Why ++i++ gives “L-value required error” in C? [duplicate]

Answer 1

The expression ++i++ is evaluated as ++(i++) which is illegal in C as the postfix increment returns a value and prefix incrementing on a that value makes no sense.

What you have is somewhat equivalent to: ++(5) which is obviously illegal as you can apply prefix increment on something that's not a l-value.

Answer 2

This is because you are doing increment to a constant..

In your case

++i => 6
++i++ => 6++ //this is not possible. Gives lvalue error
++i+10 => 6+10 => 16 //this is possible

So doing ++i++ is not good. Post increment/decrement, pre increment/decrement possible only on variables. At runtime your statement will become a constant so it gives lvalue error here.

++i => 6 => i=6;
i++ => 6 => i=7;
printf("%d",i); //7

Answer 3

Generally, you should not be doing this, as it obscures the code.
The reason you're getting the error is that the post-increment has precedent and thus returns an r-value, i.e. ++i++ == ++(i++) which cannot be incremented.

However, you can use (++i)++ since the pre-increment (apparently, on VS2010) returns i itself, an l-value which can be post-incremented.

Answer 4

Because postfix operators have higher precedence than prefix operators, so the expression ++i++ is equivalent to ++(i++), that is equivalent to ++( i + 1). The compiler gives l-value error because you are applying ++ on an expression (i++) that is not a modifiable lvalue, so not a valid expression in c according to increment operator definition.

According to Dennis M. Ritchie's book: "The C Programming Language":

2.8 Increment and Decrement Operators

(page 44)

The increment and decrement operators can only be applied to variables; an expression like (i + j)++ is illegal. The operand must be a modifiable lvalue of arithmetic or pointer type.

Related: An interesting bug one may like to know about in gcc 4.4.5 is that expression j = ++(i | i); compiles that should produce l-value error. Read: j = ++(i | i); and j = ++(i & i); should an error: lvalue?

Additionally, modifying same variable more then once in an expression without an intervening sequence point causes which is undefined behavior in c and c++. To understand read Explain these undefined behaviors in i = i++ + ++i;.

Answer 5

this line:

printf("%d",++i++)

==>

printf("%d",(++i)++)

And

==>

 printf("%d",(++i)); &(++i)++ ==> (++i) = (++i) +1;

See you use (++i) as a left value.