Static variable inside of a function in C

Question

What will be printed out  6 6 or 6 7  And why   void foo         static int x   5      x        printf   d   x      int main         foo        foo        return 0

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Let s just read the Wikipedia article on Static Variables        Static local variables  variables declared as static inside a function are statically allocated while having the same scope as automatic local variables  Hence whatever values the function puts into its static local variables during one call will still be present when the function is called again

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In C  11 at least  when the expression used to initialize a local static variable is not a  constexpr   cannot be evaluated by the compiler   then initialization must happen during the first call to the function  The simplest example is to directly use a parameter to intialize the local static variable  Thus the compiler must emit code to guess whether the call is the first one or not  which in turn requires a local boolean variable  I ve compiled such example and checked this is true by seeing the assembly code  The example can be like this   void f  int p       static const int first p   p     cout  lt  lt   first p       lt  lt  p  lt  lt  endl      void main        f 1   f 2   f 3       of course  when the expresion is  constexpr   then this is not required and the variable can be initialized on program load by using a value stored by the compiler in the output assembly code

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Output  6 7  Reason  static variable is initialised only once  unlike auto variable  and further definition of static variable would be bypassed during runtime  And if it is not initialised manually  it is initialised by value 0 automatically  So    void foo         static int x   5     assigns value of 5 only once     x        printf   d   x      int main         foo       x   6     foo       x   7     return 0

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That is the same as having the following program   static int x   5   void foo         x        printf   d   x      int main          foo         foo         return 0      All that the static keyword does in that program is it tells the compiler  essentially   hey  I have a variable here that I don t want anyone else accessing  don t tell anyone else it exists    Inside a method  the static keyword tells the compiler the same as above  but also   don t tell anyone that this exists outside of this function  it should only be accessible inside this function    I hope this helps

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There are two issues here  lifetime and scope   The scope of variable is where the variable name can be seen   Here  x is visible only inside function foo     The lifetime of a variable is the period over which it exists   If x were defined without the keyword static  the lifetime would be from the entry into foo   to the return from foo    so it would be re-initialized to 5 on every call   The keyword static acts to extend the lifetime of a variable to the lifetime of the programme  e g  initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo

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6 and 7 Because static variable intialise only once  So 5   becomes 6 at 1st call       6   becomes 7 at 2nd call      Note-when 2nd call occurs it takes x value is 6 instead of 5 because x is static variable

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6 7  compiler arranges that static variable initialization does not happen each time the function is entered

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Output  6 7  Reason  The declaration of x is inside foo but the x 5 initialization takes place outside of foo   What we need to understand here is that   static int x   5    is not the same as  static int x  x   5    Other answers have used the important words here  scope and lifetime  and pointed out that the scope of x is from the point of its declaration in the function foo to the end of the function foo  For example I checked by moving the declaration to the end of the function  and that makes x undeclared at the x    statement   So the static int x  scope  part of the statement actually applies where you read it  somewhere INSIDE the function and only from there onwards  not above it inside the function   However the x   5  lifetime  part of the statement is initialization of the variable and happening OUTSIDE of the function as part of the program loading  Variable x is born with a value of 5 when the program loads   I read this in one of the comments   Also  this doesn t address the really confusing part  which is the fact that the initializer is skipped on subsequent calls   It is skipped on all calls  Initialization of the variable is outside of the function code proper   The value of 5 is theoretically set regardless of whether or not foo is called at all  although a compiler might optimize the function away if you don t call it anywhere  The value of 5 should be in the variable before foo is ever called   Inside of foo  the statement static int x   5  is unlikely to be generating any code at all   I found the address x uses when I put a function foo into a program of mine  and then  correctly  guessed that the same location would be used if I ran the program again  The partial screen capture below shows that x has the value 5 even before the first call to foo

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The output will be 6 7  A static variable  whether inside a function or not  is initialized exactly once  before any function in that translation unit executes  After that  it retains its value until modified

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You will get 6 7 printed as  as is easily tested  and here s the reason  When foo is first called  the static variable x is initialized to 5  Then it is incremented to 6 and printed   Now for the next call to foo  The program skips the static variable initialization  and instead uses the value 6 which was assigned to x the last time around  The execution proceeds as normal  giving you the value 7

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6 7   x is a global variable that is visible only from foo    5 is its initial value  as stored in the  data section of the code  Any subsequent modification overwrite previous value  There is no assignment code generated in the function body

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A static variable inside a function has a lifespan as long as your program runs  It won t be allocated every time your function is called and deallocated when your function returns

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Vadiklk   Why      Reason is that static variable is initialized only once  and maintains its value throughout the program  means  you can use static variable between function calls  also it can be used to count  how many times a function is called    main        static int var   5     printf   d   var--      if var        main         and answer is 5 4 3 2 1 and not  5 5 5 5 5 5       infinite loop  as you are expecting  again  reason is static variable is initialized once  when next time main   is called it will not be initialize to 5 because it is already initialized in the program So we can change the value but can not reinitialized  Thats how static variable works    or you can consider as per storage  static variables are stored on Data Section of a program and variables which are stored in Data Section are initialized once  and before initialization they are kept in BSS section   In turn Auto local  variables are stored on Stack and all the variables on stack  reinitialized all time when function is called as new FAR function activation record  is created for that   okay for more understanding  do the above example without  static  and let you know what will be the output  That make you to understand the difference between these two   Thanks Javed

[c] Static variable inside of a function in C

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