C - 为什么 puts 行为异常，而 printf 在使用 scanf 时却没有？

Question

I was testing some code earlier that looked like this:

char input;
int data;

while(scanf("%c %d", &input, &data) != EOF) {
    puts("foo");
}

And when I read it with the following input:

c 8
c 10
c 15
c 18
^D

This was my output:

foo
foo
foo
foo
foo
foo
foo
foo

Although I was only really expecting to have foo printed once; that is, once the EOF signal was sent. I also tried running it outside my IDE and manually in the terminal (because CLion has issues with EOF), and when I ran it manually through the terminal, foo would print once after the first input, and then double print each time after.

However, when I use printf instead of puts inside the loop, my code runs as expected. I want to know why puts was causing an issue here? I imagine it has something to do with the IO stream, but I am not exactly sure why this is happening.

Is this undefined behavior? If so, why is that happening?

EDIT : Someone in the comments said that they were not getting the same issue. I've attached a screenshot of my code + terminal. I am also running gcc-9.4 if that makes any difference.

enter image description here

Answer 1

The root of the problem is that %c does not skip over leading whitespace.

Here's what's happening - your input stream contains the following sequence:

{ 'c', ' ', 8, '\n', 'c', ' ', 10, '\n', 'c', ' ', 15, '\n', 'c', ' ', 18, '\n' }

The first call to scanf reads 'c' into input and 8 into data and returns 2 (for two successful conversions and assignments). The next call to scanf reads '\\n' into input and tries to read 'c' into data - that's a matching failure, so data isn't assigned. However, since '\\n' was successfully read into input , scanf returns 1 (which is not EOF ). The next call to scanf reads ' ' into input and 10 into data .

Lather, rinse, repeat. Depending on where you are in the sequence, scanf will return 2 , 1 , or 0 , none of which are the same as EOF . This is why you get multiple foo outputs per line, because you're not reading what you think you're reading.

There are two things you need to do to fix this:

1. Put a blank in front of the %c specifier - this will tell scanf to skip over leading whitespace;

2. Instead of testing against EOF , test against 2 - you are expecting 2 successful conversions and assignments per input line:

while ( scanf( " %c %d", &input, &data ) == 2 )
  puts( "foo" );

Answer 2

Don't compare to EOF, count the number of valid conversions. In particular, on the input you show:

scanf("%c %d", &input, &data) 

will read the first c into input , assign the value 8 to data , and return 2. The next character in the input stream is \\n , so the second scanf stores that in input and is unable to get an integer value out of c , so it does not write anything to data and returns 1. The next call to scanf writes c to input and sets data to 10. Rinse and repeat.

You probably meant to write:

while(scanf(" %c %d", &input, &data) == 2) {

(note the space before the %c ).

Answer 3

After the first line, "%c %d" is consuming the prior line's '\\n' and then chokes on 'c' as an int , necessitating another loop to process the line .

• Use a leading space to consume whitespace.

• Check results.

---

// while(scanf("%c %d", &input, &data) != EOF) {
int count;
while((count = scanf(" %c %d", &input, &data)) != EOF) {
  printf("count:%d input:%c data:%d\n", 
      count, count >= 1 ? input : '?', count >= 2 ? data : -1);
}