I was going through K&R C book and got through this --> operator in the precedence table. So, I wondered if there was a similar operator i.e., <-- and wrote the following program:
#include<stdio.h>
void main()
{
int x = 5;
while(0 <-- x)
printf("%d",x);
}
It worked perfectly fine. So why isn't <-- is not considered as an operator? (as it is not in the precedence table!) and what is it's precedence?
--> is not one operator, it is two; (post) decrement and less than. C is whitespace agnostic for the most part, so:
x --> y
/* is the same as */
x-- > y
<-- is the same idea:
x <-- y
/* is the same as */
x < --y
Perhaps you are confusing -> with -->. -> dereferences a pointer to get at a member of the type it refers to.
typedef struct
{
int x;
} foo;
int main(void)
{
foo f = {1};
foo *fp = &f;
printf("%d", fp->x);
return 0;
}
<- is simply not an operator at all.
This is not an operator but two operators: < and --. The code is identical to
#include<stdio.h>
void main()
{
int x = 5;
while(0 < --x)
printf("%d",x);
}
As I see it, that is just a "less than" < operator followed by decreasing the variable x (--). It is not one operator, but two. And -- has precedence over <.
