Initializer as Full Expression

[Shao Miller]

I've noticed that each initializer is a full expression with a sequence
point at its end in C1X. But the order of evaluation and side effects
are also unspecified. Are these unsequenced or indeterminately
sequenced? Is each initializer in an initializer-list a full
expression, separate from its peer initializers? If so, what can we
expect from:

int i = 1;
struct foo {
int j;
int k;
} bar = { ++i, i + 5 };

?

My preference would be for indeterminately sequenced. Thanks for any
insight.

- Shao Miller

 

[Shao Miller]

I've noticed that each initializer is a full expression with a sequence
point at its end in C1X. But the order of evaluation and side effects
are also unspecified. Are these unsequenced or indeterminately
sequenced? Is each initializer in an initializer-list a full
expression, separate from its peer initializers? If so, what can we
expect from:

int i = 1;
struct foo {
int j;
int k;
} bar = { ++i, i + 5 };

?

My preference would be for indeterminately sequenced. Thanks for any
insight.

- Shao Miller

To be clear, please allow me to change that:

int i = 1;
struct foo {
int j;
int k;
} bar = { .j = ++i, .k = i + 5 };

 

[Shao Miller]

To be clear, please allow me to change that:

int i = 1;
struct foo {
   int j;
   int k;

} bar = { .j = ++i, .k = i + 5 };

Oops! Keith already mentioned this in another thread!

 

[Shao Miller]

Oops!  Keith already mentioned this in another thread!

Oh, re-oops! His mention wasn't exactly the same, but extremely
related.