我最近从大学开始从事C编程(入门课程),现在我们正在做有关患者数据库的最终检查。
我需要将数据从文本文件读取到struct数组(大小为10000)。该文件包含2个字符串数组(个人识别字符串(10个数字,以'-'分隔)和名称字符串),1个包含照片参考的int数组和1个包含每个患者的照片参考数量的整数。我已经尝试过fscanf,但是只要我尝试读取,该程序就会挂起,当我使用fgets时,它会读取整行,并将整数从photo引用数组存储到我的名称数组中(中间一个)。我想知道应该如何去做,我花了很多天试图找出解决方案,但似乎没有任何效果。这是我的文本文件的样子:
123456-1234 Name Name [1, 2, 3, 4]
234567-2345 Name2 Name2 [1, 2]
345678-3456 Name3 Name3 []
这是我的write_to_file函数,该函数在程序退出时写入文件:
void write_to_file(Patient reg[], int *pNr_of_patients){
FILE *fp;
fp=fopen("file.txt","w");
if(*pNr_of_patients>0){
int i,j;
for(i=0;i<*pNr_of_patients;i++){
fprintf(fp,"%s\t%s\t[",reg[i].pers_nr,reg[i].name);
for(j=0;j<reg[i].nr_of_ref-1;j++){
fprintf(fp,"%d, ",reg[i].photo_ref[j]);
}
if(reg[i].photo_ref[j]==0){
fprintf(fp,"]");
}else{
fprintf(fp,"%d]",reg[i].photo_ref[j]);
}
fprintf(fp,"\n");
}
fclose(fp);
}
}
这是我的read_from_file函数,缺少用于读取结尾的int数组值的代码:
编辑:我添加了一个for循环,以从名称字符串中删除以“ [”开头的字符,现在我只需要知道如何将结构末尾的数组值读取到结构的photo引用数组中即可。
void read_from_file(Patient reg[],int *pNr_of_patients){
FILE *fp;
fp=fopen("file.txt","r");
if(fp!=NULL){
reg[*pNr_of_patients].nr_of_ref=0;
int i=0, pos;
while(fgets(reg[*pNr_of_patients].pers_nr,13,fp)!=NULL){
reg[*pNr_of_patients].pers_nr[strlen(reg[*pNr_of_patients].pers_nr)-1]='\0';
fgets(reg[*pNr_of_patients].name,31,fp);
reg[*pNr_of_patients].name[strlen(reg[*pNr_of_patients].name)-1]='\0';
for(pos=0;pos<30;pos++){
if(reg[*pNr_of_patients].name[pos]=='['){
reg[*pNr_of_patients].name[pos]='\0';
}
}
(*pNr_of_patients)++;
}
fclose(fp);
}else{
printf("File does not exist\n");
}
}
这是我的Patient结构的样子:
struct patient{
char pers_nr[12], name[30];
int photo_ref[10], nr_of_ref;
};
typedef struct patient Patient;
在main中调用read_from_file:
int main(void){
Patient patient_register[10000];
int nr_of_patients=0;
read_from_file(patient_register,&nr_of_patients);
database_management(patient_register,&nr_of_patients); //this is where I fill all the data into the array before writing to the file at the end
write_to_file(patient_register,&nr_of_patients);
return 0;
}
最佳答案
我认为扫描输入是C语言中最难的一项。这就是为什么存在诸如cs50之类的库的原因,以简化为新C用户阅读输入的过程。无论如何,我构建了解决方案,但是我重新设计了您的功能。
第一种解决方案是从一行中读取单个Patient
。它不使用sscanf
设置errno的唯一标准调用是对strtol
的调用,该调用用于转换数字。
第二个函数使用sscanf
和一些疯狂的格式字符串构造来保持缓冲区溢出的安全。
所有这些都归结为如何构造输入流以及您对它的信任程度。
#include <stdio.h>
#include <assert.h>
#include <stddef.h>
#include <string.h>
#include <ctype.h>
#include <errno.h>
#include <stdlib.h>
#include <limits.h>
struct patient{
char pers_nr[12];
char name[30];
int photo_ref[10];
size_t nr_of_ref;
};
typedef struct patient Patient;
int patient_read_from_line_1(const char line[], Patient *p)
{
assert(line != NULL);
assert(p != NULL);
// check the first 12 characters ----------
// first 6 chars must be numbers
for (int i = 0; i < 6; ++i) {
if (!isdigit(line[i])) {
return -__LINE__;
}
}
// followed by a single '-'
if (line[6] != '-') {
return -__LINE__;
}
// followed by 4 numbers
for (int i = 7; i < 7 + 4; ++i) {
if (!isdigit(line[i])) {
return -__LINE__;
}
}
// followed by a space
if (line[7 + 4] != ' ') {
return -__LINE__;
}
// read up first field ---------------------
// cool first field checks out
memcpy(p->pers_nr, line, 11);
p->pers_nr[11] = '\0';
line += 12;
// let's omit spaces
while (line[0] == ' ') {
line++;
}
// read up second field --------------------------
// now we should read a two strings separated by a space
// so we should read up until a second space
if (!isalpha(*line)) {
return -__LINE__;
}
const char *pnt_first_space = strchr(line, ' ');
if (pnt_first_space == NULL) {
return -__LINE__;
}
const char *pnt_another_space = strchr(pnt_first_space + 1, ' ');
if (pnt_another_space == NULL) {
return -__LINE__;
}
const size_t name_to_read_length = pnt_another_space - line;
if (name_to_read_length > sizeof(p->name)) {
return -__LINE__;
}
memcpy(p->name, line, name_to_read_length);
p->name[name_to_read_length] = '\0';
// buh two fields done, now the array
line += name_to_read_length;
// let's omit the spaces
while (line[0] == ' ') {
line++;
}
// read up array -----------------------------------
// array
if (line[0] != '[') {
return -__LINE__;
}
line++;
for (size_t numscnt = 0;; ++numscnt) {
if (numscnt >= sizeof(p->photo_ref)/sizeof(*p->photo_ref)) {
return -__LINE__;
}
char *pnt;
errno = 0;
long num = strtol(line, &pnt, 10);
if (errno) {
return -__LINE__;
}
if (!(INT_MIN < num && num < INT_MAX)) {
return -__LINE__;
}
p->photo_ref[numscnt] = num;
line = pnt;
// omit spaces
while (*line == ' ') line++;
// now we should get a comma
if (line[0] != ',') {
// if don't get a comma, we need to get a ]
if (line[0] == ']') {
// cool
++line;
// but remember to save the count
p->nr_of_ref = numscnt + 1;
// cool
break;
}
return -__LINE__;
}
++line;
// omit spaces
while (*line == ' ') line++;
// start again
}
// this needs to be end of line or newline
if (line[0] != '\0' && line[0] != '\n') {
return -__LINE__;
}
// success!
