Why does not the input of a string work?

Question

#include <stdio.h>
int main()
{
        char temp[1024];
        if(getchar() != 'y')
        {
                printf("no options\n");
                return 1;
        }
        scanf(temp, "%s");
        printf("%s", temp);
}

I get snippet as below. I just want twice input from user. but the first input works, however the second directly skips and the printf("%s", temp); printed unexpected characters. How can I resolve the problem.. thanx

It is better to be consistent and if you return any value from main, then return a valid value (usually 0) from the end of the function too. — Jonathan Leffler, Dec 24 '09 at 05:24
please refer to the usage of the scanf function, the passed arguments has to be in reversed manned... scanf( "%s", temp ); — wrapperm, Dec 24 '09 at 07:54

score 6 · Accepted Answer · answered Dec 24 '09 at 04:37

6

The first parameter to scanf is the format, and the second is the buffer. You have it backwards. Try scanf( "%s", temp );.

answered Dec 24 '09 at 04:37

Asher Dunn

2,384
16
12

1

Or, as we like to say. you have it bass-ackward :-) You should also be aware that scanf should not be used for input in robust applications since there's no way to protect from buffer overruns (use fgets and sscanf instead as one option). But, since you're at the beginner level, you can just file that advice for latter (no insult intended, just pointing out that you probably have no need yet for it). – paxdiablo Dec 24 '09 at 04:43
paxdiable: respectfully disagree with that last sentence. Unless the safe functions are significantly more complicated, why not teach beginners to use safe functions first, and dangerous ones only second? Get 'em during the habit-forming stage! – itowlson Dec 24 '09 at 04:48
@itowlson, as you wish, I've put a safer version in my answer. – paxdiablo Dec 24 '09 at 05:19

paxdiablo · Answer 2 · 2010-09-17T02:03:49.140

Others have already given you the actual answer, and you should accept one of those but feel free to upvote me if you like this sage advice :-)

Use of gets should never be contemplated if you're looking for a robust application. That's because there's no way to guard against a buffer overflow which could render your program insecure.

I prefer a little function like getLine() in the following program. It uses fgets which can be protected from oveflow and is a robust solution.

#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;
}

// Test program for getLine().

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

    rc = getLine ("Enter string> ", buff, sizeof(buff));
    if (rc == NO_INPUT) {
        printf ("No input\n");
        return 1;
    }

    if (rc == TOO_LONG) {
        printf ("Input too long\n");
        return 1;
    }

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

    return 0;
}

Sample runs with 'hello', CTRLD, and a string that's too big:

pax> ./qq
Enter string> hello
OK [hello]

pax> ./qq
Enter string>
No input

pax> ./qq
Enter string> dfgdfgjdjgdfhggh
Input too long

pax> _

score 1 · Answer 3 · answered Dec 24 '09 at 04:40

You need to switch the parameters to scanf.

#include <cstdio>

using namespace std;

int main()
{
    char buf[100];
    while (true)
    {
        if (scanf("%s",buf) == EOF)
        {
            printf("fail");
            return 1;
        }
    }
}

Why does not the input of a string work?

3 Answers3

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