I've seen code like this:
char str[1024] = {0, };
and suspect that it is similar to doing this:
char str[1024];
str[0] = '\0';
But I couldn't find anything on it so I'm not sure.
What is this (called) and what does it do?
Disclaimer: I'm aware this might have been asked and answered before, but searching for {0, } is astonishingly hard. If you can point out a duplicate, I'll happily delete this question.
No, they are not the same.
This statement
char str[1024] = {0, };
initializes the first element to the given value 0, and all other elements are to be initialized as if they have static storage, in this case, with a value 0. Syntactically this is analogous to using
char str[1024] = {0};
Quoting C11, chapter 6.7.9, p21
If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.
and, from p10 (emphasis mine)
If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static or thread storage duration is not initialized explicitly, then:
if it has pointer type, it is initialized to a null pointer;
if it has arithmetic type, it is initialized to (positive or unsigned) zero;
if it is an aggregate, every member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;
if it is a union, the first named member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;
On the other hand
char str[1024];
str[0] = '\0';
only initializes the first element, and the remaining elements remains unitialized, containing indeterminate values.
The {0, } initializer is the same as the more common {0} initializer.
The trailing comma is allowed by the syntax but it makes no difference.
6.7.9--Initialization:
initializer:
assignment-expression
{ initializer-list }
{ initializer-list , }
initializer-list:
designationopt initializer
initializer-list , designationopt initializer
designation:
designator-list =
designator-list:
designator
designator-list designator
designator:
[ constant-expression ]
. identifier
The semantics are such that the 0, which is syntactically required because at least one initializer-list item is syntactically required (it's a kind of a arbitrary requirement: compilers frequently support an empty {} as well) initializes the first subobject recursively (6.7.9p17):
Each brace-enclosed initializer list has an associated current object. When no designations are present, subobjects of the current object are initialized in order according to the type of the current object: array elements in increasing subscript order, structure members in declaration order, and the first named member of a union.148) In contrast, a designation causes the following initializer to begin initialization of the subobject described by the designator. Initialization then continues forward in order, beginning with the next subobject after that described by the designator.
and the rest is initialized as it would be if the whole object had static storage duration (6.7.9p19,6.7.9p21). This practically means to 0 (as with memset(,0,) although with the caveat that paddings need not be initialized and that 0-initialized pointers need not necessarily be "all-bits zero".
As far as I know, compilers on usual platforms (where pointers are all-bits-zero) just mostly do what they would do with memset(,0,).
This "universal" zero initialization works because the first 0 will recursively hit a scalar type (number on pointer) which can be invariably initialized with the 0 initializer. The default "as-with-static-storage-duration" then initialization applies to the rest.
A perhaps slightly more interesting of the trailing initializer comma doing nothing would be:
int main()
{
char one[]={0,}; //<the comma doesn't introduce another member
_Static_assert(sizeof(one)==1,""); //holds
}
This seems like a hole in my knowledge. As far as I am aware, in C99 if you initialise a single element of a struct and no others, the others are zero initialised. Does the following code zero initialise all the members of a struct though?
typedef struct
{
int foo;
int bar;
char* foos;
double dar;
} some_struct_t;
some_struct_t mystructs[100] = {};
Update: There are some comments indicating that this syntax is an extension. If that is the case, is there any way of doing this that is pure C99 compliant?
As per C11, chapter §6.7.9, Initialization syntax, (for the sake of completeness, same mentioned in chapter §6.7.8 in C99)
initializer:
assignment-expression
{ initializer-list }
{ initializer-list , }
initializer-list:
designationopt initializer
initializer-list , designationopt initializer
designation:
designator-list =
designator-list:
designator
designator-list designator
designator:
[ constant-expression ]
. identifier
Which implies, the brace closed initializer list should have at minimum one initializer element (object).
In your code, the empty initializer list
some_struct_t mystructs[100] = {}; //empty list
is not a valid pure C syntax; it's a compiler extension.
You need to mention a single element in the list to make it standard conforming, like
some_struct_t mystructs[100] = {0};
which meets the criteria, from paragraph 21 of same standard(s),
If there are fewer initializers in a brace-enclosed list than there are elements or members
of an aggregate, or fewer characters in a string literal used to initialize an array of known
size than there are elements in the array, the remainder of the aggregate shall be
initialized implicitly the same as objects that have static storage duration.
