嗨，我仅限于stdio.h，stdlib.h和string.h，我需要询问用户输入-输入可以是1到6之间的任意数量的字符，但是前两个字符必须是大写字母，其余四个字符必须为0到9之间的数字。
有效输入示例：
无效输入示例：
到目前为止，这是我的尝试（请记住，我几乎没有C的经验，这种解决方案“惯用”的可能性几乎没有，我问这个的原因是我可以学习） ：
Flightcode是一个定义为flightcode[7]的char数组，它位于另一个称为flight的结构中。我先fgets将其放入temp_array[7]中，然后strcpy将其放入flight-> flightcode中，以便附加空终止符，但我不知道有什么更好的方法那个。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_FLIGHTCODE_LEN 6
#define MAX_CITYCODE_LEN 3
#define MAX_NUM_FLIGHTS 50
#define DB_NAME "database"
typedef struct {
int month;
int day;
int hour;
int minute;
} date_time_t;
typedef struct {
char flightcode[MAX_FLIGHTCODE_LEN + 1];
date_time_t departure_dt;
char arrival_city[MAX_CITYCODE_LEN + 1];
date_time_t arrival_dt;
} flight_t;
date_time_t departure_dt;
date_time_t arrival_dt;
char * scanline(char *dest, int dest_len);
int main(){
char temp_string[100];
flight_t flight[MAX_NUM_FLIGHTS + 1];
int correct_code = 0;
printf("Enter flight code>\n");
scanline(temp_string, sizeof(flight->flightcode));
strcpy(flight->flightcode, temp_string);
while(correct_code == 0)
{
for(int i = 0; flight->flightcode[i] != '\0' && correct_code == 0; i++)
{
while((i < 2 && (flight->flightcode[i] <= 64 || flight->flightcode[i] >= 91)) || (i > 1 && (flight->flightcode[i] < 48 || flight->flightcode[i] >= 58)))
{
printf("Invalid input.\n");
scanline(temp_string, sizeof(flight->flightcode));
strcpy(flight->flightcode, temp_string);
}
if((i < 2 && (flight->flightcode[i] > 64 || flight->flightcode[i] < 91)) || (i > 1 && (flight->flightcode[i] >= 48 || flight->flightcode[i] < 58)))
{
correct_code = 1;
}
}
}
}
char * scanline(char *dest, int dest_len){
int i, ch;
i = 0;
for (ch = getchar();
ch != '\n' && ch != EOF && i < dest_len -1; ch = getchar())
dest[i++] = ch;
dest[i] = '\0';
while (ch != '\n' && ch != EOF)
ch = getchar();
return (dest);
}
答案 0 :(得分：2)
扫描集和%n指定符可用于解析输入。
格式字符串"%n%2[A-Z]%n%4[0-9]%n"在三个位置使用%n指定符来捕获处理的字符数。如果字符集中在大写字母集中，则扫描集%2[A-Z]将最多扫描两个字符。如果字符是数字，%4[0-9]将最多扫描四个字符。
如果用sscanf扫描了两个值，则减去要处理的字符数，以确保有两个前导大写字符和六个或更少的总字符，并且结尾的字符为结尾的零。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_FLIGHTCODE_LEN 6
#define MAX_CITYCODE_LEN 3
#define MAX_NUM_FLIGHTS 50
#define DB_NAME "database"
typedef struct {
int month;
int day;
int hour;
int minute;
} date_time_t;
typedef struct {
char flightcode[MAX_FLIGHTCODE_LEN + 1];
date_time_t departure_dt;
char arrival_city[MAX_CITYCODE_LEN + 1];
date_time_t arrival_dt;
} flight_t;
date_time_t departure_dt;
date_time_t arrival_dt;
char * scanline(char *dest, int dest_len);
int main(){
int head = 0, leading = 0, tail = 0;
int correct_code = 0;
int result = 0;
char temp_string[100];
char upper[3] = "";
char digits[5] = "";
flight_t flight[MAX_NUM_FLIGHTS + 1];
do {
printf("Enter flight code>\n");
scanline(temp_string, sizeof(temp_string));
if ( 0 < ( result = sscanf ( temp_string, "%n%2[A-Z]%n%4[0-9]%n", &head, upper, &leading, digits, &tail))) {
if ( 1 == result && 0 == temp_string[leading]) {
correct_code = 1;
break;
}
if ( 2 == result && 2 == leading - head && 7 > tail - head && 0 == temp_string[tail]) {
correct_code = 1;
}
else {
printf ( "invalid input\n");
}
}
else {
printf ( "invalid input\n");
}
} while(correct_code == 0);
printf ( "Input is: %s\n", temp_string);
strcpy(flight->flightcode, temp_string);
return 0;
}
char * scanline(char *dest, int dest_len){
int i, ch;
i = 0;
for (ch = getchar(); ch != '\n' && ch != EOF && i < dest_len -1; ch = getchar()) {
dest[i++] = ch;
}
dest[i] = '\0';
while (ch != '\n' && ch != EOF) {
ch = getchar();
}
return dest;
}
答案 1 :(得分：1)
第一件事，意识到您的问题文本缺少一个问题。此外，您的问题标题没有意义。
无论如何，这是一个可能的，故意非常丑陋的解决方案。方法：您想要执行X，因此您编写了执行X的代码。让我们从scanline()开始：
int scanline(char *dest, int dest_len)
{
int i = 0;
int ch;
while (1) {
// Read
ch = fgetc(stdin);
// Check
if (ch == EOF)
break;
if (ch == '\n')
break;
if (i >= dest_len - 1)
break;
// Use
dest[i] = ch;
++i;
}
dest[i] = 0;
// Is the string finished? Ok!
