# c++ – Is uninitialized local variable the fastest random number generator?

## The Question :

329 people think this question is useful

I know the uninitialized local variable is undefined behaviour(UB), and also the value may have trap representations which may affect further operation, but sometimes I want to use the random number only for visual representation and will not further use them in other part of program, for example, set something with random color in a visual effect, for example:

void updateEffect(){
for(int i=0;i<1000;i++){
int r;
int g;
int b;
star[i].setColor(r%255,g%255,b%255);
bool isVisible;
star[i].setVisible(isVisible);
}
}



is it that faster than

void updateEffect(){
for(int i=0;i<1000;i++){
star[i].setColor(rand()%255,rand()%255,rand()%255);
star[i].setVisible(rand()%2==0?true:false);
}
}



and also faster than other random number generator?

• +1 This is a perfectly legitimate question. It’s true that in practice, uninitialized values might be kindof random. The fact that they aren’t particularly and that it’s UB doesn’t make asking that bad.
• @imallett: Absolutely. This is a good question, and at least one old Z80 (Amstrad / ZX Spectrum) game in times past used its program as the data to set up its terrain. So there are even precedents. Can’t do that these days. Modern operating systems take away all the fun.
• Surely the main problem is that it isn’t random.
• In fact, there’s an example of a uninitialized variable being used as a random value, see the Debian RNG disaster (Example 4 in this article).
• In practice – and believe me, I do a lot of debugging on various architectures – your solution can do two things: either read uninitialized registers or uninitialized memory. Now while “uninitialized” means random in a certain manner, in practice it will most likely contain a) zeroes, b) repeating or consistent values (in case of reading memory formerly occupied by digital media) or c) consistent garbage with a limited value set (in case of reading memory formerly occupied by encoded digital data). None of those are real entropy sources.

301 people think this answer is useful

As others have noted, this is Undefined Behavior (UB).

In practice, it will (probably) actually (kind of) work. Reading from an uninitialized register on x86[-64] architectures will indeed produce garbage results, and probably won’t do anything bad (as opposed to e.g. Itanium, where registers can be flagged as invalid, so that reads propagate errors like NaN).

There are two main problems though:

1. It won’t be particularly random. In this case, you’re reading from the stack, so you’ll get whatever was there previously. Which might be effectively random, completely structured, the password you entered ten minutes ago, or your grandmother’s cookie recipe.

2. It’s Bad (capital ‘B’) practice to let things like this creep into your code. Technically, the compiler could insert reformat_hdd(); every time you read an undefined variable. It won’t, but you shouldn’t do it anyway. Don’t do unsafe things. The fewer exceptions you make, the safer you are from accidental mistakes all the time.

The more pressing issue with UB is that it makes your entire program’s behavior undefined. Modern compilers can use this to elide huge swaths of your code or even go back in time. Playing with UB is like a Victorian engineer dismantling a live nuclear reactor. There’s a zillion things to go wrong, and you probably won’t know half of the underlying principles or implemented technology. It might be okay, but you still shouldn’t let it happen. Look at the other nice answers for details.

Also, I’d fire you.

214 people think this answer is useful

Let me say this clearly: we do not invoke undefined behavior in our programs. It is never ever a good idea, period. There are rare exceptions to this rule; for example, if you are a library implementer implementing offsetof. If your case falls under such an exception you likely know this already. In this case we know using uninitialized automatic variables is undefined behavior.

Compilers have become very aggressive with optimizations around undefined behavior and we can find many cases where undefined behavior has lead to security flaws. The most infamous case is probably the Linux kernel null pointer check removal which I mention in my answer to C++ compilation bug? where a compiler optimization around undefined behavior turned a finite loop into an infinite one.

We can read CERT’s Dangerous Optimizations and the Loss of Causality (video) which says, amongst other things:

Increasingly, compiler writers are taking advantage of undefined behaviors in the C and C++ programming languages to improve optimizations.

Frequently, these optimizations are interfering with the ability of developers to perform cause-effect analysis on their source code, that is, analyzing the dependence of downstream results on prior results.

Consequently, these optimizations are eliminating causality in software and are increasing the probability of software faults, defects, and vulnerabilities.

Specifically with respect to indeterminate values, the C standard defect report 451: Instability of uninitialized automatic variables makes for some interesting reading. It has not been resolved yet but introduces the concept of wobbly values which means the indeterminatness of a value may propagate through the program and can have different indeterminate values at different points in the program.

I don’t know of any examples where this happens but at this point we can’t rule it out.

