This question already has answers here:
Why are global and static variables initialized to their default values?
(5 answers)
Closed 6 years ago.
#include <stdio.h>
int a[100];
int main(){
printf("%d",a[5]);
return 0;
}
Does the above code always print '0' or is it compiler specific? I'm using gcc compiler and I got the output as '0'.
Yes, all members of a are guaranteed to be initialised to 0.
From section 3.5.7 of the C89 standard
If an object that has static storage duration is not initialized
explicitly, it is initialized implicitly as if every member that has
arithmetic type were assigned 0 and every member that has pointer type
were assigned a null pointer constant.
"Global variables" are defined at file scope, outside any function. All variables that are defined at file scope and all variables that are declared with the keyword static have something called static storage duration. This means that they will be allocated in a separate part of the memory and exist throughout the whole lifetime of the program.
It also means that they are guaranteed to be initialized to zero on any C compiler.
From the current C standard C11 6.7.9/10:
"... 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;"
Practically, this means that if you initialize your global variable to a given value, it will have that value and it will be allocated in a memory segment usually referred to as .data. If you don't give it a value, it will be allocated in another segment called .bss. Globals will never be allocated on the stack.
Yes.
Any global variable is initialized to the default value of that type.
0 is the default value and is automatically casted to any type.
If it is a pointer, 0 becomes NULL
Global variables get there space in the data segment which is zeroed out.
It is not compiler specific but defined in the C standard.
So it will always print 0.
File scope objects declared without explicit initializers are initialized by 0 by default (and to NULL for pointers).
Non-static objects at block scope declared without explicit initializers are left uninitialized.
Are the globle variable always initalized to zero in C?
Yes and It's defined in the C standard.
It is not compiler specific. The code will always print 0.
Related
This question already has answers here:
Is un-initialized integer always default to 0 in c?
(4 answers)
(Why) is using an uninitialized variable undefined behavior?
(7 answers)
Closed 4 years ago.
What will happen when we try to print the uninitialized variable in c that may be global or local? In C also variables automatically initialized by compiler with default values?
Here is my code
#include<stdio.h>
int b;
int main()
{
float a;
printf("%f %d",a,b);
}
I got the output as
0.000000 0
Please can any one explain this?
Global variables, which has static storage duration, are implicitly initialized.
Local variables, having automatic storage duration are not initialized unless done explicitly. However, if local variable has static storage class, it is initialized.
Quoting C11,
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 orunsigned) zero;
— if it is an aggregate, every member is initialized (recursively) according to these rules;
— if it is a union, the first named member is initialized (recursively) according to these rules.
This question already has an answer here:
C struct automatic initialization values, array initializations
(1 answer)
Closed 4 years ago.
Consider the following code:
void func()
{
int p;
...
if (p > MAX) {
struct my_struct s;
...
/* here we access the contents 's' as '&s' */
}
}
In this snippet s is on the stack. Is it guaranteed that the compiler initializes all structure fields to zero?
If a variable (struct or otherwise) is declared local to a function or a containing scope (i.e. has automatic storage duration), it is not initialized in any way. You need to explicitly set the fields in the struct.
If you initialize at least one field of a struct but not all, then the remaining fields will be initialized the same as file scope variables (i.e. variables with static storage duration), which means NULL for pointer types and 0 for numeric types.
From section 6.7.9 of the C standard:
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.
No, it's quite the opposite.
Since s is an automatic storage local scoped (i.e., block scoped) variable, unless initialized explicitly, the contents are indeterminate.
Quoting C11, chapter §6.7.9
If an object that has automatic storage duration is not initialized explicitly, its value is
indeterminate. [...].
However, if you want to zero-initialize the variable for an(y) aggregate type, you can simply use an initialization statement like
aggregate-type variable = {0};
which uses the following property from paragraph 21 of the same chapter, (emphasis mine)
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.
No, they won't be initialized at all. The structure values will end up with whatever garbage is on the stack where the structure is placed.
When I declare an int variable, and do not not assign a value to it, and then when I print it, a random number gets printed.
but when I allocate a char variable, and print it with %c format specifier, nothing is printed. So does a char variable in C have a default value like null? Do local variables in C start with a random value? Then why doesn't a char behave in this way?
