A question (not posted by myself) from https://stackoverflow.com/questions/65077630 :
Consider the following:
union { int i; char c; } x = {0};
x.c = x.i;
Does the assignment x.c = x.i result in undefined behavior?
C18 6.15.16.1/3 says:
| If the value being stored in an object is read from another object
| that overlaps in any way the storage of the first object, then the
| overlap shall be exact and the two objects shall have qualified or
| unqualified versions of a compatible type; otherwise, the behavior is
| undefined.
The objects x.c and x.i overlap, but have incompatible types, so on first glance it appears to be UB.
A question (not posted by myself) from https://stackoverflow.com/questions/65077630 :
Consider the following:
union { int i; char c; } x = {0};
x.c = x.i;
Does the assignment x.c = x.i result in undefined behavior?
C18 6.15.16.1/3 says:
| If the value being stored in an object is read from another object
| that overlaps in any way the storage of the first object, then the
| overlap shall be exact and the two objects shall have qualified or
| unqualified versions of a compatible type; otherwise, the behavior is
| undefined.
The objects x.c and x.i overlap, but have incompatible types, so on
first glance it appears to be UB.
A question (not posted by myself) from https://stackoverflow.com/questions/65077630 :
Consider the following:
union { int i; char c; } x = {0};
x.c = x.i;
Does the assignment x.c = x.i result in undefined behavior?
C18 6.15.16.1/3 says:
| If the value being stored in an object is read from another object
| that overlaps in any way the storage of the first object, then the
| overlap shall be exact and the two objects shall have qualified or
| unqualified versions of a compatible type; otherwise, the behavior is
| undefined.
The objects x.c and x.i overlap, but have incompatible types, so on
first glance it appears to be UB.
Ian Abbott <ijabbott63@gmail.com> writes:
A question (not posted by myself) from
https://stackoverflow.com/questions/65077630 :
Consider the following:
union { int i; char c; } x = {0};
x.c = x.i;
Does the assignment x.c = x.i result in undefined behavior?
C18 6.15.16.1/3 says:
| If the value being stored in an object is read from another
| object that overlaps in any way the storage of the first object,
| then the overlap shall be exact and the two objects shall have
| qualified or unqualified versions of a compatible type;
| otherwise, the behavior is undefined.
The objects x.c and x.i overlap, but have incompatible types, so on
first glance it appears to be UB.
On first glance, yes, but I think this passage needs to be updated
to reflect its intent.
The RHS of an assignment is not an lvalue. If it starts out as an
lvalue, then lvalue conversion is applied. Logically the value is
retrieved *and then* copied into the destination object.
A simple case like this is not likely to cause problems (in the
absence of agressive optimization), but we can construct more
problematic cases where the object being copied is arbitrarily large
and the overlap might not be detectable at compile time.
I've written an answer:
https://stackoverflow.com/a/65080498/827263
Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:
Ian Abbott <ijabbott63@gmail.com> writes:
A question (not posted by myself) from
https://stackoverflow.com/questions/65077630 :
Consider the following:
union { int i; char c; } x = {0};
x.c = x.i;
Does the assignment x.c = x.i result in undefined behavior?
C18 6.15.16.1/3 says:
| If the value being stored in an object is read from another
| object that overlaps in any way the storage of the first object,
| then the overlap shall be exact and the two objects shall have
| qualified or unqualified versions of a compatible type;
| otherwise, the behavior is undefined.
The objects x.c and x.i overlap, but have incompatible types, so on
first glance it appears to be UB.
On first glance, yes, but I think this passage needs to be updated
to reflect its intent.
The RHS of an assignment is not an lvalue. If it starts out as an
lvalue, then lvalue conversion is applied. Logically the value is
retrieved *and then* copied into the destination object.
A simple case like this is not likely to cause problems (in the
absence of agressive optimization), but we can construct more
problematic cases where the object being copied is arbitrarily large
and the overlap might not be detectable at compile time.
I've written an answer:
https://stackoverflow.com/a/65080498/827263
Let me start with where we agree. I agree that 6.5.16.1 p3 deserves
some clarification. After that however my reading and yours reach
different conclusions.
First I think the passage as written clearly conveys the meaning
intended in this case, that this assignment is undefined behavior.
The value being stored was read out of x.i, and is being stored
into x.c. The types of those two objects don't mesh, and so the
assignment is undefined behavior. This assignment is about the
clearest possible case that would violate 6.5.16.1 p3, and the
passage as written conveys that, IMO with no room for argument.
