我正在完成我的任务,并且必须使用C-Free 5.0。只需要你的帮助来解决这个难题。我想实现一个时间限制,让用户在到期前输入答案。我已经尝试过这段代码,但它在scanf()函数中遇到了阻塞。有没有其他方法,如解锁输入或其他东西。我试图实现'#include <sys/select.h>
',但这个程序没有该库。
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
int main()
{
char st[10];
printf ("Please enter a line of text : ");
time_t end = time(0) + 5; //5 seconds time limit.
while(time(0) < end)
{
scanf("%s", &st);
if(st != NULL)
{
printf ("Thank you, you entered >%s<\n", st);
exit(0);
}
}
main();
}
答案 0 :(得分:1)
这是一个示例程序,展示了如何在O_NONBLOCK
文件描述符上使用stdin
标志。
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#include <unistd.h>
#include <fcntl.h>
#define INPUT_LEN 10
int main()
{
printf ("Please enter a line of text : ");
fflush(stdout);
time_t end = time(0) + 5; //5 seconds time limit.
int flags = fcntl(STDIN_FILENO, F_GETFL, 0);
fcntl(STDIN_FILENO, F_SETFL, flags | O_NONBLOCK);
char answer[INPUT_LEN];
int pos = 0;
while(time(0) < end)
{
int c = getchar();
/* 10 is new line */
if (c != EOF && c != 10 && pos < INPUT_LEN - 1)
answer[pos++] = c;
/* if new line entered we are ready */
if (c == 10)
break;
}
answer[pos] = '\0';
if(pos > 0)
printf("%s\n", answer);
else
puts("\nSorry, I got tired waiting for your input. Good bye!");
}
答案 1 :(得分:0)
由于您fcntl.h
尝试将stdin设置为非阻塞。它不漂亮(主动等待),但如果你没有select
那么这是最简单的方法:
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <fcntl.h>
#include <errno.h>
int main()
{
// get stdin flags
int flags = fcntl(0, F_GETFL, 0);
if (flags == -1) {
// fcntl unsupported
perror("fcntl");
return -1;
}
// set stdin to non-blocking
flags |= O_NONBLOCK;
if(fcntl(0, F_SETFL, flags) == -1) {
// fcntl unsupported
perror("fcntl");
return -1;
}
char st[1024] = {0}; // initialize the first character in the buffer, this is generally good practice
printf ("Please enter a line of text : ");
time_t end = time(0) + 5; //5 seconds time limit.
// while
while(time(0) < end // not timed out
&& scanf("%s", st) < 1 // not read a word
&& errno == EAGAIN); // no error, but would block
if (st[0]) // if the buffer contains something
printf ("Thank you, you entered >%s<\n", st);
return 0;
}
对代码的评论:if (st != NULL)
将始终满足,因为st
是堆栈指针。