output of program

Question

void main()

{

int *x;

x =f();

printf("%d", *x);

}

int *f()

{

int y = 7 ;

return &y;

}

Comments

it it he case of dagling pointer and here result of this execution  would be compiler dependent .....

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@hs_yadav Bro according to me it should be 7  , i have done many questions on pointer like this only plz explain detial

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You are trying to print the value of y. Here y is local to fun(). So after fun() call,  fun() gets pushed onto the stack. Y gets some address there. Then after returning the address of y, the fun() pops out of the stack which means y also is no more present. Now the address that was allocated for y is lost.

Which means x is pointing to some address which no more belongs to the given program. So when you try to access the value present in an address which is not part of your program, it gives error.

Answer 1

yes, ans is unexpexted because when function definition is completed then local variable of function will discard

Comments

Here you are not returning the value of y ..you are returning the address of y..so we can retain the value.