I'm trying format one c header file. While struct idents is equals to class variables.
test.h:
class MyClass{
public:
int var;
int var2;
}
struct myst {
int foo;
int bar;
};
Style Config:
TabWidth: 4
IndentWidth: 4
UseTab: Always
IndentAccessModifiers: true
Run:
clang-format -style="{TabWidth: 4, IndentWidth: 4, UseTab: Always, IndentAccessModifiers: true}" test.h
The following output is bad. (in struct is must be one tab not more!)
class MyClass {
public:
int var;
int var2;
}
struct myst {
int foo;
int bar;
};
Also like this output:
class MyClass {
public:
int var;
int var2;
}
struct myst {
int foo;
int bar;
};
As per this question, I do not think that clang-format can treat them any differently. To be fair, they are still fields in a larger object and so should share that similarity.
This answer might help make the double indent go away.
Also, as per the docs you can have access modifiers in structs, further suggesting that clang will not treat them any differently.
Related
If I have these structures:
typedef struct { int x; } foo;
typedef struct { foo f; } bar;
Normally you would access x through b.f.x, but is there a way to set this up so that you can access element x without referring to f?
bar b;
b.x = ...
My first intuition is that you can't since there would be a possibility for name conflicts if two sub structures both had a member x and I can't figure out what the compile error would be. However, I recall working in some frameworks where this was possible.
In C++ I worked in a framework once where bar existed, and you could access its members as member variables this->x from a different class. I'm trying to figure out how that could be done.
You can with C11:
§ 6.7.2.1 -- 11
An unnamed member whose type specifier is a structure specifier with no tag is called an
anonymous structure; an unnamed member whose type specifier is a union specifier with
no tag is called an anonymous union. The members of an anonymous structure or union
are considered to be members of the containing structure or union. This applies
recursively if the containing structure or union is also anonymous.
So this code might work:
#include <stdio.h>
typedef struct { int x; } foo;
typedef struct { foo; } bar;
int main(void)
{
bar b;
b.x = 1;
printf("%d\n", b.x);
}
The problem here is that different compilers disagree in my tests on whether a typedef is acceptable as a struct specifier with no tag The standard specifies:
§ 6.7.8 -- 3
In a declaration whose storage-class specifier is typedef, each declarator defines an
identifier to be a typedef name that denotes the type specified for the identifier in the way
described in 6.7.6. [...] A typedef declaration does not introduce a new type, only a
synonym for the type so specified.
(emphasis mine) -- But does synonym also mean a typdef-name specifier is exchangeable for a struct specifier? gcc accepts this, clang doesn't.
Of course, there's no way to express the whole member of type foo with these declarations, you sacrifice your named member f.
Concerning your doubt about name collisions, this is what gcc has to say when you put another int x inside bar:
structinherit.c:4:27: error: duplicate member 'x'
typedef struct { foo; int x; } bar;
^
To avoid ambiguity, you can just repeat the struct, possibly #defined as a macro, but of course, this looks a bit ugly:
#include <stdio.h>
typedef struct { int x; } foo;
typedef struct { struct { int x; }; } bar;
int main(void)
{
bar b;
b.x = 1;
printf("%d\n", b.x);
}
But any conforming compiler should accept this code, so stick to this version.
<opinion>This is a pity, I like the syntax accepted by gcc much better, but as the wording of the standard doesn't make it explicit to allow this, the only safe bet is to assume it's forbidden, so clang is not to blame here...</opinion>
If you want to refer to x by either b.x or b.f.x, you can use an additional anonymous union like this:
#include <stdio.h>
typedef struct { int x; } foo;
typedef struct {
union { struct { int x; }; foo f; };
} bar;
int main(void)
{
bar b;
b.f.x = 2;
b.x = 1;
printf("%d\n", b.f.x); // <-- guaranteed to print 1
}
This will not cause aliasing issues because of
§ 6.5.2.3 -- 6
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 completed 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
C: Highly unrecommended, but doable:
#include <stdio.h>
#define BAR_STRUCT struct { int x; }
typedef BAR_STRUCT bar;
typedef struct {
union {
bar b;
BAR_STRUCT;
};
} foo;
int main() {
foo f;
f.x = 989898;
printf("%d %d", f.b.x, f.x);
return 0;
}
Anonymous structs are a widly-spread extension in standards before C11.
C++:
The same as in C, you can do here but anonymous structs are not part of any C++ standard, but an extension.
Better use inheritance, or do not use this shortcut at all.
Of course, do not use something like #define x b.x)).
In C you can't access members of members like this.
You can however access members of an anonymous inner struct:
struct bar {
struct {
int x;
}
};
...
struct bar b;
b.x = 1;
In C++ you use inheritance:
struct foo {
int x;
};
struct bar: public foo {
};
...
struct bar b;
b.x = 1;
In C (99 and onward) you can access the common initial sub-sequence of union members, even if they weren't the last member written to1.