Real examples, not the result you expect

You are unlikely to get random values. A compiler could optimize the away the loop altogether. For example, with this simplified case:

void updateEffect(int  arr[20]){
for(int i=0;i<20;i++){
int r ;
arr[i] = r ;
}
}



clang optimizes it away (see it live):

updateEffect(int*):                     # @updateEffect(int*)
retq



or perhaps get all zeros, as with this modified case:

void updateEffect(int  arr[20]){
for(int i=0;i<20;i++){
int r ;
arr[i] = r%255 ;
}
}


updateEffect(int*):                     # @updateEffect(int*)
xorps   %xmm0, %xmm0
movups  %xmm0, 64(%rdi)
movups  %xmm0, 48(%rdi)
movups  %xmm0, 32(%rdi)
movups  %xmm0, 16(%rdi)
movups  %xmm0, (%rdi)
retq



Both of these cases are perfectly acceptable forms of undefined behavior.

Note, if we are on an Itanium we could end up with a trap value:

[…]if the register happens to hold a special not-a-thing value, reading the register traps except for a few instructions[…]

Other important notes

It is interesting to note the variance between gcc and clang noted in the UB Canaries project over how willing they are to take advantage of undefined behavior with respect to uninitialized memory. The article notes (emphasis mine):

Of course we need to be completely clear with ourselves that any such expectation has nothing to do with the language standard and everything to do with what a particular compiler happens to do, either because the providers of that compiler are unwilling to exploit that UB or just because they have not gotten around to exploiting it yet. When no real guarantee from the compiler provider exists, we like to say that as-yet unexploited UBs are time bombs: they’re waiting to go off next month or next year when the compiler gets a bit more aggressive.

As Matthieu M. points out What Every C Programmer Should Know About Undefined Behavior #2/3 is also relevant to this question. It says amongst other things (emphasis mine):

The important and scary thing to realize is that just about any optimization based on undefined behavior can start being triggered on buggy code at any time in the future. Inlining, loop unrolling, memory promotion and other optimizations will keep getting better, and a significant part of their reason for existing is to expose secondary optimizations like the ones above.

To me, this is deeply dissatisfying, partially because the compiler inevitably ends up getting blamed, but also because it means that huge bodies of C code are land mines just waiting to explode.

For completeness sake I should probably mention that implementations can choose to make undefined behavior well defined, for example gcc allows type punning through unions while in C++ this seems like undefined behavior. If this is the case the implementation should document it and this will usually not be portable.

164 people think this answer is useful

No, it’s terrible.

The behaviour of using an uninitialised variable is undefined in both C and C++, and it’s very unlikely that such a scheme would have desirable statistical properties.

If you want a “quick and dirty” random number generator, then rand() is your best bet. In its implementation, all it does is a multiplication, an addition, and a modulus.

The fastest generator I know of requires you to use a uint32_t as the type of the pseudo-random variable I, and use

I = 1664525 * I + 1013904223

to generate successive values. You can choose any initial value of I (called the seed) that takes your fancy. Obviously you can code that inline. The standard-guaranteed wraparound of an unsigned type acts as the modulus. (The numeric constants are hand-picked by that remarkable scientific programmer Donald Knuth.)

42 people think this answer is useful

Good question!

Undefined does not mean it’s random. Think about it, the values you’d get in global uninitialized variables were left there by the system or your/other applications running. Depending what your system does with no longer used memory and/or what kind of values the system and applications generate, you may get:

1. Always the same.
2. Be one of a small set of values.
3. Get values in one or more small ranges.
4. See many values dividable by 2/4/8 from pointers on 16/32/64-bit system

The values you’ll get completely depend on which non-random values are left by the system and/or applications. So, indeed there will be some noise (unless your system wipes no longer used memory), but the value pool from which you’ll draw will by no means be random.

Things get much worse for local variables because these come directly from the stack of your own program. There is a very good chance that your program will actually write these stack locations during the execution of other code. I estimate the chances for luck in this situation very low, and a ‘random’ code change you make tries this luck.

Read about randomness. As you’ll see randomness is a very specific and hard to obtain property. It’s a common mistake to think that if you just take something that’s hard to track (like your suggestion) you’ll get a random value.

32 people think this answer is useful

Many good answers, but allow me to add another and stress the point that in a deterministic computer, nothing is random. This is true for both the numbers produced by an pseudo-RNG and the seemingly “random” numbers found in areas of memory reserved for C/C++ local variables on the stack.

BUT… there is a crucial difference.

The numbers generated by a good pseudorandom generator have the properties that make them statistically similar to truly random draws. For instance, the distribution is uniform. The cycle length is long: you can get millions of random numbers before the cycle repeats itself. The sequence is not autocorrelated: for instance, you will not begin to see strange patterns emerge if you take every 2nd, 3rd, or 27th number, or if you look at specific digits in the generated numbers.