So, in this case, you have one explicit 0 and remaining implicit zero-initialization (or similar).
Because it's array initialisation, you would need
some_struct_t mystructs[100] = { 0 }; // ensure all array elements (struct) being zero initialisation
For structs/unions (and arrays) there is a rule saying that if it is partially initialized, the rest of the items that didn't get initialized explicitly by the programmer are set to zero.
So by typing some_struct_t mystructs[100] = {0}; you tell the compiler to explicitly set foo to zero. And then the rest of the struct members gets set to zero as well, implicitly.
This has nothing to do with C99, but works for all C standard versions. Although in C99/C11, a designated initializer {.foo=0} would have achieved the very same result.
Consider the following struct initialization:
#include<stdio.h>
struct bar {
int b;
int a;
int r;
};
struct foo {
struct bar bar;
};
int main(int argc, char **argv) {
struct bar b = {1, 2, 3};
struct foo f = {.bar = b, .bar.a = 5 };
// should this print "1, 5, 3", "1, 5, 0", or "0, 5, 0"?
// clang on Mac prints "1, 5, 3", while gcc on Ubuntu prints "0, 5, 0"
printf("%d, %d, %d\n", f.bar.b, f.bar.a, f.bar.r);
return 0;
}
The C11 standard seems to do a quite poor job of describing what behavior should be expected here in section 6.7.9, but seems to think it's doing a reasonable job, as I don't see any warnings regarding undefined behavior in this case either.
In practice, it seems the behavior is either not standardized or the standard is violated by at least one common compiler, with clang/llvm 8.0.0 on a Mac producing "1, 5, 3", and gcc 5.4 on Ubuntu producing "0, 5, 0".
According to the C standard, should f.bar.b and f.bar.r well defined at this point, or does this initialization result in undefined or unspecified behavior?
The C11 standard seems to do a quite poor job of describing what behavior should be expected here in section 6.7.9,
Standardese can be difficult to read, but I don't think this area of the standard is worse in that respect than should be expected.
but seems to think it's doing a reasonable job, as I don't see any warnings regarding undefined behavior in this case either.
The standard is not required to explicitly declare undefined behaviors. Indeed, the standard contains a blanket statement that wherever it does not define behavior for a given piece of code, that code's behavior is undefined. Nevertheless, I do think section 6.7.9 covers this area pretty thoroughly. The main area left open is this:
The evaluations of the initialization list expressions are indeterminately sequenced with respect to one another and thus the order in which any side effects occur is unspecified.
(C2011, 6.7.9/23)
That doesn't present any problem for your example.
In practice, it seems the behavior is either not standardized or the standard is violated by at least one common compiler, with clang/llvm on a Mac producing "1, 5, 3", and gcc on Ubuntu producing "0, 5, 0".
I'm completely prepared to believe that one or the other of those is non-conforming in this area. However, do also pay attention to compiler versions and compilation options -- they may be compiling for different versions of the standard, with or without extensions.
According to the C standard, should f.bar.b and f.bar.r well defined at this point, or does this initialization result in undefined or unspecified behavior?
If the declaration of an object has an associated initializer then the whole object is initialized, and furthermore, the resulting initial value is well-defined by the standard, subject to caveats arising from 6.7.9/23. As for the initial values required of a conforming implementation in your example, the key provisions are these:
The initialization shall occur in initializer list order, each initializer provided for a particular subobject overriding any previously listed initializer for the same subobject; all subobjects that are not initialized explicitly shall be initialized implicitly the same as objects that have static storage duration.
(C2011, 6.7.9/19; emphasis added)
Each designator list begins its description with the current object associated with the closest surrounding brace pair. Each item in the designator list (in order) specifies a particular member of its current object and changes the current object for the next designator (if any) to be that member. The current object that results at the end of the designator list is the subobject to be initialized by the following initializer.
(C2011, 6.7.9/18; emphasis added)
If the aggregate or union contains elements or members that are aggregates or unions, these rules apply recursively to the subaggregates or contained unions.
(C2011, 6.7.9/20)
Thus, given f's initializer,
struct foo f = {.bar = b, .bar.a = 5 };
we first process element .bar = b, as required by 6.7.9/19. That contains a designator list designating foo.b, of type struct bar, as the object to initialize from the following initializer. This initializer exercises the option of being "a single expression that has compatible structure or union type", per 6.7.9/13, therefore the initial value of f.bar is the value of b, subject to partial or full override by subsequent initializers.
