Using for loop inside declaration of variable

Question

Can I use for loop inside declaration a variable?

int main() {
    int a = {
        int b = 0;
        for (int i = 0; i < 5; i++) {
            b += i;
        }
        return b;
    };

    printf("%d", a);
}

Use a function/lambda.

zdf
– zdf

2020-01-18 18:46:05 +00:00
Commented Jan 18, 2020 at 18:46 — zdf
– zdf, Commented Jan 18, 2020 at 18:46

deW1 · Accepted Answer · 2020-01-18 18:49:02Z

8

You can use a lambda:

int main() {
    int a = []{
        int b = 0;
        for (int i = 0; i < 5; i++) {
            b += i;
        }
        return b;
    }();

    printf("%d", a);
}

It's important to note that you have to immediately execute it otherwise you attempt to store the lambda. Therefore the extra () at the end.

If you intent to reuse the lambda for multiple instantiations, you can store it separately like this:

int main() {
    auto doCalculation = []{
        int b = 0;
        for (int i = 0; i < 5; i++) {
            b += i;
        }
        return b;
    };

    int a = doCalculation();

    printf("%d", a);
}

If you need it in more than one scope, use a function instead.

answered Jan 18, 2020 at 18:49

deW1

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4 Comments

John Doe Over a year ago

Thanks! What about I want to have another declaration of variable above a (unsigned char x = 7) and I want to return x*b. It gets me error: void value not ignored as it ought to be and error: ‘x’ is not captured.

deW1 Over a year ago

You need to capture it e.g. [x]{ but it's hard to say without code. Please read some more about lambdas and captures, e.g. here en.cppreference.com/w/cpp/language/lambda or simply google the term, then you will find many tutorials

John Doe Over a year ago

Thanks a lot! I just didn't know the term.

A M Over a year ago

We are not talking about declaration here, even not definition. We are talking about initialization, and especially about copy initialization. Please see here: en.cppreference.com/w/cpp/language/copy_initialization . But tons of upvotes and an accepted answer, so everything perfectly OK. +1 from my end. If we would be strict: The only correct answer to the given question is simply: "No, not possible"

score 0 · Accepted Answer · 2020-01-20 01:40:17Z

0

actually has been prepared by C++ committee..
constexpr has many usefulness not yet exlpored

constexpr int b(int l) {
            int b=0;
            for (int i = 0; i < l; i++)
                b += i;
            return b;
        }

int main() {

    constexpr int a = b(5);

    printf("%d", a);
}

edited Jan 20, 2020 at 1:40

answered Jan 19, 2020 at 4:34

user12450543

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