Variable Declarations Following If Statements

Variable declarations following if statements

The C# language specification distinguishes between three types of statements (see chapter 8 for more details). In general you can have these statements:

labeled-statement - my guess that this is for the old-fashioned goto statement
declaration-statement - which would be a variable declaration
embedded-statement - which includes pretty much all the remaining statements

In the if statement the body has to be embedded-statement, which explains why the first version of the code doesn't work. Here is the syntax of if from the specification (section 8.7.1):

if ( boolean-expression ) embedded-statement
if ( boolean-expression ) embedded-statement else embedded-statement

A variable declaration is declaration-statement, so it cannot appear in the body. If you enclose the declaration in brackets, you'll get a statement block, which is an embedded-statement (and so it can appear in that position).

C - Variable Declaration in If Condition Available in Else?

You can't declare a variable like this

 if((int a = 0))

The compiler does not allow the code to run and you get an error

and if you try this

if(something_that_is_false){
        int a = 12;
    }
    else{
        do_something;
    }

again error because they are on the same level and they do not have access to their local variables.

Warning: you can use this code and runs without error

int a;
    if(a=0){
         printf("True");
    }
    else{
        printf("False");
    }

and you will see 'False' in screen because It's like writing

if(0) // and its false!

and for the last

int a;
    if(a=0){
         printf("True");
    }
    else{
        printf("False");
    }

you will see 'True' in screen because It's like writing

if(5) // any number other than zero is true!

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

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.

javascript variable declaration in the if/else loop

You can just declare let hello='' at the beginning of the code:

As let variable have the scope inside the brackets of them { }...

The let statement declares a block-scoped local variable, optionally
initializing it to a value.

What's the difference between using "let" and "var"?

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/let

var test = 4;
let hello = "";

if (test > 3) {
  hello = "hello world";
} else {
  hello = "hello gold";
}

console.log(hello);

Why is variable declared in an if-statement still in scope in else block?

Why is variable declared in an if-statement still in scope in else block?

Because the standard says that it is. It was designed so presumably because it is useful.

How is this possible?

It's unclear why it wouldn't be possible.

I always thought that scope of such variables ends with the if-block.

You assumed wrongly.

I'd be thankful for quotation from the standard if possible. [language-lawyer]

Latest draft says:

[stmt.select.general]
Selection statements choose one of several flows of control.
selection-statement:
...
if constexpr_opt ( init-statement_opt condition ) statement else statement
...

Note that the entire if (condition) statement else statement is a selection-statement.

[basic.scope.block]
Each
selection or iteration statement ([stmt.select], [stmt.iter]),
...
introduces a block scope that includes that statement or handler.

Note that the condition is directly within the if-statement and not inside a sub-statement, and thus a declaration within it extends until the end of the entire block scope, which contains both the if-sub-statement, and the else-sub-statement (not standard names for those sub-statements).

There is also a pretty clear example that demonstrates ill-formed re-declarations, and incidentally shows the scope of such declaration:

if (int x = f()) {
  int x;            // error: redeclaration of x
}
else {
  int x;            // error: redeclaration of x
}

If it's legal though, shouldn't it at least generate a warning?

Yeah, it would be nice if compilers were able to detect all provable null pointer indirections at compile time. It may be worth submitting a feature request regarding this corner case.

Are there any reasonable applications of such feature?

Sure. Following fabricated example seems reasonable to me:

if (Response r = do_request()) {
    log << "great success. content: " << r.content;
} else {
    log << "someone broke it :( error: " << r.error_code;
}

Or if you dislike implicit conversions:

if (Response r = do_request(); r.is_success()) {

Java Declaring a variable in an if-statement

When you scope a variable, it will only be available in that scope.

if (stuff) {
    int i;
    // i available here
}
// i not available here

The first type, the compiler doesn't know if you will use the variable later. it doesn't think there's anything wrong with creating a variable in that block.

In the second type, the compiler knows there is only one statement, because you didn't create a {} block. The variable that you created will definitely not be used, so the compiler is alerting you that you will never be able to use the variable that you defined there.

Why a short variable declaration in an else if statement doesn't fail to compile even though no new variable is being defined on the left side?

From the Go specs, here is how an if statement is defined:

IfStmt = "if" [ SimpleStmt ";" ] Expression Block [ "else" ( IfStmt | Block ) ] .

Later on, in the Declarations and Scope sections it is said:

An identifier declared in a block may be redeclared in an inner block. While the identifier of the inner declaration is in scope, it denotes the entity declared by the inner declaration.

Now, the if statement is an implicit block:

Each "if", "for", and "switch" statement is considered to be in its own implicit block.

Then as you can see from the IfStmt definition, after the keyword else may come:

a Block, i.e. else { /* code */ }
an IfStmt again, as in your case, i.e. else if /* statement */ { /* code */ }. This means the IfStmt is recursive, and it is an implicit block within another IfStmt (still an implicit block). Therefore it meets the condition for redeclaring the identifier.

Compare also with explicit blocks:

func foo() {
    x := 10
    {
        x := 20
        fmt.Println("Inner x:", x) // 20
    }
    fmt.Println("Outer x:", x) // 10
}

Declare variable in if statement (ANSI C)

No, you cannot do that.

What you can do is create a compound statement (anonymous or hanging block) just for the if

    {
        int variable;
        variable = some_function();
        if (variable) return 1;
    }
    /* variable is out of scope here */

Note that for this simple case you can call the function as the condition of the if (no need for an extra variable)

if (some_function()) return 1;

C++, variable declaration in 'if' expression

As of C++17 what you were trying to do is finally possible:

if (int a = Func1(), b = Func2(); a && b)
{
    // Do stuff with a and b.
}

Note the use of ; of instead of , to separate the declaration and the actual condition.

Variable Declarations Following If Statements