c - 如何用 C 编程语言将随机访问文件中的记录放入数组中？

Question

如何将随机访问文件 DATA.data 中的记录放入数组 allRecs[10] 中？

/*Reading a random access file and put records into an array*/
#include <stdio.h>

/*  somestruct structure definition */
struct somestruct{
    char namn[20];
    int artNr;
}; /*end structure somestructure*/

int main (void) {
    int i = 0;
    FILE *file; /* DATA.dat file pointer */
    /* create data with default information */
    struct somestruct rec = {"", 0};
    struct somestruct allRecs[10]; /*here we can store all the records from the file*/
    /* opens the file; exits it file cannot be opened */
    if((file = fopen( "DATA.dat", "rb")) == NULL) {
        printf("File couldn't be opened\n");
    } 
    else { 
        printf("%-16s%-6s\n", "Name", "Number");
        /* read all records from file (until eof) */
        while ( !feof( file) ) { 
            fread(&rec, sizeof(struct somestruct), 1, file);
            /* display record */
            printf("%-16s%-6d\n", rec.namn, rec.artNr);
            allRecs[i].namn = /* HOW TO PUT NAME FOR THIS RECORD IN THE STRUCTARRAY allRecs? */
            allRecs[i].artNr = /* HOW TO PUT NUMBER FOR THIS RECORD IN THE STRUCTARRAY allRecs? */
            i++;
        }/* end while*/
        fclose(file); /* close the file*/
    }/* end else */
    return 0; 
}/* end main */

Answer 1

我立即想到了两种方法。首先，您可以简单地分配，如下所示:

allRecs[i] = rec;

但是，从您的代码来看，您甚至不需要它 - 您可以直接在适当的元素中读取:

fread(&allRecs[i], sizeof(struct somestruct), 1, file);
/* display record */
printf("%-16s%-6d\n", allRecs[i].namn, allRecs[i].artNr);
i++;

顺便问一下 - 您确定该文件永远不会包含超过 10 条记录吗？因为如果这样做的话，你会遇到很多麻烦......