Why isn't there a "<

This question already has an answer here:

What is the “-->” operator in C++? 21 answers

---

--> 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 <.

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