LinuxQuestions.org - [SOLVED] output

- Programming (https://www.linuxquestions.org/questions/programming-9/)

- - output (https://www.linuxquestions.org/questions/programming-9/output-4175413822/)

Code:

#include<stdio.h>

#include<iostream.h>

void fun(int *b)

{

 int c=5;

cout<<"value of pointer in fun()"<<"\n"<<b;

b=&c;

 (*b)++;

cout<<"the value of c is"<<*b;

}

int main()

{

int a=4;

int *p=&a;

cout<<"previous value of a  is "<<*p;

cout<<"value of pointer in main :"<<"\n "<<p;



fun(p);

cout<<"after calling fun ,new value of a is "<<*p;

return 0;}

...

how is the o/p remaining the same in main() even after we change the pointer's to point to another variable in func()??

The parameter is passed by value, so this statement: 'b=&c' changes only the copied value, not the original.

PS: you really should use printf, it is much more flexible, eg:

Code:

printf ("main, beforore 'fun': a=%d, p=%p\n", a, p);

Quote:

Originally Posted by NevemTeve (Post 4714077)

The parameter is passed by value, so this statement: 'b=&c' changes only the copied value, not the original.

PS: you really should use printf, it is much more flexible, eg:

Code:

printf ("main, beforore 'fun': a=%d, p=%p\n", a, p);

thanx.i thnk i get it..but jst to make sure when i change the argument to be passed as refernce, the o/p changes..so does this actually mean when we pass a reference to a variable,another copy of that reference is not created in func()??

Code:

#include<iostream.h>

void fun(int &b)

{

 int c=5;

 b=c;

 b++;

}

int main()

{

int a=4;

int &p=a;

cout<<"previous value of a  is "<<p; 



fun(p);

cout<<"after calling fun ,new value of a is "<<p;

return 0;

}

Yes, that's how call-by-reference is different to call-by-value.