We next process the second element, .bar.a = 5. This initializes f.bar.a and only that subobject, per 6.7.9/18, overriding the initialization specified by the previous initializer per 6.7.9/19.
The result of conforming initialization thus leads to printing
1, 5, 3
GCC seems to be failing by re-initializing all of f.bar when it processes the the second initializer, instead of only f.bar.a.
In the C Standard there is written (6.7.9 Initialization)
17 Each brace-enclosed initializer list has an associated current
object. When no designations are present, subobjects of the current
object are initialized in order according to the type of the current
object: array elements in increasing subscript order, structure
members in declaration order, and the first named member of a
union.148) In contrast, a designation causes the following initializer
to begin initialization of the subobject described by the designator.
Initialization then continues forward in order, beginning with the
next subobject after that described by the designator
And
19 The initialization shall occur in initializer list order, each
initializer provided for a particular subobject overriding any
previously listed initializer for the same subobject;151) all
subobjects that are not initialized explicitly shall be initialized
implicitly the same as objects that have static storage duration.
This footnote is important
148) If the initializer list for a subaggregate or contained union
does not begin with a left brace, its subobjects are initialized as
usual, but the subaggregate or contained union does not become the
current object: current objects are associated only with
brace-enclosed initializer lists.
Thus I see neither undefined nor unspecified behavior.
In my opinion the result should look like { 1, 5, 3 }.
If to leave aside the Standard then it is reasonable at first to initialize the memory with the default initializes and then overwrite it with the explicit initializers.
The standard says…
I'm going to quote from §6.7.9 Initializers of ISO/IEC 9899:2011 (the C11 standard), the same section as Vlad from Moscow quotes in his answer:
¶16 Otherwise, the initializer for an object that has aggregate or union type shall be a brace-enclosed list of initializers for the elements or named members.
¶17 Each brace-enclosed initializer list has an associated current object. When no designations are present, subobjects of the current object are initialized in order according to the type of the current object: array elements in increasing subscript order, structure members in declaration order, and the first named member of a union.148) In contrast, a designation causes the following initializer to begin initialization of the subobject described by the designator. Initialization then continues forward in order, beginning with the next subobject after that described by the designator.149)
¶18 Each designator list begins its description with the current object associated with the closest surrounding brace pair. Each item in the designator list (in order) specifies a particular member of its current object and changes the current object for the next designator (if any) to be that member.150) The current object that results at the end of the designator list is the subobject to be initialized by the following initializer.
¶19 The initialization shall occur in initializer list order, each initializer provided for a particular subobject overriding any previously listed initializer for the same subobject;151) all subobjects that are not initialized explicitly shall be initialized implicitly the same as objects that have static storage duration.
¶20 If the aggregate or union contains elements or members that are aggregates or unions, these rules apply recursively to the subaggregates or contained unions. If the initializer of a subaggregate or contained union begins with a left brace, the initializers enclosed by that brace and its matching right brace initialize the elements or members of the subaggregate or the contained union. Otherwise, only enough initializers from the list are taken to account for the elements or members of the subaggregate or the first member of the contained union; any remaining initializers are left to initialize the next element or member of the aggregate of which the current subaggregate or contained union is a part.
¶21 If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.
148) If the initializer list for a subaggregate or contained union does not begin with a left brace, its subobjects are initialized as usual, but the subaggregate or contained union does not become the current object: current objects are associated only with brace-enclosed initializer lists.
149) After a union member is initialized, the next object is not the next member of the union; instead, it is the next subobject of an object containing the union.
150) Thus, a designator can only specify a strict subobject of the aggregate or union that is associated with the surrounding brace pair. Note, too, that each separate designator list is independent.
151) Any initializer for the subobject which is overridden and so not used to initialize that subobject might not be evaluated at all.
Interpretation
I think your code is well-formed and that GCC is handling it incorrectly and Clang is handling it correctly.
With your code modified only so that the unused argc and argv are replaced by void, running on a Mac with macOS Sierra 10.12.1, compiling with GCC 6.2.0 and with Apple's clang version 'Apple LLVM version 8.0.0 (clang-800.0.42.1)', I get the same results as you:
0, 5, 0 from GCC.
1, 5, 3 from Clang.
The key wording in the standard is:
In contrast, a designation causes the following initializer to begin initialization of the subobject described by the designator.
