Value of a non-initialized variable in C++

Question

This is pretty simple, yet I cannot explain it. I declare 4 variables, and print it out in C++:

#include <iostream>
using namespace std;

int main(){
  int a, b, c, d;

  cout << "a = " << a << endl;
  cout << "b = " << b << endl;
  cout << "c = " << c << endl;
  cout << "d = " << d << endl;

  return 0;
}

And the result is:

a = 0
b = 1
c = 0
d = 0

I thought if I don't initialize value for a variable, the program will take some random value from the memory for it. Or at least it's all equal to zero. I don't know where the difference of those values come from.

This is not because of a specific variable b, if I switch those "cout <<" lines, the second line always give the value = 1.

Can anyone explain this for me, please. Thank you so much.

The value of variable that have not be initialized is undefined in C++ — litelite, Aug 08 '17 at 19:10
I suppose those undefined vars are random values, and trying to display them is UB. — 2785528, Aug 08 '17 at 19:11
Turn on your compiler warnings for using uninitialized variables. It'll help you write code that actually works. — tadman, Aug 08 '17 at 19:11
No, I mean if I switch those cout lines, the second one always comes up with 1. That's where I'm confused. — Huyen, Aug 08 '17 at 19:15
You'd probably need assembly reading skills to understand what exactly happens in your case. I doubt anyone would be able to tell you, because the results might be different on other compilers, systems and machines. — HolyBlackCat, Aug 08 '17 at 19:32

Jesper Juhl · Accepted Answer · 2017-08-08T19:28:12.977

3

Reading an uninitialized variable is Undefined Behaviour in C++.

The compiler is free to generate whatever code it likes for your program, since you broke the rules of the language.

Your program may crash, do exactly what you expect, run forever or do any other thing you can imagine.

Now reasonable implementations of compilers won't use UB as an excuse to format your harddrive (but they could), but UB still means that your program has no meaning according to the standard and that you can and will get arbitrary results across compilers and operating systems. Don't expect that the result you see now, with your current compiler, on your current OS, will be the same in a different situation with a different compiler or on a different OS, when UB is involved.

In a nutshell; it's your responsibility to not invoke UB. The compiler is not required to warn you. And if you want a program with deterministic behaviour, you need to not invoke UB.

edited Aug 08 '17 at 19:28

answered Aug 08 '17 at 19:16

Jesper Juhl

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I get that one should not read an uninitialized variable. My question is, if I let this line 'cout << "c = " << c << endl;' instead of print b, I will have c = 1 instead. This is about the order thingy that I don't get yet. – Huyen Aug 08 '17 at 19:21
@Nguyễn Ngọc Huyền you can't reason about your program when it contains UB. Any result/effect you see is arbitrary and *not* deterministic. Stop trying to reason about what you currently observe and instead *fix* the UB - *then* you can start to try and reason about your programs behaviour. And btw `cout << "c = " << c` *is* reading an uninitialized variable --> UB. – Jesper Juhl Aug 08 '17 at 19:25
@Nguyễn Ngọc Huyền - What's "harsh"? I'm just trying to state *facts* - clear and simple. – Jesper Juhl Aug 08 '17 at 19:34

Value of a non-initialized variable in C++

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