C - read scanf until feof(stdin), don't output error

Question

I am using the scanf() function in my program to scan an integer. The way I am doing it is:

while (!feof (stdin))
if (scanf("%d", &height) != 1) { puts("Wrong input!"); return 1; }

The problem is, that after actually doing EOF, I get the wrong input and return 1, due to the scanf not returning 1. How should I solve this issue?

In the case I get EOF, I return 0 and don't do anything else. — prkist, Nov 17 '13 at 19:09
If you want to read only one integer from STDIN why do you use while(!feof(stdin)) ? — Deck, Nov 17 '13 at 19:18
http://stackoverflow.com/questions/5431941/while-feof-file-is-always-wrong — William Pursell, Nov 18 '13 at 00:42

chux - Reinstate Monica · Answer 1 · 2013-11-18T00:44:12.387

2

There are many problems with this approach. (EOF not active until after trying to read, text input like 'A' creates infinite loop, etc.) Instead :

char buf[100];
while (fgets(buf, sizeof buf, stdin) != NULL) {
  if (sscanf(buf, "%d", &height) != 1) {
    puts("Wrong input!");
    return 1;
    }
  do_something(height);
}
return 0;

edited Nov 18 '13 at 00:44

answered Nov 18 '13 at 00:36

chux - Reinstate Monica

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score 1 · Accepted Answer · answered Nov 17 '13 at 19:21

As the documentation for scanf states, it will return the number of items successfully matched and assigned. This can be the number of items you gave if completely successful or any number less than that, down to zero, if it failed to match input. In the event of EOF, it will return EOF. So, check for it!

int result = scanf("%d", &height);
if (result != EOF && result != 1) {
    // We've failed!
}

C - read scanf until feof(stdin), don't output error

2 Answers2

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