*X ||. 2. If TRANS = 'N' and m < n: find the minimum norm solution of an underdetermined system A * X = B. 3. If TRANS = 'C' and m >= n: find the minimum norm solution of an undetermined system A**H * X = B. 4. If TRANS = 'C' and m < n: find the least squares solution of an overdetermined system, i.e., solve the least squares problem minimize || B - A**H * X ||. Several right hand side vectors b and solution vectors x can be handled in a single call; they are stored as the columns of the M-by-NRHS right hand side matrix B and the N-by-NRHS solution matrix X. ARGUMENTS
TRANS (input) CHARACTER = 'N': the linear system involves A; = 'C': the linear system involves A**H. M (input) INTEGER The number of rows of the matrix A. M >= 0. N (input) INTEGER The number of columns of the matrix A. N >= 0. NRHS (input) INTEGER The number of right hand sides, i.e., the number of columns of the matrices B and X. NRHS >= 0. A (input/output) COMPLEX*16 array, dimension (LDA,N) On entry, the M-by-N matrix A. if M >= N, A is overwritten by details of its QR factorization as returned by ZGEQRF; if M < N, A is overwritten by details of its LQ factorization as returned by ZGELQF. LDA (input) INTEGER The leading dimension of the array A. LDA >= max(1,M). B (input/output) COMPLEX*16 array, dimension (LDB,NRHS) On entry, the matrix B of right hand side vectors, stored columnwise; B is M-by-NRHS if TRANS = 'N', or N-by-NRHS if TRANS = 'C'. On exit, B is overwritten by the solution vectors, stored columnwise: if TRANS = 'N' and m >= n, rows 1 to n of B contain the least squares solution vectors; the residual sum of squares for the solution in each column is given by the sum of squares of elements N+1 to M in that column; if TRANS = 'N' and m < n, rows 1 to N of B contain the minimum norm solution vectors; if TRANS = 'C' and m >= n, rows 1 to M of B contain the minimum norm solution vectors; if TRANS = 'C' and m < n, rows 1 to M of B contain the least squares solution vectors; the residual sum of squares for the solution in each column is given by the sum of squares of elements M+1 to N in that column. LDB (input) INTEGER The leading dimension of the array B. LDB >= MAX(1,M,N). WORK (workspace/output) COMPLEX*16 array, dimension (LWORK) On exit, if INFO = 0, WORK(1) returns the optimal LWORK. LWORK (input) INTEGER The dimension of the array WORK. LWORK >= max( 1, MN + max( MN, NRHS ) ). For optimal performance, LWORK >= max( 1, MN + max( MN, NRHS )*NB ). where MN = min(M,N) and NB is the optimum block size. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA. INFO (output) INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value LAPACK version 3.0 15 June 2000 ZGELS(l)
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