How to print the values that are stored in the array using pointer and function in c

Question

I am new to the pointer in c, I have done the following simple array programme using the pointer.

#include<stdio.h>
void disp(int *);
void show(int a);

int main()
{
    int i;
    int marks[]={55,65,75,56,78,78,90};
    for(i=0;i<7;i++)
        disp(&marks[i]);
    return 0;
}
void disp(int *n)
{
    show((int) &n);
}
void show(int a)
{
    printf("%d",*(&a));
}

I want to get all these values that are stored in the array as output but I only get the memory number of these stored value in the array. plz, help me how to get the array values as output.

Answer 1

I guess you want to play with pointers.

Please note that void show(int a) expects int value. Therefor you do not have to do anything to a to print it. *(&a) is equivalent to a . &a gets the address of a and * dereference the pointer.

Of course it is possible that pointer entering disp(int *n) is passed down the road and dereferenced later. This is ilustrated by calling show1 function inside disp .

#include <stdio.h>
#include <string.h>

void disp(int *);  //  function disp receives the address on int value 

void show(int a);
void show1(int *a); // function show1 will receive the address of n

int main()
{
    int i;
    int marks[]={55,65,75,56,78,78,90};

    for(i=0;i<7;i++)  // 7 since you want to print all elements

        disp( &marks[i] );

    return 0;
}

void disp(int *n)
{
    show(*n); // show expects the 'int' value therefore we have to dereference the pointer. 
    show1(n); // function show1 will receive the address of n and will dereference the pointer inside the function
}

void show(int a)
{
    printf("%d ",a);
}

void show1(int *n) // show1 gives the output of the value that is stored in address n
{
    printf("%d\n",*n);  // dereference the address n to print the value
}

Output:

55 55
65 65
75 75
56 56
78 78
78 78
90 90

Answer 2

& always gives you the memory address of the variable. So &n is giving you the memory address of the variable n .

If you want the value of a pointer, use * . To get the value stored by the pointer n , you want to use (int)*n . Of course you don't need the cast at all, just *n .

I recommend going over some tutorials in C/C++ pointer basics. Pointers are a basic skill you want to have a solid foundation of.

Answer 3

If you just wanted to print all the elements of that array, you just need to do

for(i=0;i<7;i++)
    printf("%d", marks[i]))

Note that there are 7 elements in marks , the exit condition in the loop should be either i<7 or i<=6 and not i<6 .

---

You are sending the address of a variable to the function disp() as n .

The n in disp() acts similar to a local variable of that function except that it gets its value from the function calling it.

So n is stored somewhere in the memory and hence has an address. This address is what you get when you do &n . So what you see is probably meaningless (because they are allocated on stack memory ).

You do an explicit type cast to int on this address with (int) &n whose value you then pass to show() . Read about explicit type casting here .

In show() , you first take the address of a with &a and then find the value in that address with *(&a) which is the same thing as a (think along the lines of -(-a) ie, taking the negative of a number whose negative is in turn found which is the same number you started off with).

Answer 4

&--> get me the address.

*--> get me the value at that address

disp(&marks[i])--> get the address of variable "marks[i]" and pass it to *n(disp(int *n)), so that when *n is executed it will get the value at the address it has been assigned.

show((int) &n)--> get address of the variable that stores the address of the variable marks[i].

To make it work as intended, it must be show(*n)--> pass the value at the address pointed to by n. (Typecasting is not needed as you're passing an int value to show())