4 ms·

struct A { int a; int b; A() : b(1), a(b) { } }; . Here, the b field is used uninitialized because the order of member initializers in the mem

struct A { int a; int b; A() : b(1), a(b) { } }; . Here, the b field is used uninitialized because the order of member initializers in the member initializer list is irrelevant; what matters is the order of declarations in the class definition. (A related warning, -Wreorder, can be used to warn when the order of member initializers does not match the declaration order.) I don't get it. Why b? Do they mean a?

dundarious 4y ago
The order of initialization is the order of definition, i.e., a then b (because the definition order is `int a; int b;`). So initialization happens like so: // implicitly, a = undefined; // implicitly, b = undefined; a = b; b = 1; So b is used when its value is undefined. The undefined value is assigned to a.
- BruceEel 4y ago
  What confused me in their wording is that they seem to imply that the problem is: b would be left uninitialized, rather than: a depends on uninitialized b. Your explanation, OTOH, is straightforward to read.
  - dundarious 4y ago
    Understandable. The original is correct that b is used uninitialized, but they don't clarify that it is a that is left uninitialized.
    - tialaramex 4y ago
      Mmm? The consequence of reading the uninitialized variable is that your program's behaviour is Undefined. There's no point worrying about this other variable named 'a', it's already game over. It's not called "Temporarily unusual" behaviour or "Reliably crashing" behaviour or even "I bet it'll probably just skip that and do the next thing" behaviour, it's Undefined for a reason.
      - dundarious 4y ago
        The distinction is helpful so that people properly understand what's going on with and what's wrong with the code in question (the context is helping someone who did not fully understand). You're talking about the consequences in terms of compiler output and code execution, and yes, the usage is the problem.
- nicoburns 4y ago
  Seems like that really ought to be made a compile error!
vitus 4y ago
In this case, a is the first value to be initialized, but it's using the value of `b` (which is uninitialized).
swatcoder 4y ago
It’s tricky wording. When they say “b is used uninitialized”, they’re referring to it’s usage in that snippet. It’s uninitialized when referenced as a value for “a”. It’s pretty easy to read that text as suggesting that b would be uninitialized in subsequent code, but that’s (not true, as you suspect, and) not what it’s talking about.

dundarious 4y ago

The order of initialization is the order of definition, i.e., a then b (because the definition order is `int a; int b;`). So initialization happens like so: // implicitly, a = undefined; // implicitly, b = undefined; a = b; b = 1; So b is used when its value is undefined. The undefined value is assigned to a.

BruceEel 4y ago
What confused me in their wording is that they seem to imply that the problem is: b would be left uninitialized, rather than: a depends on uninitialized b. Your explanation, OTOH, is straightforward to read.
- dundarious 4y ago
  Understandable. The original is correct that b is used uninitialized, but they don't clarify that it is a that is left uninitialized.
  - tialaramex 4y ago
    Mmm? The consequence of reading the uninitialized variable is that your program's behaviour is Undefined. There's no point worrying about this other variable named 'a', it's already game over. It's not called "Temporarily unusual" behaviour or "Reliably crashing" behaviour or even "I bet it'll probably just skip that and do the next thing" behaviour, it's Undefined for a reason.
    - dundarious 4y ago
      The distinction is helpful so that people properly understand what's going on with and what's wrong with the code in question (the context is helping someone who did not fully understand). You're talking about the consequences in terms of compiler output and code execution, and yes, the usage is the problem.
nicoburns 4y ago
Seems like that really ought to be made a compile error!

vitus 4y ago

In this case, a is the first value to be initialized, but it's using the value of `b` (which is uninitialized).

swatcoder 4y ago

It’s tricky wording. When they say “b is used uninitialized”, they’re referring to it’s usage in that snippet. It’s uninitialized when referenced as a value for “a”. It’s pretty easy to read that text as suggesting that b would be uninitialized in subsequent code, but that’s (not true, as you suspect, and) not what it’s talking about.