There is no default value which is assigned to it. Some values are not printable and you can assume that random value is one of them, so that is the reason why you are not able to see the result.
The C99 standard says that:
If an object that has automatic storage duration is not initialized
explicitly, its value is indeterminate.
On a side note:
As per C99
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;
if it is a union, the first named member is initialized (recursively) according to these rules.
Automatic variables that are not initialized have indeterminate value, we can see this by going to the draft C99 standard section 6.7.8 Initialization:
If an object that has automatic storage duration is not initialized explicitly, its value is
indeterminate.
using an indeterminate value is undefined behavior. The definition for indeterminate values is as follows:
either an unspecified value or a trap representation
It just may be the case that the values you are ending up with for char are not printable.
There is not default value assigned. Never trust in a default value you didn't assign.
I have a question about the initialization of static variables in C. I know if we declare a global static variable that by default the value is 0. For example:
static int a; //although we do not initialize it, the value of a is 0
but what about the following data structure:
typedef struct
{
int a;
int b;
int c;
} Hello;
static Hello hello[3];
are all of the members in each struct of hello[0], hello[1], hello[2] initialized as 0?
Yes, all members are initialized for objects with static storage. See 6.7.8/10 in the C99 Standard (PDF document)
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;
— if it is a union, the first named member is initialized (recursively) according to these
rules.
To initialize everything in an object, whether it's static or not, to 0, I like to use the universal zero initializer
sometype identifier0 = {0};
someothertype identifier1[SOMESIZE] = {0};
anytype identifier2[SIZE1][SIZE2][SIZE3] = {0};
There is no partial initialization in C. An object either is fully initialized (to 0 of the right kind in the absence of a different value) or not initialized at all.
If you want partial initialization, you can't initialize to begin with.
int a[2]; // uninitialized
int b[2] = {42}; // b[0] == 42; b[1] == 0;
a[0] = -1; // reading a[1] invokes UB
Yes, they are, as long they have static or thread storage duration.
C11 (n1570), § 6.7.9 Initialization #10
If an object that has static or thread storage duration is not initialized
explicitly, then:
[...]
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;
[...]
Yes, file-scope static variables are initialized to zero, including all members of structures, arrays, etc.
See this question for reference (I'll vote to close this as a duplicate, too).
Edit: this question is getting much better answers, so I'm voting to close that question as a duplicate of this, instead.
For reference, here is the C FAQ link from that question's accepted answer, although of course the C99 and C11 standards linked here are canonical.
I would add that static variables (or arrays) are classified into two types.
Initialized are the ones that are given value from code at compile time. These are usually stored in DS though this is compiler specific.
The other type is uninitialized statics which are initialized at run time and are stored into BSS segment though again this is compiler specific.
BSS
For the ones who don't want to read the standard, it's also mentioned in https://en.cppreference.com/w/c/language/initialization :
Implicit initialization
If an initializer is not provided:
objects with automatic storage duration are initialized to indeterminate values (which may be trap representations)
objects with static and thread-local storage duration are zero-initialized
What should the code print? 0 or any garbage value or will it depend on the compiler?
#include <stdio.h>
int a;
int main()
{
printf("%d\n",a);
return 0;
}
the answer is 0. Global variables are initialized to zero.
I would say your code might output anything or simply anything can happen because your code invokes Undefined Behaviour as per C99.
You don't have a prototype for printf in scope.
J.2 Undefined behavior
— For call to a function without a function prototype in scope where the function is defined with a function prototype, either the prototype ends with an ellipsis or the types of the arguments after promotion are not compatible with the types of the parameters (6.5.2.2).
If the question is about initialization of global variables then a would be initialized to 0 because it has static storage duration.
I found on C99 standard, Section 6.7.8.10, Initialization:
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;
— if it is a union, the first named member is initialized (recursively) according to these
rules.
Section 6.2.4.3 defines:
An object whose identifier is declared with external or internal linkage, or with the
storage-class specifier static has static storage duration. Its lifetime is the entire
execution of the program and its stored value is initialized only once, prior to program
startup.
In other words, globals are initialized as 0. Automatic variables (i.e. non-static locals) are not automatically initialized.
without automatic variable [generally what we use in function in most cases] all other variable's value is assigned to 0
Global variables are initialized as 0. Automatic variables (i.e. non-static locals) are not automatically initialized.