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-expressionwhile in C++ it is:
logical-OR-expression ? expression :
assignment-expressionHence, 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.