Scanf (% d, ..) in a loop when the given char does not block once

Question

I wrote this code:

char str[10]; while(1) { scanf("%d",str); } 

if a char is given (for example, 'a'), the loop just continues without stopping and asking for more input (scanf does not block suddenly)

why?:-)

Answer 1

Because a not a decimal integer. scanf will try to read it, but not execute or advance its internal pointer, so it will endlessly try to read the same a as a decimal and fail.

consider the following program:

 int d; char c; scanf("%d", &d); scanf("%c", &c); 

if you enter a , the first scanf will fail, and the second will read ' a ' in c . This should explain the problem :)

Answer 2

Since it cannot force 'a' into integer form, therefore it "returns it to the stream" (there is no real stream there, just a manner of speaking). Then your endless loop makes him try the same procedure again. Infinitely.

Answer 3

Your format string is %d not %s . Scanf() searches for an integer, not a string. Your loop holds forever because it says while(1) , so it never ends.

Answer 4

He may be trying to read int , but then he notices that there is no int in the data stream, so he returns immediately. If it does not consume the input stream, it will do it again.

Answer 5

input buffer has a....

your scanf trying to read int:

  a ....
      ^
      oops: cannot be int.  stop scanfing

and in the next loop, a still exists.

scanf is generally difficult to use correctly, but you should always check its return value:

 int chk; chk = scanf(...); /* pseudo-code */ if (chk != EXPECTED_VALUE) { /* HANDLE ERROR! */ } 

Answer 6

To fix this, you need to check that scanf () actually reads the values ​​you need. scanf () returns the number of successfully read items. So, save the return value, check if it is 1. If it is 0, then 0 elements have been read, and you need to clear the stream before trying again. (edit: Well, maybe it’s not necessary to empty it, but at least go past the offensive data in the stream)

As others said, he tries to read an integer, sees 'a' , cannot read an integer, but then 'a' is still in the stream when he tries again.