Why is it allowed to use a variable in its own declaration?

Asked Sep 30 '19 at 20:13

Active Sep 30 '19 at 20:34

Viewed 48 times

int main(){
    int i = i;
    return i;
}

There are similar questions, but no answer. Maybe I've not searched well enough.

Why the standard doesn't prohibit such expressions? What could possibly be the reason? The expression clearly doesn't make any sense.

Conclusion:

int i = (i=1) + 1; is a valid expression with a defined behavior.
The example of a recursive lambda with the help of std::function is also a very valid use.
Thus, such expressions should not be prohibited because it would be unwise to oblige the compiler to do the analysis(though any decent compiler does it)

edited Sep 30 '19 at 20:34

asked Sep 30 '19 at 20:13

Hrisip

What makes you think this is permitted? – ggorlen Sep 30 '19 at 20:15
@ggorlen I'm not going to get an error in many compilers – Hrisip Sep 30 '19 at 20:16
The compiler doesn't give an error for many UB things, though, right? See this answer: https://stackoverflow.com/a/54200564/6243352 – ggorlen Sep 30 '19 at 20:17
@ggorlen It is allowed per the standard. Has very few valid use cases though. – NathanOliver Sep 30 '19 at 20:18
@NathanOliver Yeah, that's what the dupe indicates--good point. – ggorlen Sep 30 '19 at 20:19
alright, since `int x = (x=5) + 1;` is an expression with defined behaviour, such expression can't be prohibited. – Hrisip Sep 30 '19 at 20:20
1

One valid use case is a recursive lambda. For a lambda to be able to call itself the right hand side of the `=` needs to be able to capture left hand side. – NathanOliver Sep 30 '19 at 20:20
@NathanOliver can you give an example with the lambda that compiles? – Hrisip Sep 30 '19 at 20:22
1

@DanA. See: http://coliru.stacked-crooked.com/a/ceaba42c9afb7030 – NathanOliver Sep 30 '19 at 20:24
It does need a `std::function` to get it to actually compile. Otherwise you get a circular reference. – NathanOliver Sep 30 '19 at 20:24

Why is it allowed to use a variable in its own declaration?

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