CODEWITHSUNDEEP
c++cpointersglobal-variables
Shan
Shan

Reputation: 5242

What's the difference between global and local pointers?

I have a small code something like this:

 #include <stdio.h>

    int *ptr1;
    int *ptr2;


  void some_function(void)
    {

    int B = 5;
    ptr2 = &B;
    }

  main (){
    int C, D;
    int A =10;
    int *ptr3;
    ptr1= &A;

    ptr3=ptr2;

    some_function();

    C = *ptr1 + *ptr2;

    printf("Sum of the numbers C= %d\n",C);

    some_function();
    D = *ptr1 + *ptr3;

    printf("Sum of the numbers D= %d\n",D);

  }

Why dont I get the result for D but get the result for C? I get the result for the print statement Sum of the numbers C=15 but nothing for the last print statement for D. What is the difference between local and global pointers ( I mean both ptr1 and ptr2 are defined global and ptr3 is define local)? Is the pointer assignment ptr3=ptr2 valid? are there any significant difference with pointer to local variable Vs. pointer to global variable ?

Upvotes: 0

Views: 1062

Answers (2)

Grijesh Chauhan
Grijesh Chauhan

Reputation: 58291

ptr3=ptr2; coped before the call of function some_function(); so wrong address assigned to ptr3 that is NULL because ptr2 is 0 (by-default value of global variable).

Do like this: 

some_function();
ptr3=ptr2;
D = *ptr1 + *ptr3;
printf("Sum of the numbers D= %d\n",D);

Also, following code running correct because ptr2 value is used after call of some_function(); function.

some_function();
C = *ptr1 + *ptr2;
printf("Sum of the numbers C= %d\n",C)

Ooh!, Scope of variable ptr2 points to local to some_function(). 

void some_function(void)
    {

    int B = 5;
    ptr2 = &B;
    }

Do like this: 

void some_function(void)
    {

    int* B = calloc(1,sizeof(int));
    *B = 5; 
    ptr2 = B;
    }

Also don't forget to free(ptr2) after print statement;

Upvotes: 2

ouah
ouah

Reputation: 145919

ptr3 = ptr2;

ptr3 is initialized with the value of ptr2 which is NULL. It is NULL because you didn't initialize it and it has static storage duration.

then you do

D = *ptr1 + *ptr3;

You are dereferencing ptr3 which is a null pointer: this is undefined behavior. The value of ptr2 may have changed during your program, but ptr3 is still a null pointer.

EDIT: For the sake of completeness (well, sort of):

void some_function(void)
{
    int B = 5;
    ptr2 = &B;
}

B object lifetime ends when the functions some_function exits. You cannot access B through ptr2 after some_function has returned. It means:

C = *ptr1 + *ptr2;

also invokes undefined behavior because you are dereferencing ptr2 which is an invalid pointer in this statement.

Upvotes: 7

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