L-Value, Pointer arithmetic [duplicate]

Question

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Why ++i++ gives "L-value required error" in C? [duplicate] (5 answers)

Closed 8 years ago.

I am looking for an explanation of how lines L1 and L2 in the code snippet below differ w.r.t l-values, i.e, Why am I getting the: C2105 error in L1, but not in L2?

*s = 'a';
printf("%c\n", *s );
//printf("%c\n", ++(*s)++ ); //L1 //error C2105: '++' needs l-value
printf("%c\n", (++(*s))++);  //L2
printf("%c\n", (*s) );

Note: I got the above result when the code was compiled as a .cpp file. Now, on compilation as .c file, I get the same error C2105 on both lines L1 and L2. Why does L2 compile in C++, and not in C is another mystery :(.

If its of any help, I'm using Visual C++ Express Edition.

Answer 1

Compiler see ++(*s)++ as ++((*s)++), as post-increment has higher precedence than pre-increment. After post-incrementation, (*s)++ becomes an r-value and it can't be further pre-incremented (here).
And yes it is not a case of UB (at least in C).
And also read this answer.

For L2 in C++ not giving error because
C++11: 5.3.2 Increment and decrement says:

The operand of prefix ++ is modified by adding 1, or set to true if it is bool (this use is deprecated). The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type or a pointer to a completely-defined object type. The result is the updated operand; it is an lvalue, and it is a bit-field if the operand is a bit-field. If x is not of type bool, the expression ++x is equivalent to x+=1.

C++11:5.2.6 Increment and decrement says:

The value of a postfix ++ expression is the value of its operand. [ Note: the value obtained is a copy of the original value —end note ] The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type or a pointer to a complete object type. The value of the operand object is modified by adding 1 to it, unless the object is of type bool, in which case it is set to true. The value computation of the ++ expression is sequenced before the modification of the operand object. With respect to an indeterminately-sequenced function call, the operation of postfix ++ is a single evaluation. [ Note: Therefore, a function call shall not intervene between the lvalue-to-rvalue conversion and the side effect associated with any single postfix ++ operator. —end note ] The result is a prvalue. The type of the result is the cv-unqualified version of the type of the operand.

and also on MSDN site it is stated that:

The operands to postfix increment and postfix decrement operators must be modifiable (not const) l-values of arithmetic or pointer type. The type of the result is the same as that of the postfix-expression, but it is no longer an l-value.

Answer 2

For completeness and quotes from C++ documentation:

Looking at L2, the prefix-increment/decrement returns an l-value. Hence, no error when performing the post-increment. From the C++ documentation for prefix-increment/decrement:

The result is an l-value of the same type as the operand.

Looking at L1, it becomes:

++( ( *s )++ )

... after operand precedence. The post-increment operator by definition evaluates the expression (returning an r-value) and then mutates it. From C++ documentation for post-increment/decrement:

The type of the result is the same as that of the postfix-expression, but it is no longer an l-value.

... and you cannot prefix-increment/decrement an r-value, hence error.

References:

• Postfix Operand Doc.
• Prefix Operand Doc.