In contrast, the “random” numbers left behind on the stack have none of these properties. Their values and their apparent randomness depend entirely on how the program is constructed, how it is compiled, and how it is optimized by the compiler. By way of example, here is a variation of your idea as a self-contained program:

#include <stdio.h>

notrandom()
{
int r, g, b;

printf("R=%d, G=%d, B=%d", r&amp;255, g&amp;255, b&amp;255);
}

int main(int argc, char *argv[])
{
int i;
for (i = 0; i < 10; i++)
{
notrandom();
printf("\n");
}

return 0;
}



When I compile this code with GCC on a Linux machine and run it, it turns out to be rather unpleasantly deterministic:

R=0, G=19, B=0
R=130, G=16, B=255
R=130, G=16, B=255
R=130, G=16, B=255
R=130, G=16, B=255
R=130, G=16, B=255
R=130, G=16, B=255
R=130, G=16, B=255
R=130, G=16, B=255
R=130, G=16, B=255



If you looked at the compiled code with a disassembler, you could reconstruct what was going on, in detail. The first call to notrandom() used an area of the stack that was not used by this program previously; who knows what was in there. But after that call to notrandom(), there is a call to printf() (which the GCC compiler actually optimizes to a call to putchar(), but never mind) and that overwrites the stack. So the next and subsequent times, when notrandom() is called, the stack will contain stale data from the execution of putchar(), and since putchar() is always called with the same arguments, this stale data will always be the same, too.

So there is absolutely nothing random about this behavior, nor do the numbers obtained this way have any of the desirable properties of a well-written pseudorandom number generator. In fact, in most real-life scenarios, their values will be repetitive and highly correlated.

Indeed, as others, I would also seriously consider firing someone who tried to pass off this idea as a “high performance RNG”.

29 people think this answer is useful

Undefined behavior means that the authors of compilers are free to ignore the problem because programmers will never have a right to complain whatever happens.

While in theory when entering UB land anything can happen (including a daemon flying off your nose) what normally means is that compiler authors just won’t care and, for local variables, the value will be whatever is in the stack memory at that point.

This also means that often the content will be “strange” but fixed or slightly random or variable but with a clear evident pattern (e.g. increasing values at each iteration).

For sure you cannot expect it being a decent random generator.

28 people think this answer is useful

Undefined behaviour is undefined. It doesn’t mean that you get an undefined value, it means that the the program can do anything and still meet the language specification.

A good optimizing compiler should take

void updateEffect(){
for(int i=0;i<1000;i++){
int r;
int g;
int b;
star[i].setColor(r%255,g%255,b%255);
bool isVisible;
star[i].setVisible(isVisible);
}
}



and compile it to a noop. This is certainly faster than any alternative. It has the downside that it will not do anything, but such is the downside of undefined behaviour.

18 people think this answer is useful

Not mentioned yet, but code paths that invoke undefined behavior are allowed to do whatever the compiler wants, e.g.

void updateEffect(){}



Which is certainly faster than your correct loop, and because of UB, is perfectly conformant.

18 people think this answer is useful

Because of security reasons, new memory assigned to a program has to be cleaned, otherwise the information could be used, and passwords could leak from one application into another. Only when you reuse memory, you get different values than 0. And it is very likely, that on a stack the previous value is just fixed, because the previous use of that memory is fixed.

13 people think this answer is useful

Your particular code example would probably not do what you are expecting. While technically each iteration of the loop re-creates the local variables for the r, g, and b values, in practice it’s the exact same memory space on the stack. Hence it won’t get re-randomized with each iteration, and you will end up assigning the same 3 values for each of the 1000 colors, regardless of how random the r, g, and b are individually and initially.

Indeed, if it did work, I would be very curious as to what’s re-randomizing it. The only thing I can think of would be an interleaved interrupt that piggypacked atop that stack, highly unlikely. Perhaps internal optimization that kept those as register variables rather than as true memory locations, where the registers get re-used further down in the loop, would do the trick, too, especially if the set visibility function is particularly register-hungry. Still, far from random.

12 people think this answer is useful

As most of people here mentioned undefined behavior. Undefined also means that you may get some valid integer value (luckily) and in this case this will be faster (as rand function call is not made). But don’t practically use it. I am sure this will terrible results as luck is not with you all the time.

12 people think this answer is useful

A_Function_that_use_a_lot_the_Stack();
updateEffect();



If the function A_Function_that_use_a_lot_the_Stack() make always the same initialization it leaves the stack with the same data on it. That data is what we get calling updateEffect(): always same value!.

11 people think this answer is useful

I performed a very simple test, and it wasn’t random at all.

#include <stdio.h>

int main() {

int a;
printf("%d\n", a);
return 0;
}



Every time I ran the program, it printed the same number (32767 in my case) — you can’t get much less random than that. This is presumably whatever the startup code in the runtime library left on the stack. Since it uses the same startup code every time the program runs, and nothing else varies in the program between runs, the results are perfectly consistent.

