Why int b; has 16 value in shell/terminal?

Question

#include <iostream>

using namespace std;

int main()
{
    int a;
    int b;
    
    cout<<a<<endl;
    cout<<b;

}

Why b has 16 value when i compile this? I cant understand why b has 16 not 0. Int doesnt have 0 as default value?

define variable in c++ without initialize in future it will give undefined value. — N0ll_Boy, May 28 '21 at 07:59
You have stumbled upon something super important, and not to forget in the future. When you declare a variable, its value is basically random. This is why you should always initialise a variable before using its value. — , May 28 '21 at 08:01
@N0ll_Boy the behavior is undefined, which is different from having undefined values. For example on some architectures like Itanium, accessing an uninitialized variable will give you a segfault — phuclv, May 28 '21 at 08:20

Bathsheba · Answer 1 · 2021-05-28T08:05:28.480

4

Absolutely not. C++ does not initialise variables unless you ask it to. (Setting a variable to 0 for example is at least one instruction: typically reg XOR reg. That could be wasteful.) The behaviour of reading an uninitialised int is undefined.

(Note you can do some things with uninitialised variables, such as setting a pointer or reference to one, computing sizeof(a), and using decltype(a). But passing it by value to a function is undefined behaviour. That often trips up even professional programmers.)

edited May 28 '21 at 08:05

answered May 28 '21 at 07:58

Bathsheba

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so i have to write all the time for example: int a = 0;? – lolorek May 28 '21 at 08:00
3

Again, absolutely not. Only if you need to read the value before you write something sensible to it. – Bathsheba May 28 '21 at 08:00
I did this: int a; int b; cout< – lolorek May 28 '21 at 08:03
@lolorek well you don't **have to**, but it is advisable to always initialize variabels. That doesn't mean you should initialize everything with `0`, but rather only declare a variable when you can initialize it with something meaningful. Not always this is possible, but thats the exceptions – 463035818_is_not_an_ai May 28 '21 at 08:03
2

`a += 1` is a read of an uninitialised `a`. You are incrementing the value. – Bathsheba May 28 '21 at 08:04
i did this int a; a + 10; cout< – lolorek May 28 '21 at 08:07
2

@lolorek: `a + 10` is an uninitialised read of `a`. Can you not see that? Did you mean `a = 10`? – Bathsheba May 28 '21 at 08:08
1

The best thing to do is not declare `int a` until you have a value for it. `int a = 0;` can be problematic, e.g. if you want to divide by `a`. – Caleth May 28 '21 at 09:08

Why int b; has 16 value in shell/terminal?

1 Answers1