Why is it displaying the rest of the line?

Question

linux

char str_[32];
fgets(str_, 30, stdin);
printf("%s\n", str_);

If you enter lines of smaller length, then everything seems to be fine. True, sometimes there is a request for a string several times in a row without output. I don't understand why either.
Well, the main problem is that if I mean a longer string, then it will cut it off. Displays the required amount and immediately below displays everything that is left. How to explain it??? ubuntu?

Answer 1

printf("%s\n", str_);
I think the problem is in this place.
Try outputting like this:
And here's how fgets() works:

The fgets() function reads at most num-1 characters from the input stream and places them in the character array pointed to by str. Characters are read until a newline or EOF is read, or until the specified limit is reached. At the end of reading characters, a null character is placed immediately after the last character. The newline character is preserved and becomes part of the array addressed by the str element.

Answer 2

char str_[32] = {0};

Answer 3

How to explain it

It looks like your code is running in a loop. You type a line and press enter -- fgets reads the first 30 characters, leaving the rest in the input buffer -- printf prints them out, the loop starts again, fgets reads the next 30 characters -- whatever was left in the buffer from the previous read. If you want to flush the buffer use fflush(stdin).

Answer 4

void PrintStr(const char *str)
{
   int size = strlen(str);
   for(int i = 0; i < size; ++i)
      std::cout << str[i];
}