I have a specific example below, which works perfectly fine if integers are inputted (see output1), when I try to scan a char using %d specifier in scanf function call I get the output2 below.
So, my question is if input a char I hope the type specifier should convert it to an equivalent int value, if not a junk value, even in the either case it should print/segfault. But, here I'm getting continuous prints which I feel is wrong as scanf is getting bypassed every single time. I'm pretty unsure what's happening in the background and would like to know the same.
#include <stdio.h>
int main()
{
int a;
while (1){
printf("enter a number:");
scanf("%d", &a);
printf("entered number is %d\n", a);
}
return 0;
}
Output1:
> enter a number:1
> entered number is 1
> enter a number:
> 3
> entered number is 3
> enter a number:4
> entered number is 4
> enter a number:5
> entered number is 5 enter a number:
Output2: for input a
> enter a number:entered number is 32767
> enter a number:entered number is 32767
> enter a number:entered number is 32767
> enter a number:entered number is 32767
> enter a number:entered number is 32767
> enter a number:entered number is 32767
> enter a number:entered number is 32767
> enter a number:entered number is 32767
> enter a number:entered number is 32767
PS: I know this is a stupid question of asking what happens in an invalid case where a type specifier unintended (%d in this case) is used for different type, but I would like to know what happens in the background, if any. Thanks
You may check scanf as @Some programmer dude. You may compare the count arguments succesfully filled (Thanks to @chux)
In your case, scanf didn't find any integer value, reached the end of the input and returned EOF, keeping the
a
variable untouched.On failure it'll return EOF (read http://www.cplusplus.com/reference/cstdio/scanf/#return).
For characters, you better use
getch()
or at least, ask for"%c"
on scanf:The "junk" value you recieve is what was in your program memory, because
a
is not initialized.