Why does scanf get stuck in an infinite loop on invalid input? [duplicate]

Answer 1

As others have mentioned, if scanf can't parse the input, it leaves it unscanned.

Generally scanf is a poor choice for interactive input because of this kind of behavior, and because it doesn't match the line-at-a-time interface experienced by the user.

You are better off reading one line into a buffer using fgets. Then parse that line using sscanf. If you don't like the input, throw the whole line away and read another one.

Something like this:

#include <stdio.h>

int main(void)
{
  char line[256];

  int arg1;
  int isint;

  while (1) {
    printf("Give attacking argument:");
    fgets(line, sizeof line, stdin);
    isint = sscanf(line, "%d",&arg1);
    if (isint) break;

    printf("You did not enter a number.Please enter an argument's number\n");
  }

  printf("Thanks for entering %d\n", arg1);

  return 0;
}

(For production code you'll want to handle long lines, check return codes, also check for trailing garbage after the number, etc.)

Actually, an even better approach would be to not use scanf if you just want to read an integer, and instead use strtol. That gives you a handy pointer to the character just after the number, and you can check that it's whitespace or nul.

Answer 2

When scanf is confronted with a non-digit it will not consume any input and return that zero integers were read. The non-digit will stay in the input for the next call to scanf that will behave the same as the first call, etc.

In answer to your question below. You could use fgetc to parse at least one character, but this will give the error messages for every character already typed. Typically I think you want to skip until a newline. To this end you could use fgets as suggested by poolie. Or you could add the following after scanf.

int ch;
if (isint == 0)
while ((ch = fgetc(stdin)) != EOF && ch != '\n')
{
     /* Skip characters */
}

P.S: In your case it is probably better to put it just before the first printf in the loop.