if (ch == '\n' || ch == EOF)
return 1;
// Otherwise discard the rest of the line. Not ok!
while (ch != '\n' && ch != EOF)
ch = fgetc(stdin);
return 0;
}
我知道这很丑陋，但是我认为弄清楚文件输入中涉及的三个步骤会很有帮助：读取，检查，使用。请注意，如果该行达到要求的字符数（比终止符所能容纳的缓冲区大小小1个字符），它将返回true。
然后您要检查是否：
scanline()成功让我们为此编写代码：
int main(void)
{
flight_t flight;
while (1) {
printf("Enter flight code>\n");
if (!scanline(flight.flightcode, sizeof(flight.flightcode))) {
printf("Too many characters.\n");
continue;
}
int i = 0;
if (flight.flightcode[i] == 0) {
printf("Empty input.\n");
continue;
}
if (flight.flightcode[i] < 'A' || flight.flightcode[i] > 'Z') {
printf("Character %d is not upper case.\n", i);
continue;
}
i++;
if (flight.flightcode[i] == 0)
break;
if (flight.flightcode[i] < 'A' || flight.flightcode[i] > 'Z') {
printf("Character %d is not upper case.\n", i);
continue;
}
i++;
if (flight.flightcode[i] == 0)
break;
if (flight.flightcode[i] < '0' || flight.flightcode[i] > '9') {
printf("Character %d is not a digit.\n", i);
continue;
}
i++;
if (flight.flightcode[i] == 0)
break;
if (flight.flightcode[i] < '0' || flight.flightcode[i] > '9') {
printf("Character %d is not a digit.\n", i);
continue;
}
i++;
if (flight.flightcode[i] == 0)
break;
if (flight.flightcode[i] < '0' || flight.flightcode[i] > '9') {
printf("Character %d is not a digit.\n", i);
continue;
}
i++;
if (flight.flightcode[i] == 0)
break;
if (flight.flightcode[i] < '0' || flight.flightcode[i] > '9') {
printf("Character %d is not a digit.\n", i);
continue;
}
i++;
if (flight.flightcode[i] == 0)
break;
}
}
一些评论：
correct_code设置为1。如果要遍历字符，则必须检查是否有错误并退出循环。 MAX_FLIGHTCODE_LEN，并且可以使用任意数量的字母和数字。当然MAX_FLIGHTCODE_LEN应该等于它们的总和！<ctype.h>的无用要求，并同时使用<stdbool.h>，这使程序员的意图更加清晰。答案 2 :(得分：1)
您的功能scanline的功能不超过标准功能fgets。我建议改用标准功能。删除结尾的换行符'\n'很容易。
我将支票分为3部分：
建议的代码：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_FLIGHTCODE_LEN 6
#define MAX_CITYCODE_LEN 3
#define MAX_NUM_FLIGHTS 50
#define DB_NAME "database"
typedef struct {
int month;
int day;
int hour;
int minute;
} date_time_t;
typedef struct {
char flightcode[MAX_FLIGHTCODE_LEN + 1];
date_time_t departure_dt;
char arrival_city[MAX_CITYCODE_LEN + 1];
date_time_t arrival_dt;
} flight_t;
date_time_t departure_dt;
date_time_t arrival_dt;
int main(void){
char temp_string[100];
flight_t flight[MAX_NUM_FLIGHTS + 1];
int correct_code;
size_t len;
int i;
do
{
/* we first assume the code is correct and set this to 0 on any error */
correct_code = 1;
printf("Enter flight code>\n");
if(fgets(temp_string, sizeof(temp_string), stdin) == NULL)
{
if(feof(stdin)) fprintf(stderr, "no input (EOF)\n");
else perror("fgets");
correct_code = 0;
temp_string[0] = '\0';
}
if(correct_code)
{
len = strlen(temp_string);
/* cut off newline
* Use a loop to handle CR and LF just in case Windows might leave more than one character */
while((len > 0) &&
((temp_string[len - 1] == '\n') ||
(temp_string[len - 1] == '\r')))
{
len--;
temp_string[len] == '\0';
}
if(len > MAX_FLIGHTCODE_LEN)
{
correct_code = 0;
fprintf(stderr, "Input must not be longer than %d characters.\n", MAX_FLIGHTCODE_LEN);
}
if(len == 0)
{
correct_code = 0;
fprintf(stderr, "Empty input.\n");
}
}
/* check first two letters */