In your initializer, you have:
struct foo f = { .bar = b, .bar.a = 5 };
The first part of the initializer, .bar = b, clearly initializes the bar subobject. At that point, .bar.b has the value 1, .bar.a has the value 2, .bar.r has the value 3. If you omit the , .bar.a = 5 portion of the initializer, the compilers agree.
When you include the , .bar.a = 5, the designator causes the following initialize to begin intialization of the subobject described by the designator — and the designator is .bar.a so the initialization 5 initializes .bar.a. The compilers agree on this; both set .bar.a to 5. But the subobject designated by .bar was previously initialized, so the initializer for .bar.a only affects the .a element; it should not override any other element.
If the initializer is extended with with , 19, then the 19 is not a designation, but it initializes the subobject after the previous designation, which is .bar.r. Both the compilers agree with this.
This test code, a minor variant on your code, illustrates:
#include <stdio.h>
struct bar
{
int b;
int a;
int r;
};
struct foo
{
struct bar bar;
};
static inline void foobar(struct foo f)
{
printf("%d, %d, %d\n", f.bar.b, f.bar.a, f.bar.r);
}
int main(void)
{
struct bar b = {1, 2, 3};
struct foo f0 = {.bar = b, .bar.a = 5 };
struct foo f1 = {.bar = b, .bar.a = 5, 19 };
struct foo f2 = {.bar = b };
foobar(f0);
foobar(f1);
foobar(f2);
return 0;
}
The output from GCC is:
0, 5, 0
0, 5, 19
1, 2, 3
The output from Clang is:
1, 5, 3
1, 5, 19
1, 2, 3
Note that even with no warnings specifically enabled, clang gripes about this code:
$ clang -O3 -g -std=c11 so-4092-0714.c -o so-4092-0714
so-4092-0714.c:21:36: warning: subobject initialization overrides initialization of other fields within its
enclosing subobject [-Winitializer-overrides]
struct foo f0 = {.bar = b, .bar.a = 5 };
^~~~~~
so-4092-0714.c:21:29: note: previous initialization is here
struct foo f0 = {.bar = b, .bar.a = 5 };
^
so-4092-0714.c:22:36: warning: subobject initialization overrides initialization of other fields within its
enclosing subobject [-Winitializer-overrides]
struct foo f1 = {.bar = b, .bar.a = 5, 19 };
^~~~~~
so-4092-0714.c:22:29: note: previous initialization is here
struct foo f1 = {.bar = b, .bar.a = 5, 19 };
^
2 warnings generated.
$
As I said, I think Clang is initializing these structures correctly, even if it complains more than necessary while doing so.
This is not undefined behavior.
From section 6.7.9 of the C standard:
19 The initialization shall occur in initializer list order, each initializer provided for a particular subobject overriding
any previously listed initializer for the same subobject; all
subobjects that are not initialized explicitly shall be initialized
implicitly the same as objects that have static storage duration.
So when there is a conflict among designated initializers, the last one listed takes precedence.
In your example, you initialize .bar, then .bar.b. Both of these initialize .bar, so the second one is used. So .bar is initialized, along with its subfield .bar.b, but not .bar.a or .bar.r. And because some fields are initialized but not all, the others are initialized to 0:
21 If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in
a string literal used to initialize an array of known size than
there are elements in the array, the remainder of the
aggregate shall be initialized implicitly the same as objects that
have static storage duration.
This means that the correct behavior is to output "0,5,0". So gcc is conforming and the Mac compiler is not.
Since this question was posted, the C18 standard has been released, and includes some additional clarifications that make the answer completely clear.
A clarification request similar to this question had been asked of the standards body as early as 2012, with some changes to the language being briefly discussed that might make the meaning clearer...ultimately it was decided that the C11 language was already correct, but that an example should be added to clarify.
Example 12 in section 6.7.9 of the last publicly available C17 draft demonstrates the correct behavior when a subobject is fully and then partially initialized, noting that any trailing values of the larger subobject not explicitly overwritten should survive (rather than be overwritten by default values), "because implicit initialization does not override explicit initialization."
So the correct behavior is to print "1, 5, 3".
