C++ casting question (structs)
In my C++ application I have the definition:
struct socksddr_storage ss;
struct sockaddr *sa;
what is the different between the following:
sa = (struct sockaddr *)&ss;
and
sa = (sockaddr *)&ss;
???
开发者_运维技巧Any help?
There is no difference between the two. The typecast with the struct keyword present harkens back to the days of C, where the type of a struct included the keyword "struct" and is why the phrase "typedef struct" became popular. C++ eliminated the need for this keyword in this context, but kept the syntax alive for backward compatibility.
There is no difference in C++ as the struct
keyword is optional. In C, the second form is not legal.
The difference is that (struct sockaddr*)
is a valid cast in both C and C++, where is (sockaddr*)
is valid in C++ but not in C. Both casts achieve the same result.
As others already said, you don't need to write the struct
prefix in c++ as the compiler is able to determine the correct type by the name only. But I suggest to use reinterpret_cast in this situation, as the c style cast is the big-hammer cast which casts types and const specifiers in one go.
struct socksddr_storage ss;
const struct socksddr_storage ss2;
struct sockaddr *sa;
sa = reinterpret_cast<sockaddr *>(&ss); // Works like the C Cast
sa = reinterpret_cast<sockaddr *>(&ss2); // Fails, because ss2 is a constant,
// but sa is a pointer to a no const memory location. The C cast will allow
// this assignment, but this can lead to runtime errors since the linker
// can place ss2 into read-only memory.
I cannot say definite thing as I don't know your actual context, but if you
see an error when you write (sockaddr*)
and don't when (struct sockaddr*)
,
my answer may apply.
struct sockaddr
in (struct sockaddr*)
is an elaborated type specifier.
Sometimes this is needed.
For example,
void f( void* p )
{
int stat( char const*, struct stat* );
struct stat *a = (struct stat*) p; // cannot omit this struct
}
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