Second I think whether the RHS is an lvalue has no bearing on the
question. The cited paragraph does not mention anything about
lvalues; it speaks only of "the value being stored". Consider a
slight variation:
x.c = (printf( "hello world\n" ), x.i);
The RHS of this assignment is not an lvalue. But the /value/
being stored was read from x.i. As I read the Standard this
assignment too is undefined behavior, and IMO the Standard does
convey that intention.
Here is another variation:
x.c = 0 ? printf( "hello world\n" ) : x.i;
Once again the value being stored was read from x.i, and again
as I read the Standard this assignment too is undefined behavior,
and IMO the Standard does convey that intention.
Now let's look at an example you give in your stackoverflow answer.
Quoting a whole paragraph:
The passage says that the value "is read from another
object". That's ambiguous. Must name of the object be the
entire RHS expression, or can it be just a subexpression?
If the latter, then x.c = x.i + 1; would have undefined
behavior, which in my opinion would be absurd.
You left out an important part: "the value /being stored/".
Here the value being stored is x.i + 1. That /value/ was not
read from x.i; it was formed by an addition operation after
reading x.i. Ostensibly this assignment would not be undefined
behavior. I say "ostensibly" because actually I think this case
is not clear, and may have been intended to be undefined behavior
along with the others. And I don't see anything absurd about
having it be undefined behavior, or even surprising if it were
discovered that this case had been meant to be undefined behavior
all along.
Disclaimer: I haven't yet done any research into Defect Reports
or committee meeting notes to see if this question has been
addressed there. I suspect it has, but I just haven't looked
yet.
On 01/12/2020 11:39, Tim Rentsch wrote:
Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:
Ian Abbott <ijabbott63@gmail.com> writes:
A question (not posted by myself) from
https://stackoverflow.com/questions/65077630 :
Consider the following:
union { int i; char c; } x = {0};
x.c = x.i;
Does the assignment x.c = x.i result in undefined behavior?
C18 6.15.16.1/3 says:
| If the value being stored in an object is read from another
| object that overlaps in any way the storage of the first object,
| then the overlap shall be exact and the two objects shall have
| qualified or unqualified versions of a compatible type;
| otherwise, the behavior is undefined.
The objects x.c and x.i overlap, but have incompatible types, so on
first glance it appears to be UB.
On first glance, yes, but I think this passage needs to be updated
to reflect its intent.
The RHS of an assignment is not an lvalue. If it starts out as an
lvalue, then lvalue conversion is applied. Logically the value is
retrieved *and then* copied into the destination object.
A simple case like this is not likely to cause problems (in the
absence of agressive optimization), but we can construct more
problematic cases where the object being copied is arbitrarily large
and the overlap might not be detectable at compile time.
I've written an answer:
https://stackoverflow.com/a/65080498/827263
Let me start with where we agree. I agree that 6.5.16.1 p3 deserves
some clarification. After that however my reading and yours reach
different conclusions.
First I think the passage as written clearly conveys the meaning
intended in this case, that this assignment is undefined behavior.
The value being stored was read out of x.i, and is being stored
into x.c. The types of those two objects don't mesh, and so the
assignment is undefined behavior. This assignment is about the
clearest possible case that would violate 6.5.16.1 p3, and the
passage as written conveys that, IMO with no room for argument.
Second I think whether the RHS is an lvalue has no bearing on the
question. The cited paragraph does not mention anything about
lvalues; it speaks only of "the value being stored". Consider a
slight variation:
x.c = (printf( "hello world\n" ), x.i);
The RHS of this assignment is not an lvalue. But the /value/
being stored was read from x.i. As I read the Standard this
assignment too is undefined behavior, and IMO the Standard does
convey that intention.
Here is another variation:
x.c = 0 ? printf( "hello world\n" ) : x.i;
Once again the value being stored was read from x.i, and again
as I read the Standard this assignment too is undefined behavior,
and IMO the Standard does convey that intention.
Now let's look at an example you give in your stackoverflow answer.
Quoting a whole paragraph:
The passage says that the value "is read from another
object". That's ambiguous. Must name of the object be the
entire RHS expression, or can it be just a subexpression?
If the latter, then x.c = x.i + 1; would have undefined
behavior, which in my opinion would be absurd.
if
x.c = x.i + 1;
has defined behaviour then
x.c = x.i + 0;
also has defined behaviour.
So IMO, consistency requires that we tighten the requirement so that
x.c = any expression that uses x.i;
has undefined behaviour.
However I still have reservations because:
{int const i = x.i; x.c = i;}
is surely OK. However any optimising compiler will concatenate that to
x.c = x.i;
Note the braces limit the scope of i.
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