What happen exactly when we pass the value to function

I wrote to understand basic of pass by value

Code:

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#include<stdio.h>
/* function declaration */
void pass(int x, int y);
void pass(int x, int y)
{
        printf ( "\n x : %d", x);
        printf ( "\n y : %d", y);       
}
 
int main () {


  /* local variable definition */
  int a = 100;
  int b = 200;
 
  printf("Before pass, value of a : %d\n", a );
  printf("Before pass, value of b : %d\n", b );
 
  /* calling a function to pass the values */
  pass(a, b);
 
  printf("\n After pass, value of a : %d\n", a );
  printf("\n After pass, value of b : %d\n", b );
 
  return 0;
}

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Before pass, value of a : 100
Before pass, value of b : 200


x : 100
y : 200
After pass, value of a : 100


After pass, value of b : 200

I am not sure but i think when we call function value of x and y will replace with a and b

we are passing value 100 and 200 to the function

Is it the whole process of pass by value ?

Each time you call the 'pass' function, two variables called x and y with static scope are created. As you pass the value of 'a' to 'x' and 'b' to 'y', you essentially end up creating two aliases aka duplicates of a and b. This is essentially all about pass by value. If I were to clone you and experiment on your cloned version, it is obvious that it'll have no effect on the 'real' you. Same with creating copies of variables in a pass by value situation.

Passing by Value vs. by Reference Visual Explanation with animations and diagrams

Quote:

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Originally Posted by Zeus_

Passing by Value vs. by Reference Visual Explanation with animations and diagrams

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thank you so much

what exactly happen call by reference

Code:

---

#include<stdio.h>
/* function declaration */
void passbyrefrence(int *x, int *y);
void passbyrefrence(int *x, int *y)
{
  printf("\n After pass, value of x : %d\n", *x );
  printf("\n After pass, value of y : %d\n", *y );
 
  printf("\n After pass, address of x : %p \n", x );
  printf("\n After pass, address of y : %p \n", y );


  printf ("\n");
}
 
 
int main () {


  /* local variable definition */
  int a = 100;
  int b = 200;
 
  printf("Before pass, value of a : %d\n", a );
  printf("Before pass, value of b : %d\n", b );
 
  printf("Before pass, address of a : %p \n", &a );
  printf("Before pass, address  of b : %p \n", &b );


  passbyrefrence(&a, &b);
 
  printf("After pass, value of a : %d\n", a );
  printf("After pass, value of b : %d\n", b );
 
  printf("after  pass, address of a : %p \n", &a );
  printf("after pass, address  of b : %p \n", &b );
 
  return 0;
}

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Before pass, value of a : 100
Before pass, value of b : 200
Before pass, address of a : 0061FF2C
Before pass, address of b : 0061FF28


After pass, value of x : 100


After pass, value of y : 200


After pass, address of x : 0061FF2C


After pass, address of y : 0061FF28


After pass, value of a : 100
After pass, value of b : 200
after pass, address of a : 0061FF2C
after pass, address of b : 0061FF28

definition say we pass parameter by reference

Code:

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passbyrefrence(&a, &b);

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What is meaning of this line ?

does it pass the address of a and b or it pass the content of a and b to x and b at the address a and address of b

Quote:

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does it pass the address of a and b or it pass the content of a and b to x and b at the address a and address of b

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The former – it passes the address, not the content.

Quote:

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What is meaning of this line ?

does it pass the address of a and b or it pass the content of a and b to x and b at the address a and address of b

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& is the 'reference' operator. Your function expects integer pointers. Now, if you have learnt what pointers are, then you'd be correct if you thought that they are just a variable that hold a memory address. Put the dots together and you have your answer. You pass the address of 'a' to 'x' which expects an address and the same with 'b' to 'y'.

Take a look at the website I shared above, it'll clear all your doubts or misconceptions about pass by value and pass by reference. The take from here is, pass by value doesn't carry back the changes you do to the clone, but pass by reference affects you TOO if any change is done on your clone.

Quote:

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Originally Posted by Guest

The former – it passes the address, not the content.

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All arguments are passed by value in c language. ?
or
arguments get passed by value and/or reference in c language?

If you insist on discussing semantics, you won't get a universal answer.

Code:

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void foo(int a); // passing an integer
void foo(int* a); // passing a pointer to an integer

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You know these two, you know how they work/differ, time to move on.