how to read scanf with spaces

Question

I'm having a weird problem

i'm trying to read a string from a console with scanf()

like this

scanf("%[^\n]",string1);

but it doesnt read anything. it just skips the entire scanf.

I'm trying it in gcc compiler

Worser

I see you didn't take my advice in your other question ( #2719095 ) about avoiding scanf completely? Sigh. – Seigniory

Answer 1

Trying to use scanf to read strings with spaces can bring unnecessary problems of buffer overflow and stray newlines staying in the input buffer to be read later. gets() is often suggested as a solution to this, however,

From the manpage:

Never use gets(). Because it is impossible to tell without knowing the data in advance how many characters gets() will read, and because gets() will continue to store characters past the end of the buffer, it is extremely dangerous to use. It has been used to break computer security. Use fgets() instead.

So instead of using gets, use fgets with the STDIN stream to read strings from the keyboard

Marbles

Answer 2

That should work fine, so something else is going wrong. As hobbs suggests, you might have a newline on the input, in which case this won't match anything. It also won't consume a newline, so if you do this in a loop, the first call will get up to the newline and then the next call will get nothing. If you want to read the newline, you need another call, or use a space in the format string to skip whitespace. Its also a good idea to check the return value of scanf to see if it actually matched any format specifiers.

Also, you probably want to specify a maximum length in order to avoid overflowing the buffer. So you want something like:

char buffer[100];
if (scanf(" %99[^\n]", buffer) == 1) {
    /* read something into buffer */

This will skip (ignore) any blank lines and whitespace on the beginning of a line and read up to 99 characters of input up to and not including a newline. Trailing or embedded whitespace will not be skipped, only leading whitespace.

Skylight

Answer 3

I'll bet your scanf call is inside a loop. I'll bet it works the first time you call it. I'll bet it only fails on the second and later times.

The first time, it will read until it reaches a newline character. The newline character will remain unread. (Odds are that the library internally does read it and calls ungetc to unread it, but that doesn't matter, because from your program's point of view the newline is unread.)

The second time, it will read until it reaches a newline character. That newline character is still waiting at the front of the line and scanf will read all 0 of the characters that are waiting ahead of it.

The third time ... the same.

You probably want this:

if (scanf("%99[^\n]%*c", buffer) == 1) {

Edit: I accidentally copied and pasted from another answer instead of from the question, before inserting the %*c as intended. This resulting line of code will behave strangely if you have a line of input longer than 100 bytes, because the %*c will eat an ordinary byte instead of the newline.

However, notice how dangerous it would be to do this:

scanf("%[^n]%*c", string1);

because there, if you have a line of input longer than your buffer, the input will walk all over your other variables and stack and everything. This is called buffer overflow (even if the overflow goes onto the stack).

Ume

Answer 4

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

char *text(int n);

int main()
{
char str[10];
printf("enter username : ");
scanf(text(9),str);
printf("username = %s",str);

return 0;
}

char *text(int n)
{
fflush(stdin);fflush(stdout);
char str[50]="%",buf[50],st2[10]="[^\n]s";
char *s;itoa(n,buf,10);
// n == -1 no buffer protection
if(n != -1) strcat(str,buf);
strcat(str,st2);s=strdup(str);
fflush(stdin);fflush(stdout);
return s;
}

Osteology