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I see a difference between global and static variables These two variables do the same thing. But the theory says that these two are different
1Code:#include <stdio.h>
static int i = 0;
void m() {
static int j = 0;
j++;
printf("%d \n", j);
}
int main() {
m();
m();
i++;
printf("%d \n", i);
i++;
printf("%d \n", i);
return 0;
}
2
1
2
i is not a "true" global variable: because of the use of the static keyword, it is a file scope variable that is restricted to the current translation unit (i.e., source file + headers), i.e., another file scope variable named i in a different source file would refer to a different object. We say that the static keyword causes the identifier i to have internal linkage.
If you omitted the static keyword, then i would have external linkage, i.e., unless it is shadowed by another variable named i in a local scope, all variables named i would refer to this object.
With or without the static keyword, the object referred to by i has static storage duration, i.e., its lifetime is that of the program.
j, on the other hand, is a static local variable. Like i, it has static storage duration. Unlike i, it is local, i.e., the only place where j refers to that particular object is in the scope of m().
A little hard to understandQuote:
Originally Posted by laserlight
What is the similarity and difference between global and static variable ?
similarity
Both will variables will initialize once they will not create and destroy each time
Both variables will alive as long as the program is running
difference
We always declare the global variable outside the main function, whereas the static variable can be local or global.
Depending on the compiler (usually older ones), global variables may not be initialized, unless explicitly initialized, while static variables are initialized to zero by default.
A compiler that doesn't initialize global variables is non-standard and should be used with care (or, if possible, discarded in favor of one that does follow the relevant standard(s)).Quote:
Originally Posted by rcgldr
When and where is it stated that global variables are initialized to zero if there is not a initial value used?Quote:
Originally Posted by christop
Tim S.
Since ANSI/c89Quote:
Originally Posted by stahta01
6.7.9p10
Edit:Quote:
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;
Just realised that 6.2.4p3 is also relevant
Quote:
An object whose identifier is declared without the storage-class specifier _Thread_local, and either 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.
So, it says nothing about global variables!Quote:
Originally Posted by Hodor
Tim S.
It ultimately stems from the fact that in a multi-user machine, memory initially allocated to a new process is zeroed by the OS, lest it contain information from a previous process, possibly run by a different user.
Yeah it does. 6.2.4p3Quote:
Originally Posted by stahta01
"global" variables are initialised as if they were declared static
Edit: It's talking about storage duration here, not linkage. "globals" in C have static storage duration and are therefore initialised as per 6.7.9
Quote:
Originally Posted by stahta01
By older compilers, I meant ones that aren't c89 compliant. This would include most 16 bit compilers for X86 (like Microsoft (although the last 16 bit tool set was released in 1994, I don't think it's c89 compliant)). You would need to be running on a 16 bit or 32 bit OS, perhaps using a virtual machine, in order to run these old compilers and 16 bit programs. Rarer still would be compilers for the old 68000 systems (Amiga, Atari ST, Macintosh).Quote:
Originally Posted by Hodor