In C11, you can have anonymous union members. So:
typedef struct { int x; } foo;
typedef struct {
union {
foo f;
int x;
};
} bar;
Yes, that applies to structures. But according to the standard:
A structure pointer, suitably converted, points to the first member.
A union pointer, suitably converted, points to any union member.
So their location in memory is the same.
This is not possible in C. In C++ however you can use inheritance which is probably what you were thinking about.
In C++, you can use inheritance and member name conflicts are sort of resolvable with :: and treating the base classes as members.
struct foo { int x; };
struct bar : foo { };
struct foo1 { int x; };
struct bar1 : foo1 { char const* x; };
bar b;
bar1 b1;
int main()
{
return b.x + b1.foo1::x;
}
In standard C, it's impossible, however several compilers (gcc, clang, tinycc) support a similar thing as an extension (usually accessible with -fms-extensions (on gcc also with -fplan9-extensions which is a superset of -fms-extensions)), which allows you to do:
struct foo { int x; };
struct bar { struct foo; };
struct bar b = { 42 };
int main()
{
return b.x;
}
However, there's no resolution for conflicting member names with it, AFAIK.
In C++, it is possible in two ways. The first is to use inheritence. The second is for bar to contain a reference member named x (int &x), and constructors that initialise x to refer to f.x.
In C, it is not possible.
Since the C standard guarantees that there isn't padding before the first member of a struct, there isn't padding before the foo in bar, and there isn't padding before the x in foo. So, a raw memory access to the start of bar will access bar::foo::x.
You could do something like this:
#include <stdio.h>
#include <stdlib.h>
typedef struct _foo
{
int x;
} foo;
typedef struct _bar
{
foo f;
} bar;
int main()
{
bar b;
int val = 10;
// Setting the value:
memcpy(&b, &val, sizeof(int));
printf("%d\n", b.f.x);
b.f.x = 100;
// Reading the value:
memcpy(&val, &b, sizeof(int));
printf("%d\n", val);
return 0;
}
As others have noted, C++ offers a more elegant way of doing this through inheritance.
I have a struct that might have 2 or 4 members depending on a preprocessor definition.
struct foo {
int m1;
int m2;
#ifdef MORE_MEMBERS
int m3;
int m4;
#endif
}
I have functions that take foo.m3 and foo.m4 as arguments. Because of that, I get a compilation error when MORE_MEMBERS isn't defined.
I'm trying to avoid adding #ifdef at every function call. Is there some way to make m3 and m4 be aliases/dummies when MORE_MEMBERS isn't defined? I'd like to still keep the struct size equal to sizeof(int) * 2.
I agree with Iharob, it's horrible code. It's not too clear what you're after, but it sounds like you could use a union:
struct foo
{
int a;
int b;
};
union bar
{
struct foo m1m2;
struct foo m3m4;
};
void f(union bar *x)
{
printf("%d", x->m1m2.a);
printf("%d", x->m3m4.b);
}
I am a little bit surprised the following code does not compile. Can you shed some light on it? (header file)
enum CarType_e {
CAR_BMW = 0,
CAR_KIA,
CAR_HONDA,
CAR_FORD,
CAR_MERCEDES
};
int build_car(CarType_e type);
and then I get the following error:
In file included from car.c:19:0:
car.h:35:16: error: unknown type name ‘CarType_e’
This is gcc version 4.7.3
You have two choices, use
typedef enum {
CAR_BMW = 0,
CAR_KIA,
CAR_HONDA,
CAR_FORD,
CAR_MERCEDES
} CarType_e;
Or, you can use -
int build_car(enum CarType_e type);
I tested both with gcc on linux.
In C, custom types (enums, unions, and structs) each have their own naming scope.
You need to write int build_car(enum CarType_e type);.
This also means there is no point in the _e suffix.
Alternatively, you can use typedef.
You need to say enum CarType_e wherever you use it:
int build_car(enum CarType_e type);
Or typedef the enum.
typedef enum {
CAR_BMW = 0,
CAR_KIA,
CAR_HONDA,
CAR_FORD,
CAR_MERCEDES
} CarType_e;
You need to tell the compiler that you are using an enum by specifying its tag along with the enum keyword. Otherwise how would the compiler resolve an ambiguity like
struct foo { int x; char y; };
enum foo { CAR, BIKE, ROCKET };
union foo { int x; char y; };
int build_car (foo x); /* A struct foo or an enum foo or a union foo? */
Note that structs, enums and unions all have a separate tag name space, so I am free to reuse the same tag name for each of them as shown above.
Requiring the tag name as in
int build_car (enum foo x);
int build_bike (struct foo x);
int build_rocket (union foo x);
makes this non-ambiguous for the compiler.
And don't bother with typedefs for structs/unions; they are useless and only the unenlightened would consider them. Yes, that's a strong opinion. All they do is save you from writing struct in a few places--that's some information you rather want to know about.
