What is the scope of a Java variable in a block?

Question

I know in C++ variables have block scope, for example, the following code works in C++:

void foo(){
    int a = 0;
    for (int i = 0; i < 10; ++i){
        int a = 1; // Redefine a here.
    }
}

But this snippet doesn't work in Java, it reports "duplicate local variable a", does it mean Java variables don't have BLOCK scope?

What do you mean by "doesn't work"? A compiler error or warning, or ...? Please try to be precise with your descriptions. — Péter Török, Oct 05 '10 at 08:53
Good extended discussion about this can be found here: http://stackoverflow.com/questions/141140/why-does-java-not-have-block-scoped-variable-declarations — Andriy Sholokh, Oct 05 '10 at 08:53

score 10 · Answer 1 · answered Oct 05 '10 at 08:52

10

They have block scope. 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.

answered Oct 05 '10 at 08:52

Yakov Galka

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score 4 · Accepted Answer · answered Oct 05 '10 at 08:53

4

java variables do have a block scope but if you notice int a is already defined in scope

all subscopes are in scope of the uppermost curly braces. Hence you get a duplicate variable error.

answered Oct 05 '10 at 08:53

sushil bharwani

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codaddict · Answer 3 · 2010-10-05T09:04:59.373

Section §14.4.2 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.

score 3 · Answer 4 · answered Oct 05 '10 at 08:55

The previous answers already stated the reason, but I just want to show that this is still allowed:

void foo(){
    for(int i = 0; i < 10; ++i){
        int a = 1;
    }
    int a = 0;
}

In this case the a inside the loop doesn't hide the outer a, so it's valid.

Also IMHO it should be this way in C++ too, it's less confusing and prevents accidental declaration of variable with same name.

score 2 · Answer 5 · answered Oct 05 '10 at 08:54

It does, but it's nested, so the "a" you defined in foo() is available in all blocks within foo.

Here is an example of what you're looking for:

void foo(){
    {
        int a = 0;
        // Do something with a
    }
    for(int i = 0; i < 10; ++i){
        int a = 1; //define a here.
    }
}

What is the scope of a Java variable in a block?

5 Answers5

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