I have following struct:
typedef struct my_struct {
int a;
int b;
int *c;
} my_struct;
is:
my_struct n = (my_struct) { .b = 3 };
equivalent to:
my_struct n = (my_struct) { .a = 0, .b = 3, .c = NULL };
What about:
my_struct n = (my_struct) { .b = 3, 0 };
They shall be initialized as if they were static, we can find this in the draft C99 standard section 6.7.8 Initialization paragraph 19 says (emphasis mine):
The initialization shall occur in initializer list order, each
initializer provided for a particular subobject overriding any
previously listed initializer for the same subobject;132) all
subobjects that are not initialized explicitly shall be initialized
implicitly the same as objects that have static storage duration.
If the following initializer is not a designator then it will pick up with the next field after that designator, which is covered in paragraph 17:
Each brace-enclosed initializer list has an associated current object.
When no designations are present, subobjects of the current object are
initialized in order according to the type of the current object:
array elements in increasing subscript order, structure members in
declaration order, and the first named member of a union.129) In
contrast, a designation causes the following initializer to begin
initialization of the subobject described by the designator.
Initialization then continues forward in order, beginning with the
next subobject after that described by the designator.130)
This applies recursively to subaggregates as per paragraph 20:
If the aggregate or union contains elements or members that are
aggregates or unions, these rules apply recursively to the
subaggregates or contained unions
The rules for initializing objects of static duration are found in section 6.7.8 paragraph 10:
If an object that has automatic storage duration is not initialized
explicitly, its value is indeterminate. If an object that has static
storage duration is not initialized explicitly, then:
— if it has pointer type, it is initialized to a null pointer;
— if it has arithmetic type, it is initialized to (positive or unsigned) zero;
— if it is an aggregate, every member is initialized (recursively) according to these rules;
[...]
Is my_struct n = (my_struct) { .b = 3 }; equivalent to my_struct n = (my_struct) { .a = 0, .b = 3, .c = NULL };?
Yes. A compound literal may fail to provide full initialization, in which case any unimitialized members initialize to zero (NULL is case of pointer member) by default.
What about my_struct n = (my_struct) { .b = 3, 0 };?
Member b and c will be initialized to 3 and 0 respectively while a will be initialized to 0 by default.
all:
In C language:
struct A
{
int a;
int b;
};
A aa = {0};
This statement initialize aa.a only or initialize the whole struture? Or the behavior depends on Compiler?
Thanks in advance!
One more ex:
struct A
{
int a;
int b;
};
struct A aa = {5};
This will initialize the whole structure but aa.b will be initialized to 0.
If you initialize only few members of a structure then all other members of that will be automatically initialized to 0.
From the standard (N1570)
6.7.9 Initialization
...
10 If an object that has automatic storage duration is not initialized explicitly, its value is
indeterminate. If an object that has static or thread storage duration is not initialized
explicitly, then:
— if it has pointer type, it is initialized to a null pointer;
— if it has arithmetic type, it is initialized to (positive or unsigned) zero;
— if it is an aggregate, every member is initialized (recursively) according to these rules,
and any padding is initialized to zero bits;
— if it is a union, the first named member is initialized (recursively) according to these
rules, and any padding is initialized to zero bits;
...
21 If there are fewer initializers in a brace-enclosed list than there are elements or members
of an aggregate, or fewer characters in a string literal used to initialize an array of known
size than there are elements in the array, the remainder of the aggregate shall be
initialized implicitly the same as objects that have static storage duration.
So in your example aa.a would be explicitly initialized to 0 due to the initializer, while aa.b would be implicitly initialized to 0 because of the clauses above.
According to the C99 standard, section 6.7.8.17, only the first member (in declaration order, i.e. the a field) is initialized explicitly in your example:
Each brace-enclosed initializer list has an associated current object. When no designations are present, subobjects of the current object are initialized in order according to the type of the current object: array elements in increasing subscript order, structure members in declaration order, and the first named member of a union.
By "designation" the standard means
A aa = {.b = 0};
This is a more precise way of letting programmers decide what fields get initialized.
This statement initialize aa.a only or initialize the whole struture?
Have a look at the example below , initializing the structure with {3} , is similar to initializing with {3,0}.
Hence in your program when you initialize with {0} , you are actually initializing both a and b (the whole structure) , with {0,0}
#include<stdio.h>
typedef struct A
{
int a;
int b;
}a;
int main()
{
a aa = {3};
printf("\na1 = %d",aa.a);
printf("\nb1 = %d",aa.b);
a bb = {3,0};
printf("\na2 = %d",bb.a);
printf("\nb2 = %d",bb.b);
}