CODEWITHSUNDEEP
c++initialization
Miljan Rakita
Miljan Rakita

Reputation: 1533

Why do I need to initialize an int variable to 0?

I just made this program which asks to enter number between 5 and 10 and then it counts the sum of the numbers which are entered. Here is the code:

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
    int a,i,c;
    cout << "Enter the number between 5 and 10" << endl;
    cin >> a;
    if (a < 5 || a > 10)
    {
        cout << "Wrong number" << endl;
        system("PAUSE");
        return 0;
    }
    for(i=1; i<=a; i++)
    {
        c=c+i;
    }
    cout << "The sum of the first " << a << " numbers are " << c << endl;
    system("PAUSE");
    return 0;
}

If I enter 5, it should display

The sum of the first 5 numbers are 15

but it displays

The sum of the first 5 numbers are 2293687

When I set c to 0, it works correctly.

So what is the difference?

Upvotes: 6

Views: 28414

Answers (5)

masoud
masoud

Reputation: 56479

Because C++ doesn't automatically set it zero for you. So, you should initialize it yourself:

int c = 0;

An uninitialized variable has a random number such as 2293687, -21, 99999, ... (If it doesn't invoke undefined behavior when reading it)

Also, static variables will be set to their default value. In this case 0.

Upvotes: 21

John Dibling
John Dibling

Reputation: 101456

Because when you do:

int a,i,c;

thus instantiating and initializing c, you haven't said what you want it initialized to. The rules here are somewhat complex, but what it boils down to is two things:

  1. For integral types, if you don't specify an initializer, the variable's value is indeterminate
  2. When you try to read an uninitialized variable, you evoke Undefined Behavior

Upvotes: 0

Paul Roub
Paul Roub

Reputation: 36438

Non-static variables are, by definition, uninitialized - their initial values are undefined.

On another compiler, you might get the right answer, another wrong answer, or a different answer each time.

C/C++ don't do extra work (initialization to zero involves at least an instruction or two) that you didn't ask them to do.

Upvotes: 5

haccks
haccks

Reputation: 106012

The sum of the first 5 numbers are 2293687

This is because without initializing c you are getting value previous stored at that location (garbage value). This will make yor program's behavior undefined. You must have to initialize c before using it in your program. 

int c= 0;

Upvotes: 0

juanchopanza
juanchopanza

Reputation: 227418

If you don't set c to 0, it can take any value (technically, an indeterminate value). If you then do this

c = c + i;

then you are adding the value of i to something that could be anything. Technically, this is undefined behaviour. What happens in practice is that you cannot rely on the result of that calculation.

In C++, non-static or global built-in types have no initialization performed when "default initialized". In order to zero-initialize an int, you need to be explicit:

int i = 0;

or you can use value initialization:

int i{};
int j = int();

Upvotes: 8

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