for(i = 0; (i < 2) && (i < len) && correct_code; i++)
{
/* you could use function isupper when you make sure the locale is set to "C" */
if((temp_string[i] < 'A') || (temp_string[i] > 'Z'))
{
correct_code = 0;
fprintf(stderr, "first two characters must be uppercase letters. Found '%c' at position %d\n", temp_string[i], i);
}
}
/* check digits starting from 3rd character */
for(i = 2; (i < MAX_FLIGHTCODE_LEN) && (i < len) && correct_code; i++)
{
/* you could use function isdigit here */
if((temp_string[i] < '0') || (temp_string[i] > '9'))
{
correct_code = 0;
fprintf(stderr, "Third to last characters must be digits. Found '%c' at position %d\n", temp_string[i], i);
}
}
if(correct_code)
{
/* we already checked that length is not more than MAX_FLIGHTCODE_LEN, so we don't need strncpy to avoid buffer overflow */
strcpy(flight->flightcode, temp_string);
printf("Valid code: %s\n", flight->flightcode);
}
else
{
fprintf(stderr, "Invalid code.\n");
}
} while(!correct_code);
return 0;
}
答案 3 :(得分：1)
您有一个要求与scanf可以轻松完成的工作不完全匹配，因此我会远离它，而将fgets用作主要读取实用程序。
但是，由于可接受的大写字母和数字字符的数量不受固定的限制，因此我将使用基于状态机的自定义解析器。这可能不是最优雅也不有效的方法，但是它简单，健壮且易于维护。
仅出于演示目的，我在第一个大写字母之前允许使用空白字符，在最后一位数字之后使用空格。因此，以下代码接受此正则表达式模式[ \t]*[A-Z]{1,maxupper}[0-9]{0,maxdigit}\s*之后的任意长行，条件是该代码接收大小至少为maxupper+maxupper+1的缓冲区。它返回一个指向缓冲区成功的指针，否则返回NULL。
正如您所说的，您不能使用ctype宏，我已经定义了与我所使用的等效的ASCII（或从ASCII衍生的任何字符集）。
#define TRUE 1
#define FALSE 0
inline int isupper(int c) {
return c >= 'A' && c <= 'Z'; // only for ASCII and derived
}
inline int isdigit(char c) {
return c >= '0' && c <= '9'; // guarantee per standard
}
inline int isblank(int c) {
return c == ' ' || c == '\t';
}
inline int isspace(int c) {
static const char spaces[] = " \t\r\n\v";
for(const char *s=spaces; *s != '\0'; s++) {
if (c == *s) return TRUE;
}
return FALSE;
}
char *get_string(char *buffer, int maxupper, int maxdigit, FILE *fd) {
char buf[16]; // any size >=2 will fit
char *cur = buffer;
int state = 0, uppersize=0, digitsize=0;
for (;;) { // allow lines longer than buf
if (NULL == fgets(buf, sizeof(buf), fd)) {
*cur = '\0'; // EOF: do not forget the terminating NULL
return state >= 1 ? buffer : NULL; // must have at least 1 char
}
for (char *b=buf; *b!='\0'; b++) {
switch(state) {
case 0: // spaces before first uppercase
if (isblank(*b)) break;
state++;
case 1: // first uppercase
if (! isupper(*b)) {
state = 5; // must read up to \n
break;
}
state++;
case 2: // process uppercase chars
if (! isupper(*b)) {
if (uppersize > 0) state++;
else {
state = 5; // must read up to \n
break;
}
}
else {
if (uppersize >= maxupper) {
state = 5; // must read up to \n
break;
}
*cur++ = *b;
uppersize++;
break;
}
case 3: // process digit chars
if (! isdigit(*b)) {
state++;
}
else {
if (digitsize >= maxdigit) {
state = 5; // must read up to \n
break;
}
*cur++ = *b;
digitsize++;
break;
}
case 4: // allow spaces after last digit
if ('\n' == *b) {
*cur = '\0';
return buffer;
}
if (! isspace(*b)) state++
break;
case 5: // on error clean end of line
if ('\n' == *b) return NULL;
}
}
}
}
然后在您的代码中，您只需这样调用即可：
...
printf("Enter flight code>\n");
if (NULL == get_string(flight->flightcode, 2, 4, stdin)) {
// process the error
...
}
...