[riazrizvi]

Curly braces designate scope in C like languages:

    int i = 100;
    std::cout << i << std::endl;
    if (1) {
        int i = 200;
        std::cout << i << std::endl;
    }
    std::cout << i << std::endl;

giving

    100
    200
    100

They don't seem to serve such a purpose in Pytov. Eyeballing the source, it looks like you'd get this:

    i = 100
    print (i)
    if (True) {
        i = 200
        print (i)
    }
    print (i)

giving

    100
    200
    200

So what's the point?

[pansa2]

> it looks like you'd get this

Yes, you would - variables in Python have function-scope, not block-scope.

It doesn’t look like this project changes anything about Python except trivial bits of syntax.

[quietbritishjim]

> Curly braces designate scope in C

It's not the curly braces that create the scope as such. It's the block, and the curly braces just happen to create the block. In C++ you can even do this, for example (I was surprised to find it doesn't work in C):

    int i = 1;
    if (true) int i = 2;
    std::cout << i << '\n';
    // Prints 1

In any case, the semanics of blocks (whether they create scopes) is totally orthogonal to the syntax of blocks. You can just as easily imagine the reverse: the current colon-indent syntax where blocks do create scopes.

Another issue, which other replies have alluded to but not quite spelt out. You'd need to add a syntax to distinguish between regular assignment and declaration. In C you can still choose to assign to the outer-scope variable from a nested scope of you wish, so you don't want i=1 in a nested scope to always declare a new variable that shadows a variable in the outer scope.

Oh and one more thing. Your snippet (or equivalent) won't work in all C-style braces languages: in C# it is a compilation error to shadow a variable from an outer scope in the same function (presumably because it's a common source of bugs and extremely easy to just choose a different name for one of the variables).

[deaddodo]

Not sure if you're aware, but you could just write that as:

    int i = 100;
    std::cout << i << std::endl;
    {
        int i = 200;
        std::cout << i << std::endl;
    }
    std::cout << i << std::endl;

And since PyTov is supposed to be C-like, this might be a better example:

    #include <stdio.h>
    
    int main(void) {
        int i = 100;
        printf("%i\n", i);
        {
            int i = 200;
            printf("%i\n", i);
        }
        printf("%i\n", i);
        
        return 0;
    }

Also, the same would occur in C(/C++) if you didn't shadow the variable in the block scope and just called:

    i = 200

[zzo38computer]

So would you in C if you do not redeclare the variable inside of the block. Also in JavaScript (although in JavaScript the type is not declared; in C it is). I don't know how it is working in Python, though.