return 0;
}
// ok, ok, ok, let's use sscanf
int patient_read_from_line_2(const char line[], Patient *p)
{
assert(line != NULL);
assert(p != NULL);
int ret;
int pos;
// read up first fiedl and half of the second ------------------
ret = sscanf(line, "%12s %30[^ ] %n", p->pers_nr, p->name, &pos);
if (ret != 2) {
return -__LINE__;
}
line += pos;
// read up another half of the second field -------------------
const size_t cur_name_len = strlen(p->name);
p->name[cur_name_len] = ' ';
char tmp[20];
ret = snprintf(tmp, 20, "%%%d[^ ] [%%n", (int)(sizeof(p->name) - cur_name_len - 1));
if (ret < 0) {
return -__LINE__;
}
ret = sscanf(line, tmp, &p->name[cur_name_len + 1], &pos);
if (ret != 1) {
return -__LINE__;
}
line += pos;
// read up array *sigh* -------------------------------------------
for (p->nr_of_ref = 0;; ++p->nr_of_ref) {
if (p->nr_of_ref >= sizeof(p->photo_ref)/sizeof(*p->photo_ref)) {
return -__LINE__;
}
ret = sscanf(line, " %d%1s%n", &p->photo_ref[p->nr_of_ref], tmp, &pos);
if (ret == 0) {
// hm...
if (line[0] == ']') {
// ach all ok, empty numbers list;
line++;
p->nr_of_ref++;
break;
}
return -__LINE__;
}
if (ret != 2) {
return -__LINE__;
}
line += pos;
if (tmp[0] != ',') {
if (tmp[0] == ']') {
// whoa! success
p->nr_of_ref++;
// cool
break;
}
return -__LINE__;
}
}
// so what's left? - EOF or newline
if (line[0] != '\0' && line[0] != '\n') {
return -__LINE__;
}
// success!
return 0;
}
long patient_read_from_file(FILE *fp, Patient patients[], size_t patients_len)
{
size_t patients_cnt = 0;
char line[256];
// for each line in file
while (fgets(line, sizeof(line), fp) != NULL) {
const int ret = patient_read_from_line_2(line, &patients[patients_cnt]);
if (ret < 0) {
// hanle reading error
return ret;
}
patients_cnt++;
if (patients_cnt > patients_len) {
// no more memory in patients left
return -__LINE__;
}
}
return patients_cnt;
}
void patient_fprintln(FILE *f, const Patient *p)
{
fprintf(f, "%s %s [", p->pers_nr, p->name);
for (size_t i = 0; i < p->nr_of_ref; ++i) {
fprintf(f, "%d", p->photo_ref[i]);
if (i + 1 != p->nr_of_ref) {
fprintf(f, ",");
}
}
fprintf(f, "]\n");
}
int main()
{
FILE *fp;
fp = stdin; // fopen("file.txt","r");
if (fp == NULL) {
return -__LINE__;
}
Patient patients[3];
const long patients_cnt = patient_read_from_file(fp, patients, sizeof(patients)/sizeof(*patients));
if (patients_cnt < 0) {
fprintf(stderr, "patient_read_from_file error %ld\n", patients_cnt);
return patients_cnt;
}
fclose(fp);
printf("Readed %d patients:\n", patients_cnt);
for (size_t i = 0; i < patients_cnt; ++i) {
patient_fprintln(stdout, &patients[i]);
}
return 0;
}
实时版本位于onlinedbg。
这可以简化为100%。这有100%的错误。这只是为了说明人们有时使用什么方法(
strtol
,memcpy
,sscanf
,isdigit
,isalpha
)从输入中读取内容。另外,我将长度修饰符指定为scanf(sscanf(..., "%12s"
)以处理溢出(希望如此)。尝试始终检查scanf和其他标准函数的返回值(也许检查snprintf
返回值有点太多,但是,让我们保持一致)。有所不同,在某些平台上,%n
scanf修饰符碰巧不起作用。也可以将其构建为使用malloc
,realloc
和free
使用动态分配,这两种方式都是在线读取(基本上等于编写GNU getline的自定义版本),从输入中读取字符串,从中读取int的数组患者的投入和动态分配。关于c - 从文本文件读取到结构数组时遇到麻烦,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/52652687/