Preeti Yadav

 Profile Answers by Preeti Yadav Questions by Preeti Yadav

• Apr 26th, 2006

 

in first case a variable will be create in memeory with the default base type value (depending upon compiler 2 compiler) bcoz it is not initialized. in second case the variable will be created in the memory with the value retuned by the function int() (if int is a user define function) the second statement should have been int *i = new int();

mohit12379

 Profile Answers by mohit12379 Questions by mohit12379

• Jul 28th, 2006

 

@PRITI YADAV HAV U TRIED THE FOLOWING STATEMENT EVER

   int iVar2 = int();

PLS TRY IT....

FOR UR KIND INFO :- int * ivar2 = new int(); and int iVar2 = int(); are two different statement... thas why i have asked this que...., DNT MIX IT.

ramachandra

• Sep 13th, 2006

 

Preeti Yadav is right, Ivar1 is not initialized but in the second case compiler initilizes the Ivar2 to sero. create a console program and use "COUT" to print Ivar1 and Ivar2 and observe that Ivar1 prints junk value and Ivar2 value will be zero

Shiva Shankar Anumula

• Oct 15th, 2006

 

int iVar1;
int iVar2 = int();

In first case, memory fpr iVar1 is allocated and it is not intialized.

 second case is used to intialize the '0' to ivar2

Ashish

• Nov 7th, 2006

 

int iVar1 does not initalize the iVar1. Its C style of declation of integer.

whereas in C++ everything is implemented using classes, even basic built-in data types.

int iVar2 = int();

so int() is nothing but constructor that creates one object of int type and assigned to iVar2. (means assignment operator is overloaded). try int(5)... i.e.

int iVar2 = int(5);

in this case iVar2 is initalized to 5.

Anil

• May 21st, 2007

 

depends

if use in main both will get initialize to '0'.
ie: global

if use inside a function (local) iVar1 will be garbage, iVar2 will be initialised to '0';

mucdull

 Profile Answers by mucdull 

• Nov 17th, 2008

 

int   iVar1;   // iVar1 is never initialized
int   iVar2 = int();   // iVar2 will always be initialized with some initial value

The latter should be used because use of uninitialized variable is dangerous and can cause runtime check failure for some C++ compiler.