I have 2 headers that need to define 2 similar structures. One structure should be aligned to 32bytes and other is similar but doesn't need to align to 32bytes so I can save some memory here. One is included in the other. Here is main_header.h
struct mainStruct
{
int a;
int b;
int c;
char pad[20];
};
Following is sub_header.h
#include "main_header.h"
struct subStruct
{
int a;
int b;
int c;
};
Now what I wish to do is define subStruct in a way that, whenever i change mainStruct other than pad[20] all other fields should be updated in subStruct. That is, if tomorrow, I make mainStruct like this:
struct mainStruct
{
int a;
int b;
int c;
int d;
char pad[16];
};
then, automatically it should be reflected in sub_header.h as
struct subStruct
{
int a;
int b;
int c;
int d;
};
Is there an efficient way of doing it using some kind of Macros or other preprocessor directives??
And subStruct should always be derived from mainStruct and not other way around, this is very important.
Thanks in advance.
Did you consider the other way around; including subStruct as an element in mainStruct? Like this:
struct sub {
int a, b, c;
};
struct main {
struct sub head;
char dummy[16];
};
True, it does take an extra layer of syntax to get at a, b or c, but the memory layout and performance should be the same.
Otherwise, there's nothing preventing you from using a macro, as you suggest yourself:
#define SUB_FIELDS int a; int b; int c;
struct main {
SUB_FIELDS
char dummy[16];
};
struct sub {
SUB_FIELDS
};
Well, that's why C++ was initially invented. But if you wish to stay with plain C (there could be reasons for such decision) you may check GLib / GObject type system implemented as a set of macros over plain C.
I've seen C structs declared several different ways before. Why is that and what, if anything, does each do different?
For example:
struct foo {
short a;
int b;
float c;
};
typedef struct {
short d;
int e;
float f;
} bar;
typedef struct _baz {
short a;
int b;
float c;
} baz;
int main (int argc, char const *argv[])
{
struct foo a;
bar b;
baz c;
return 0;
}
Well, the obvious difference is demonstrated in your main:
struct foo a;
bar b;
baz c;
The first declaration is of an un-typedefed struct and needs the struct keyword to use. The second is of a typedefed anonymous struct, and so we use the typedef name. The third combines both the first and the second: your example uses baz (which is conveniently short) but could just as easily use struct _baz to the same effect.
Update: larsmans' answer mentions a more common case where you have to use at least struct x { } to make a linked list. The second case wouldn't be possible here (unless you abandon sanity and use a void * instead) because the struct is anonymous, and the typedef doesn't happen until the struct is defined, giving you no way to make a (type-safe) pointer to the struct type itself. The first version works fine for this use, but the third is generally preferred in my experience. Give him some rep for that.
A more subtle difference is in namespace placement. In C, struct tags are placed in a separate namespace from other names, but typedef names aren't. So the following is legal:
struct test {
// contents
};
struct test *test() {
// contents
}
But the following is not, because it would be ambiguous what the name test is:
typedef struct {
// contents
} test;
test *test() {
// contents
}
typedef makes the name shorter (always a plus), but it puts it in the same namespace as your variables and functions. Usually this isn't an issue, but it is a subtle difference beyond the simple shortening.
It's largely a matter of personal preference. I like to give new types a name starting with a capital letter and omit the struct, so I usually write typedef struct { ... } Foo. That means I cannot then write struct Foo.
The exception is when a struct contains a pointer to its own type, e.g.
typedef struct Node {
// ...
struct Node *next;
} Node;
In this case you need to also declare the struct Node type, since the typedef is not in scope within the struct definition. Note that both names may be the same (I'm not sure where the underscore convention originated, but I guess older C compilers couldn't handle typedef struct X X;).
All your uses are syntactically correct. I prefer the following usage
/* forward declare all structs and typedefs */
typedef struct foo foo;
.
.
/* declare the struct itself */
struct foo {
short a;
int b;
foo* next;
};
Observe that this easily allows to use the typedef already inside the declaration of the struct itself, and that even for struct that reference each other mutually.
The confusion comes about because some of the declarations are in fact declaring up to three C constructs. You need to keep in mind the difference between:
A typedef declaration,
A struct definition, and
A struct declaration.
They are all very different C constructs. They all do different things; but you can combine them into the one compound construct, if you want to.
Let's look at each declaration in turn.
struct foo {
short a;
int b;
float c;
};
Here we are using the most basic struct definition syntax. We are defining a C type and give it the name foo in the tag namespace. It can later be used to declare variables of that type using the following syntax:
struct foo myFoo; // Declare a struct variable of type foo.
This next declaration gives the type another name (alias) in the global namespace. Let's break it down into its components using the previous basic declaration.
typedef foo bar; // Declare bar as a variable type, the alias of foo.
bar myBar; // No need for the "struct" keyword
Now just replace "foo" with the the struct's definition and voila!
typedef struct {
short d;
int e;
float f;
} bar;
typedef struct _baz {
short a;
int b;
float c;
} baz;
The above syntax is equivalent to the following sequence of declarations.
struct _baz {
short a;
int b;
float c;
}
typedef _baz baz; // Declare baz as an alias for _baz.
baz myBaz; // Is the same as: struct _baz myBaz;