Error: lvalue required in this simple C code? (Ternary with assignment?)

Question

I have :

#include<stdio.h>

int main()
{
 int a=5,b=6;
 (a>b)?b=a:b=b;    // Here is the error
 return 0;
}

But if I replace :

(a>b)?b=a:b=b;       // Error
with   
(a>b)?(b=a):(b=b);   // No-Error

I understand the lvalue is a value to which something can be assigned and how is it different from rvalue, but why is the extra parenthesis making the difference.

Answer 1

Actually, in C, this code

(a>b)?b=a:b=b; 

is parsed by many compilers as

((a>b)?b=a:b)=b;

which is an error, as the expression ((a>b)?b=a:b) evaluates to an rvalue which you try to assign with b which results in an error. Trying to assign an rvalue is an error. If it is not parsed that way, then its simply a syntax error. But a C compiler is NOT allowed to parse it as:

((a>b)?b=a:(b=b)); //not allowed to parse by C language

Because the grammar of C does not allow a compiler to parse the code as above.

But what you've written (the original code) is correct as C++.

Here the grammars of C and C++ differ a lot. And because of that difference you see both languages treat the expression differently. That is, the conditional expression in C++ is different from the conditional expression in C .

Wikipedia has very good and correct explanation for this:

The binding of operators in C and C++ is specified (in the corresponding Standards) by a factored language grammar, rather than a precedence table. This creates some subtle conflicts. For example, in C, the syntax for a conditional expression is:

logical-OR-expression ? expression : conditional-expression

while in C++ it is:

logical-OR-expression ? expression : assignment-expression

Hence, the expression:

e = a < d ? a++ : a = d

is parsed differently in the two languages. In C, this expression is a syntax error, but many compilers parse it as:

e = ((a < d ? a++ : a) = d)

which is a semantic error, since the result of the conditional-expression (which might be a++) is not an lvalue. In C++, it is parsed as:

e = (a < d ? a++ : (a = d))

which is a valid expression.