Why the ternary operator does not have blocks? In other words, why the following code does not work and reports error for {} braces?
int main()
{
int i = 1;
(i==1)?{printf("Hello\n")}:{printf("World\n")};
return 0;
}
EDIT
Perhaps the question is misunderstood. It was: why blocks are not supported? Why only single expression?
Why this is not allowed to work?
int main()
{
int i = 1;
(i==1)?{printf("Hello\n");printf("World\n");}:{printf("Bye\n");printf("World\n");};
return 0;
}
One reason could be that ternary are often used for conditional assignment on left side and blocks will have no such returns or it will get confusing with multiple statements inside block.
To quote C11 standard, chapter §6.5.15, the syntax of the conditional operator is
conditional-expression:
logical-OR-expression
logical-OR-expression ? expression : conditional-expression
Where, the second and third operands are expression, not statements.
Just to elaborate,
One of the following shall hold for the second and third operands:
— both operands have arithmetic type;
— both operands have the same structure or union type;
— both operands have void type;
— both operands are pointers to qualified or unqualified versions of compatible types;
— one operand is a pointer and the other is a null pointer constant; or
— one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void.
Edit:
To answer the question
Why only single expression?
again, quoting the standard,
....the result is the value of the second or third operand (whichever is evaluated), converted to the type described below.
Block of statements, will not give a value. The evaluation of an expression can.
The ternary operator consists of expressions. There is no such a kind of expression that uses braces.
You can write simply
( i == 1 ) ? printf("Hello\n") : printf("World\n");
It semss that the only case when braces can be present in an expression is the use of a compound literal. For example
struct A { int x; int y; } a = { 1, 2 };
a = a.x < a.y ? ( struct A ){ a.y, a.x } : ( struct A ){ ++a.x, --a.y };
As for this statement
(i==1)?{printf("Hello\n");printf("World\n");}:{printf("Bye\n");printf("World\n");};
then it can be rewritten the following way using the comma operator
i == 1 ? ( printf("Hello\n"), printf("World\n") ) : ( printf("Bye\n"), printf("World\n") );
Or even like
i == 1 ? printf("Hello\n"), printf("World\n") : ( printf("Bye\n"), printf("World\n") );
Shortly answering your question if you need a code block then use the if-else statement instead of the ternary operator. Though the if-else statement may not be used in expressions. On the other hand it is desirable for readeability of the code that expressions would not be too compound.
As any operator the ternary operator is used in expressions and returns some evaluated value. For example as an expression it can be used as initializer or in assignments.
The ternary operator expects an expression for each part, and {...} is not an expression, but a statement.
To expand on your edit, the result of a ternary operator is an expression (but not an lvalue as you suggest), and statement blocks can't evaluate to a value.
For example, this doesn't make sense:
int x = (i==1)?{printf("Hello\n");printf("World\n");}:{printf("Bye\n");printf("World\n");};
But you could do this:
int x = (i==1)?(printf("Hello\n"), printf("World\n")):(printf("Bye\n"), printf("World\n"));
In which case, the comma operator would cause the last value in each subexpression to be returned.