CODEWITHSUNDEEP
c
Ashwyn
Ashwyn

Reputation: 669

What is the difference between the two syntax?

Why

(a?b:c)=5;

in 'C' shows lvalue required while

*(a?&b:&c)=5;

is perfectly fine? What is the difference between the two?

Assuming a=1, for first case, it gives b=5 and second case it gives, *(&b)=5.`

What i am not able to understand is: what difference does it make if we write b=5 or *(&b)=5?

Upvotes: 0

Views: 154

Answers (1)

Nicol Bolas
Nicol Bolas

Reputation: 474436

what difference does it make if we write b=5 or *(&b)=5?

The second one gets a pointer to b and then dereferences that pointer, storing 5 into the obtained pointer.

However, your question seems to be ignoring the real question: why it works in the second case but not the first. And that has to do with expressions in C and how they're dealt with.

The result of the ?: operator is a value; specifically, it is an rvalue. Loosely defined, an rvalue is a value that can't go on the left-hand side of an assignment. They are so named because they're values that go on the right-hand side of an assignment. For example, the expression "5" is an rvalue expression. You can't do 5 = 40; that's nonsense.

A named variable is an lvalue. You can put an lvalue on the left-hand side of an equation. The expression a is an lvalue expression (assuming that a is a variable name that is in scope).

However, the expression (a + b) is an rvalue expression, not an lvalue. And with good reason; you can no more do (a + b) = 30; than you could (5 + 10) = 30;.

The result of the ?: operator is an rvalue expression, just as with the + operator above. This explains why (a?b:c) = 5; doesn't work; you're trying to assign a value to an rvalue expression. That's illegal.

Now, let's look at the second case. We know that ?: results in an rvalue expression. Now, that explains what the classification of the expression is, but what about the type of the expression's result? Well, assuming that b and c are both ints, the type of (a?b:c) is also int.

However, when you do (a?&b:&c), the type of this expression is int*, not int. It is a pointer to an integer. It's an rvalue expression of type "pointer to int." It will return either the address of b, or the address of c.

When you have an int*, and you dereference it with the * operator, what do you get? You get an lvalue expression of type int. Since (a?&b:&c) is an rvalue expression of type int*, if you dereference it, you will get an lvalue expression of type int. This lvalue will refer to either b or c, depending on the contents of a.

To put it simply, it works exactly like this:

int *ptr = NULL;
if(a)
  ptr = &b;
else
  ptr = &c;

*ptr = 5;

You get a pointer, which could point to any particular memory location, then you store something in the location being pointed to. It's that simple.

Upvotes: 6

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