union usage

[j0mbolar]

The following was labeled as invoking undefined behavior
but I don't see that happening at all:

struct ints {
int a;
int b;
};

union merge {
float f;
struct ints typ;
};

union merge i;
union merge j;

i.f = 1;
j = i; /* means j.f = 1 ? */

i.typ.a = i.typ.b = 0; /* we assign here, we are not reading the value
of i.type.b */
j = i; /* means j.typ.a = j.typ.b = 0 ? */
This is all valid, right?

--
j0mbolar

[Eric Sosman]

j0mbolar wrote:

The following was labeled as invoking undefined behavior
but I don't see that happening at all:

struct ints {
int a;
int b;
};

union merge {
float f;
struct ints typ;
};

union merge i;
union merge j;

i.f = 1;
j = i; /* means j.f = 1 ? */

i.typ.a = i.typ.b = 0; /* we assign here, we are not reading the value
of i.type.b */
j = i; /* means j.typ.a = j.typ.b = 0 ? */
This is all valid, right?

Looks all right to me.

--
Eric Sosman
es*****@acm-dot-org.invalid

[jamesonang]

i also think it works right

[S.Tobias]

j0mbolar <j0******@engineer.com> wrote:

The following was labeled as invoking undefined behavior
but I don't see that happening at all:

struct ints {
int a;
int b;
};

union merge {
float f;
struct ints typ;
};

union merge i;
union merge j;

i.f = 1;
j = i; /* means j.f = 1 ? */

i.typ.a = i.typ.b = 0; /* we assign here, we are not reading the value
of i.type.b */
Isn't the value of the object `i.typ' in the above line modified twice
without an intervening sequence point (cf. 6.5#2)?
j = i; /* means j.typ.a = j.typ.b = 0 ? */
This is all valid, right?

--
Stan Tobias
mailx `echo si***@FamOuS.BedBuG.pAlS.INVALID | sed s/[[:upper:]]//g`

[Flash Gordon]

S.Tobias wrote:

j0mbolar <j0******@engineer.com> wrote:

The following was labeled as invoking undefined behavior
but I don't see that happening at all:

struct ints {
int a;
int b;
};

union merge {
float f;
struct ints typ;
};

union merge i;
union merge j;

i.f = 1;
j = i; /* means j.f = 1 ? */

i.typ.a = i.typ.b = 0; /* we assign here, we are not reading the value
of i.type.b */

Isn't the value of the object `i.typ' in the above line modified twice
without an intervening sequence point (cf. 6.5#2)?

Since you are modifying different fields within the struct, I can't see
that that would apply.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.

[S.Tobias]

Flash Gordon <sp**@flash-gordon.me.uk> wrote:

S.Tobias wrote:

j0mbolar <j0******@engineer.com> wrote:

The following was labeled as invoking undefined behavior
but I don't see that happening at all:

struct ints {
int a;
int b;
};

union merge {
float f;
struct ints typ;
};

union merge i;
union merge j;

i.f = 1;
j = i; /* means j.f = 1 ? */

i.typ.a = i.typ.b = 0; /* we assign here, we are not reading the value
of i.type.b */

Isn't the value of the object `i.typ' in the above line modified twice
without an intervening sequence point (cf. 6.5#2)?

Since you are modifying different fields within the struct, I can't see
that that would apply.

[blushing] Right, I'm sorry, I thought `i.typ' was a union; must buy glasses.

--
Stan Tobias
mailx `echo si***@FamOuS.BedBuG.pAlS.INVALID | sed s/[[:upper:]]//g`