Pointers

Question

Can we pass pointers in the arguments of a function? If yes then do we pass the value the pointer holds or the address of the pointer?

Answer 1

Yes we can pass pointers as an argument of a function

Pointers itself contains address of any data item value.

For example int a = 10;   int *b =&a;

here 'b' contains address of 'a' and at address of 'a' contains the value = 10.

if we simply pass ' &b ' it contains address and if we pass ' *b ' then it contains value at the address stored in 'b'

Comments

@priyanka17 Thanx, it was very helpful, can you please also illustrtae through an example

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example: you can check call by value and call by reference program for better understanding

add(int a, int b)

{ int c = a+b ;

 return c;

}

add1(int *a1 , int *b1)

{ *a1++;// a1 gets updated

 *b1++;//b1 gets updated

 int c=*a1 + *b1 ;

 return c;

}

void main()

{

 int a, b, a1, b1, c1, c2;

 a=1; b=2; a1=3; b1=4;

c1= add(a,b);

printf(a,b,c1);// output a=1 b=2 c=3

c2= add( &a, &b );

printf(a1,b1,c2);// output a1=4 b1=5 c2=9

}

Answer 2

We of course can - otherwise printf function can never work because the first argument to it is a "pointer to char". And when we pass a pointer, its value is being passed - but the value being a pointer is expected to be an address of some other object. 

char *s = "hello";
printf(s);
char a[2] = {'a', '\0'};
printf(a);
char b = '\0';
printf(&b);
int d = 9;
printf("%d", d); // here address of the string literal "%d" is passed to printf