如何打印指针指向的值？

Question

I found the following code in my textbook:

#include<stdio.h>

void disp( int *k)
{
printf("%d",*k);
}

int main( )
{
 int i ;
 int marks[ ] = { 55, 65, 75, 56, 78, 78, 90 } ;
 for ( i = 0 ; i <= 6 ; i++ )
 disp ( &marks[i] ) ;
 return 0;
}

 }

The code works just fine, but I have doubts regarding the logic:

1. I am sending the address of variables of the array. But in the disp function I am using a pointer variable as the argument and printing the value of the pointer. So the type of argument sent from the main function should mismatch with the argument of disp . So how does it work?

2. I tried to do the same by changing the disp function as

    void disp( int (&k)) { printf("%d",*k); } 

   but I am getting an error. What should I do to make it work by taking an address as an argument, ie void disp(int &k) ?

Answer 1

1) I am sending the address of variables of the array. But in the disp function I am using a pointer variable as argument and printing the value of the pointer.

Understand that a pointer is an address. So &marks[i] is an int* . And you're not printing the value of the pointer, but the value it points to.

printf("%d",*k);

the *k dereferences the pointer and gives the pointed-to value.

void disp( int (&k))

is invalid syntax in C, &k is not a valid identifier.

Answer 2

When you do: -

int *k = &marks[i];

The above statement is broken into: -

int *k;   -> Integer Pointer
k = &marks[i]; --> `k` points to the address of marks[i]

So, basically, k is the an integer pointer, that points to the address of the current element in your array .

So, when you print *k , it is equivalent to: - *(&marks[i]) , which dereferences the value and prints the element marks[i] .

So, in the below code, you can understand, how the whole process of pointer assignment and de-referencing takes place: -

int *k;  // Declaring integer pointer.

// Actually 'k' is equal to the `&marks[i]`. 
// '*k' is just the indication of it being a pointer
k = &marks[i];      

// Dereference pointer
*k = *(&marks[i]); -->  = marks[i]

printf("%d",*k);  --> printf("%d",marks[i]);

Also, since you cannot declare variable like: -

int &k = &marks[i];

You cannot have them as parameter in your function: -

void disp(int &k);

Because, eventually the address of array element passed is stored in this variable. So, it has to be int *k .

Answer 3

1) You understand correct that in the for loop the address is sent to disp . But disp receives pointer as an argument... which is just an address. Inside that function you don't print the "value" of the pointer. You print the value pointed by that pointer (the value that is on that address). In the printf call "*k" you dereference the pointer - get the value pointed by that pointer.

2) You didn't change the function so that it receives address. You changed it so that it receives reference. You can also say it receives a hidden pointer - it does the same as the pointer but you don't need to "*k" to dereference it - you use it as a normal variable

Answer 4

Read up on pointers and address-of variable. And how they're used. When a function expects a pointer, you HAVE to send either a pointer or an address of a variable. Since pointers hold addresses.

The second problem you're facing is that of syntax error. Because there is no & operator that can be applied to "lvalues".

Answer 5

Well, simply i can say that you are passing an address in the function and recieving it in a pointer..Its correct as pointer can point to or hold the address. Or other way simply if you send an address but store it in a reference variable but the reference of whom, is the question to be asked. Try vice versa you'll understand the structure more better...