self doubt

Question

i am getting 4,3,4,3,8,2 is it right?????

Comments

Yup ^

vijju532 variable 'a' is static inside f() so it should retain it's value that the 1st call to f() has made to it. So when f() is called for the next time you should treat a's value as 4 and not 2.'b' is not a static variable so it will again be initialized to 2. So a becomes 8 after the multiplication.

Also the printf inside main will print the value of a that is present inside the main block. So it is 2.

Since b is not present inside main so the global b is referred. It has default value=0.

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yes.....

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thanx  @MiNiPanda

Answer 1

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Comments

if ++,-- high precedence than * , then why not increment first....a*=b++ (a=a*b++=2*3=6)

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yes. b++ is evaluated first. but, this is post-increment.

a=a*b++; // i.e., a=(a*(b++));    //b++ will take the value of b and then b is increamented.

same as, a=a*b; and b=b+1;

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If you evaluate a=a*++b; //then it will evaluate like a=(a*(++b)); // a will be multiplied with increamented value of b.

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yes...thanks

Answer 2

int a,b;   /* Here a,b are global variable whose inital value is 0 for both a,b*/
void f();
main()
{ 
    static int a =1;  /*static keyword is one time initialization, use many time*/
    f();
    a*=2;         /* line no 7 is same as a=a*2*/
    f();
    printf("%d%d",a,b);   /* printf function print the value a,b in main scope but here in main function b is not define so it will take value from global variable b */
}
void f()
{
    static int a =2; /*same use as line no 5 */
    int b =2  /* here b is define in the f function scope, it's not static so every time when b is initialized, it will assigned as b = 2 */
    a*=b++;   /* line no 15 can be written as a = a*b++ */
    printf("%d%d",a,b);   /* this printf print the values of a,b which is define in function f scope */
    
    
    
    
}
    

execution of program is as follow:

Line no 1 is executed first and assign the value 0 to both GV a,b. After that main function is executed.

In main function a is static variable assign the value 1, then call the function f().

In function F() a is static variable & b is normal variable, both assign the value 2. Line no 15 use 2 operator * and ++(post-increment). Here post-increment has high priority than * and associativity is R2L. This gives the value of a,b as 4,3.

after that comes to the main function & assigned the new value of a as 2, again call function f() and do the same things. static var a use current value 4 and variable b again initialized as 2, again execute line no 15 which gives the value of a,b as 8,3.

returning to the main function, printf function print a & b value, here b is not defined so it used the value of GV as 0  so printf gives 2,0.

so values of a & b are respectively according to their printf statements:

 

a	b
4	3
8	3
2	0