10 people think this answer is useful

You need to have a definition of what you mean by ‘random’. A sensible definition involves that the values you get should have little correlation. That’s something you can measure. It’s also not trivial to achieve in a controlled, reproducible manner. So undefined behaviour is certainly not what you are looking for.

7 people think this answer is useful

There are certain situations in which uninitialized memory may be safely read using type “unsigned char*” [e.g. a buffer returned from malloc]. Code may read such memory without having to worry about the compiler throwing causality out the window, and there are times when it may be more efficient to have code be prepared for anything memory might contain than to ensure that uninitialized data won’t be read (a commonplace example of this would be using memcpy on partially-initialized buffer rather than discretely copying all of the elements that contain meaningful data).

Even in such cases, however, one should always assume that if any combination of bytes will be particularly vexatious, reading it will always yield that pattern of bytes (and if a certain pattern would be vexatious in production, but not in development, such a pattern won’t appear until code is in production).

Reading uninitialized memory might be useful as part of a random-generation strategy in an embedded system where one can be sure the memory has never been written with substantially-non-random content since the last time the system was powered on, and if the manufacturing process used for the memory causes its power-on state to vary in semi-random fashion. Code should work even if all devices always yield the same data, but in cases where e.g. a group of nodes each need to select arbitrary unique IDs as quickly as possible, having a “not very random” generator which gives half the nodes the same initial ID might be better than not having any initial source of randomness at all.

5 people think this answer is useful

As others have said, it will be fast, but not random.

What most compilers will do for local variables is to grab some space for them on the stack, but not bother setting it to anything (the standard says they don’t need to, so why slow down the code you’re generating?).

In this case, the value you’ll get will depend on what was on previously on the stack – if you call a function before this one that has a hundred local char variables all set to ‘Q’ and then call you’re function after that returns, then you’ll probably find your “random” values behave as if you’ve memset() them all to ‘Q’s.

Importantly for your example function trying to use this, these values wont change each time you read them, they’ll be the same every time. So you’ll get a 100 stars all set to the same colour and visibility.

Also, nothing says that the compiler shouldn’t initialize these value – so a future compiler might do so.

In general: bad idea, don’t do it. (like a lot of “clever” code level optimizations really…)

3 people think this answer is useful

As others have already mentioned, this is undefined behavior (UB), but it may “work”.

Except from problems already mentioned by others, I see one other problem (disadvantage) – it will not work in any language other than C and C++. I know that this question is about C++, but if you can write code which will be good C++ and Java code and it’s not a problem then why not? Maybe some day someone will have to port it to other language and searching for bugs caused by “magic tricks” UB like this definitely will be a nightmare (especially for an inexperienced C/C++ developer).

3 people think this answer is useful

Not a good idea to rely our any logic on language undefined behaviour. In addition to whatever mentioned/discussed in this post, I would like to mention that with modern C++ approach/style such program may not be compile.

This was mentioned in my previous post which contains the advantage of auto feature and useful link for the same.

https://stackoverflow.com/a/26170069/2724703

So, if we change the above code and replace the actual types with auto, the program would not even compile.

void updateEffect(){
for(int i=0;i<1000;i++){
auto r;
auto g;
auto b;
star[i].setColor(r%255,g%255,b%255);
auto isVisible;
star[i].setVisible(isVisible);
}
}



3 people think this answer is useful

I like your way of thinking. Really outside the box. However the tradeoff is really not worth it. Memory-runtime tradeoff is a thing, including undefined behavior for runtime is not.

It must give you a very unsettling feeling to know you are using such “random” as your business logic. I woudn’t do it.

3 people think this answer is useful

Use 7757 every place you are tempted to use uninitialized variables. I picked it randomly from a list of prime numbers:

1. it is defined behavior

2. it is guaranteed to not always be 0

3. it is prime

4. it is likely to be as statistically random as uninitualized variables

5. it is likely to be faster than uninitialized variables since its value is known at compile time

1 people think this answer is useful

There is one more possibility to consider.

Modern compilers (ahem g++) are so intelligent that they go through your code to see what instructions affect state, and what don’t, and if an instruction is guaranteed to NOT affect the state, g++ will simply remove that instruction.

So here’s what will happen. g++ will definitely see that you are reading, performing arithmetic on, saving, what is essentially a garbage value, which produces more garbage. Since there is no guarantee that the new garbage is any more useful than the old one, it will simply do away with your loop. BLOOP!

This method is useful, but here’s what I would do. Combine UB (Undefined Behaviour) with rand() speed.

Of course, reduce rand()s executed, but mix them in so compiler doesn’t do anything you don’t want it to.

And I won’t fire you.