C: how does pointer arithmetic works in this case , code is below and explain the working of this ?

Question

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#include<stdio.h>
int main(){
	
	int a=10; //initilising a as 10
	int *p;   //pointer p
	
	p=&a;     //holds the address of a
	
	printf("p=%d\n*p=%d",p,*p);
	
	printf("\np=%d",p+1);
	printf("\n*p=%d",*(p+1));  //here i am facing the problem to get the value at p+1
	
	return 0;
}

What I have tried:

I was learning about pointer and I was struct with the pointer arithmetic .I have understood that the if we de p=p+1 then the address at the p pointer will be increased by 4 as it is an integer but if we want to find the value at the p+1 that is *(p+1) that is the problem I am facing with different result each time . Please explain me the working of it in the Ans and how to calculate the value . and also I am also confused that the Ans will be garbage or something else. Till now when I apply my logic according to me the value at that address is not having any value assigned to p+1.please Ans

Posted 14-May-21 19:06pm

vishal kumawat

Updated 4-Aug-21 20:51pm

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2 solutions

Solution 2

A few tips :

You should try this experiment again but this time make a an array and while you're at it, give it values :

C++

int a[ 5 ] = { 0, 1, 2, 3, 4 };

You can also move the pointer. This statement is valid : ++p; Try it and see the results. --p; is also valid, but will point to invalid memory if p equals the address of a[0].

I prefer to see addresses in hex so I use a statement like this to display them :

C++

printf( "p=%08X\n", p );

So, combining all this together and making a function out of it can result in this code :

C++

void displayPointer( int * p, int a[] )
{
    printf( "  p=%08X, value is %d\n", p, * p ); 
    printf( "p+1=%08X, value is %d\n", p+1, *(p+1) ); 

    printf( "incrementing pointer\n" );
    ++p;
    printf( "  p=%08X, value is %d\n", p, * p ); 

    printf( "decrementing pointer\n" );
    --p;
    printf( "  p=%08X, value is %d\n", p, * p ); 
}

int main()
{
    int a[ 5 ] = { 0, 1, 2, 3, 4 };
    int * p;

    p = & a[ 0 ];
    printf( "p is address of a[ 0 ] :\n" );
    displayPointer( p, a );

    p = & a[ 2 ];
    printf( "p is address of a[ 2 ] :\n" );
    displayPointer( p, a );
    return 0;
}

Try running this and observe the results.

Posted 15-May-21 8:49am

Rick York

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vishal kumawat 18-May-21 1:41am

tx @Rick York

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OriginalGriff · Accepted Answer · 2021-05-14T19:50:00

The problem is simple: a is any integer. It contains a value, which you can print.
But p + 1 move away froma completely, and advances to the next integer - which hasn't been defined, and contains no value you are aware of. What do you expect it to print? It hasn't been allocated, so it doesn't even have a default value!

Change a to a small array of values, set them all, and then use pointer arithmetic to advance through them. When you start pointing to undefined memory, you start to get a lot of problems ...