C programming

Question

#include <stdio.h> int main() {     int i = 3;     printf("%d", (++i)++);     return 0; }

why this is giving compile time error?

Comments

it should give undefined behaviour

---

@srestha why?

---

Here it is not the variable, which got post incremented. Here value got post increment.

Similarly This will also give error.
printf("%d", ++5);

---

yes, I missed that

Here i will replaced by 3

And 3 cannot be post or preincrements.

As constant term will not take pre increment or post increment

So, it will give compiler error

Answer 1

Informally, lvalue is the left side value of the statement when we try to do some sort of assignment. The assignemnt is possible only to objects. If we try to assign some values to constants, we will get the 'lvalue error'.
Here,++i outputs 4 and doing 4++ is similar as doing 4 := 4 + 1, which causes lvalue error.

Comments

@monashi

your reasoning is correct, though your explanation is not completely correct with respect to compile time.

Here,++i outputs 4 and doing 4++ is similar as doing 4 := 4 + 1, which causes lvalue error.

what you have said if performed, would be performed during runtime. At compile time l-value error is checked on the basis CFG.

---

Thanks. Can you also share some references?