Discussion
What is the output of this C code?
void main()
{
char a = 'a';
int x = (a++)++;
printf("%d\n", x);
}
Answer: C
Correct answer: Compile time error
The result of a++ is a value, not an object, so it is not an lvalue - and ++ requires a modifiable lvalue to operate on. (a++)++ therefore fails to compile. The same reasoning rules out (a + 1)++ or 5++. A single (a)++ would have been fine, leaving x equal to 97 and a equal to 'b'.
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