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6.5.2.3 Structure and union members
Constraints
1 The first operand of the . operator shall have a qualified or unqualified structure or union
type, and the second operand shall name a member of that type.
2 The first operand of the -> operator shall have type ‘‘pointer to qualified or unqualified
structure’’ or ‘‘pointer to qualified or unqualified union’’, and the second operand shall
name a member of the type pointed to.
Semantics
3 A postfix expression followed by the . operator and an identifier designates a member of
a structure or union object. The value is that of the named member,82) and is an lvalue if
the first expression is an lvalue. If the first expression has qualified type, the result has
the so-qualified version of the type of the designated member.
4 A postfix expression followed by the -> operator and an identifier designates a member
of a structure or union object. The value is that of the named member of the object to
which the first expression points, and is an lvalue.83) If the first expression is a pointer to
a qualified type, the result has the so-qualified version of the type of the designated
member.
5 One special guarantee is made in order to simplify the use of unions: if a union contains
several structures that share a common initial sequence (see below), and if the union
object currently contains one of these structures, it is permitted to inspect the common
initial part of any of them anywhere that a declaration of the complete type of the union is
visible. Two structures share a common initial sequence if corresponding members have
compatible types (and, for bit-fields, the same widths) for a sequence of one or more
initial members.
6 EXAMPLE 1 If f is a function returning a structure or union, and x is a member of that structure or
union, f().x is a valid postfix expression but is not an lvalue.
7 EXAMPLE 2 In:
struct s { int i; const int ci; };
struct s s;
const struct s cs;
volatile struct s vs;
the various members have the types:
s.i int
s.ci const int
cs.i const int
cs.ci const int
vs.i volatile int
vs.ci volatile const int
8 EXAMPLE 3 The following is a valid fragment:
union {
struct {
int alltypes;
} n;
struct {
int type;
int intnode;
} ni;
struct {
int type;
double doublenode;
} nf;
} u;
u.nf.type = 1;
u.nf.doublenode = 3.14;
/* ... */
if (u.n.alltypes == 1)
if (sin(u.nf.doublenode) == 0.0)
/* ... */
The following is not a valid fragment (because the union type is not visible within function f):
struct t1 { int m; };
struct t2 { int m; };
int f(struct t1 *p1, struct t2 *p2)
{
if (p1->m < 0)
p2->m = -p2->m;
return p1->m;
}
int g()
{
union {
struct t1 s1;
struct t2 s2;
} u;
/* ... */
return f(&u.s1, &u.s2);
}Forward references: address and indirection operators (6.5.3.2), structure and union
红色的部分为什么说“The following is not a valid fragment (because the union type is not visible within function f):”
union在函数f中是否可见与f有什么关系呀
不明白
specifiers (6.7.2.1) |
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发表于 2008-02-21 11:34
发表于 2008-02-21 15:08
