Why can't variables be declared in an if statement?

Question

The following Java code does not compile.

int a = 0;

if(a == 1) {
    int b = 0;
}

if(a == 1) {
    b = 1;
}

Why? There can be no code pa

Answer 1

you have declared b variable inside if block that is not accessible out side the if block and if you want to access then put outside if block

Answer 2

This { } defines a block scope. Anything declared between {} is local to that block. That means that you can't use them outside of the block. However Java disallows hiding a name in the outer block by a name in the inner one. This is what JLS says :

The scope of a local variable declaration in a block (§14.2) is the rest of the block in which the declaration appears, starting with its own initializer (§14.4) and including any further declarators to the right in the local variable declaration statement.

The name of a local variable v may not be redeclared as a local variable of the directly enclosing method, constructor or initializer block within the scope of v, or a compile-time error occurs.

Answer 3

{ } is used to define scope of variables.And here you declared :

if(a == 1) 
{
    int b = 0;
}

So here scope of b will be only in { }.So you are using variable b outside { }, it is giving compilation error.

You can also refer this:

http://docs.oracle.com/javase/tutorial/java/javaOO/variables.html

Answer 4

Because when b goes out of scope in the first if (a == 1) then it will be removed and no longer exists on the stack and therefore can not be used in the next if statement.

Answer 5

It is a local variable and is limited to the {} scope.

Try this here.

Answer 6

Why? There can be no code path leading to the program assigning 1 to b without declaring it first.

You are right, but the compiler doesn't know that. The compiler does not execute the code. The compiler only translates to bytecode without evaluating expressions.