读取C中的可变长度字符串用户输入

Question

我试图读取可变长度的用户输入并执行一些操作(比如在字符串中搜索子字符串).

问题是我不知道我的字符串有多大(文本很可能是3000-4000个字符).

我附加了我尝试的示例代码和输出:

char t[],p[];
int main(int argc, char** argv) {
    fflush(stdin);
    printf(" enter a string\n");
    scanf("%s",t);

    printf(" enter a pattern\n");
    scanf("%s",p);

    int m=strlen(t);
    int n =strlen(p);
    printf(" text is %s %d  pattrn is %s %d \n",t,m,p,n);
    return (EXIT_SUCCESS);
}

输出是:

enter a string
bhavya
enter a pattern
av
text is bav 3  pattrn is av 2

Answer 1

请不要永远使用像不安全的东西scanf("%s")或我个人不喜欢,gets()-有没有办法来防止缓冲区溢出了这样的事情.

您可以使用更安全的输入法,例如:

#include <stdio.h>
#include <string.h>

#define OK       0
#define NO_INPUT 1
#define TOO_LONG 2
static int getLine (char *prmpt, char *buff, size_t sz) {
    int ch, extra;

    // Get line with buffer overrun protection.
    if (prmpt != NULL) {
        printf ("%s", prmpt);
        fflush (stdout);
    }
    if (fgets (buff, sz, stdin) == NULL)
        return NO_INPUT;

    // If it was too long, there'll be no newline. In that case, we flush
    // to end of line so that excess doesn't affect the next call.
    if (buff[strlen(buff)-1] != '\n') {
        extra = 0;
        while (((ch = getchar()) != '\n') && (ch != EOF))
            extra = 1;
        return (extra == 1) ? TOO_LONG : OK;
    }

    // Otherwise remove newline and give string back to caller.
    buff[strlen(buff)-1] = '\0';
    return OK;
}

然后,您可以设置最大尺寸,它将检测线路上是否输入了太多数据,同时刷新线路的其余部分,这样就不会影响您的下一次输入操作.

您可以使用以下内容进行测试:

// Test program for getLine().

int main (void) {
    int rc;
    char buff[10];

    rc = getLine ("Enter string> ", buff, sizeof(buff));
    if (rc == NO_INPUT) {
        // Extra NL since my system doesn't output that on EOF.
        printf ("\nNo input\n");
        return 1;
    }

    if (rc == TOO_LONG) {
        printf ("Input too long [%s]\n", buff);
        return 1;
    }

    printf ("OK [%s]\n", buff);

    return 0;
}