C++ - What is difference between followin intialization.int iVar1;int iVar2 = int();and which one of t

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What is difference between followin intialization.

int iVar1;
int iVar2 = int();

and which one of two should we prefer always and why?

Total Answers and Comments: 7

Last Update: November 17, 2008 Asked by: mohit12379

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April 26, 2006 06:40:47

Preeti Yadav

Member Since: April 2006 Contribution: 30

RE: What is difference between followin intialization....

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();

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July 28, 2006 16:30:44

mohit12379

Member Since: March 2006 Contribution: 17

RE: What is difference between followin intialization....

@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.

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September 13, 2006 10:13:46

ramachandra

RE: What is difference between followin intialization....

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

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October 15, 2006 07:27:39

Shiva Shankar Anumula

RE: What is difference between followin intialization....

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

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November 07, 2006 05:20:31

Ashish

RE: What is difference between followin intialization....

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.

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May 21, 2007 15:27:31

Anil

RE: What is difference between followin intialization....

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';

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November 17, 2008 02:28:58

mucdull

Member Since: November 2008 Contribution: 6

RE: What is difference between followin intialization.int iVar1;int iVar2 = int();and which one of two should we prefer always and why?

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.

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