I am learning about scoping of variable in C.
Can anyone please explain what is going on below?
int w;
printf("\nw=%d\n", w);
w =-1;
Despite the fact that I initialized variable 'w' after 'printf', it always gets the value of "-1". This confused me, as I expect it to run sequentially. Hence, it should have printed some random value.
*** I also tried changing the value there, and it always read the written value. Hence, it did not randomly show "-1"
For experiment, I again tried the code below.
int w;
printf("\nw=%d\n", w);
w =-9;
w =-1;
Now, it reads a value of "2560". As I expect since it was not properly initialized before.
In your code
int w;
printf("\nw=%d\n", w);
invokes undefined behavior as you're trying to read the value of an uninitialized (automatic local) variable. The content of w is indeterminate at this point, and the output result is, well, undefined.
Always initialize your local variable before reading (using) the value.
Related: Quoting C11, chapter §6.7.9, Initialization
If an object that has automatic storage duration is not initialized explicitly, its value is
indeterminate. [....]
and, related to Undefined behavior, annex §J.2
The value of an object with automatic storage duration is used while it is
indeterminate
The variable in uninitialized. In "C", this means its value is "nondeterministic". In reality, the variable generally gets a value based on what's "laying around" at the memory address to which it gets assigned. In this case, its some value left on the stack.
It just so happens that often you will get consistent results across multiple runs simply due to external factors on which a program should not rely.
The compiler is optimizing the assignment of w in the first case. In the second case, it is deciding not to optimize.
In both cases, the compiler could choose to optimize out both assignments, since w is not used after they appear.
Initialize your variables before using them.
In both the above cases
int w;
printf("\nw=%d\n", w);
returns a random garbage value as we might call it which could be anything including -1 or 2560.
Blockquote
When you do not initialize a variable it can contain garbage value. Hence it's undefined behaviour and in most cases it will print random numbers as you experienced. By the way, as pointed out by others it's up to the compiler, so it may work with the expected value or it may don't work.
Related
I have this code that I'm using for something else, but, boiled it down to the root problem I think. If I enter 5 for the scanf variable when I run it, the printf out is 0,16. I don't understand why this is giving me 16 for *pScores?
#include <stdio.h>
int main(void) {
int a=0;
int sum=0;
scanf("%d",&a);
int scores[a];
int *pScores = &scores[0];
printf("%d, %d\n",scores[0],*pScores);
}
You are declaring an array
int scores[a];
and then printing out the value of scores[0] in two different ways. However, you have not stored anything into any of the elements of the scores array, so the values there are indeterminate.
Whether use of uninitialized (and therefore indeterminate) values in this way actually rises to the level of Undefined Behavior is a surprisingly deep and actually somewhat contentious question. (See the comment thread raging at the other answer.) Nevertheless, printing an uninitialized value like this isn't terribly useful. If you fill in a well-defined value to at least scores[0], I believe you'll find that both scores[0] and *pScores will print the same — that same — value.
Now, you might expect that the uninitialized value — whatever it is — would at least be consistent no matter how you print it (and I might agree with you), but when it comes to gray areas like this, and especially when a modern compiler starts leveraging every nuance of the rules in performing aggressive optimizations, the end results can be pretty surprising. When I tried your program, I got the same number printed twice (that is, I couldn't initially reproduce your result), but as suggested by Barmar in a comment, when I turned on optimization (with -O3), I started seeing conflicting results, also.
You have undefined behavior, caused by reading a variable with automatic storage duration whose value is indeterminate.
In 6.2.4 Storage durations of objects one finds the following rule
For such an object that does have a variable length array type, its lifetime extends from the declaration of the object until execution of the program leaves the scope of the declaration. If the scope is entered recursively, a new instance of the object is created
each time. The initial value of the object is indeterminate.
Then in J.2 Undefined behavior:
The behavior is undefined in the following circumstances
...
The value of an object with automatic storage duration is used while it is indeterminate.
...
Among permitted very weird outcomes when dealing with indeterminate values is that they have a different value each time you read them. The Schroedinger wavefunction does not collapse!
#include <stdio.h>
int main()
{
int a;
const int b = a;
printf("%d %d\n", a, b);
return 0;
}
The same code I tried to execute on onlinegdb.com compiler and on Ubuntu WSL. In onlinegdb.com, I got both a and b as 0 with every run, whereas in WSL it was a garbage value. I am not able to understand why garbage value is not coming onlinegdb.com
Using int a; inside a function is described by C 2018 6.2.4 5:
An object whose identifier is declared with no linkage and without the storage-class specifier static has automatic storage duration, as do some compound literals…
Paragraph 6 continues:
… The initial value of the object is indeterminate…
An indeterminate value is not an actual specific value but is a theoretical state used to describe the semantics of C. 3.19.2 says it is:
… either an unspecified value or a trap representation…
and 3.19.3 says an unspecified value is:
… valid value of the relevant type where this document imposes no requirements on which value is chosen in any instance…
That “any instance” part means that the program may behave as if a has a different value each time it is used. For example, these two statements may print different values for a:
printf("%d\n", a);
printf("%d\n", a);
Regarding const int b = a;, this is not covered explicitly by the C standard, but I have seen a committee response: When an indeterminate value is assigned (or initialized into) another object, the other object is also said to have an indeterminate value. So, after this declaration, b has an indeterminate value. The const is irrelevant; it means the source code of the program is not supposed to change b, but it cannot remedy the fact that b does not have a determined value.
Since the C standard permits any value to be used in each instance, onlinegdb.com conforms when it prints zero, and WSL conforms when it prints other values. Any int values printed for printf("%d %d\n", a, b); conform to the C standard.
Further, another provision in the C standard actually renders the entire behavior of the program undefined. C 2018 6.3.2.1 2 says:
… If the lvalue designates an object of automatic storage duration that could have been declared with the register storage class (never had its address taken), and that object is uninitialized (not declared with an initializer and no assignment to it has been performed prior to use), the behavior is undefined.
This applies to a: It could have been declared register because its address is never taken, and it is not initialized or assigned a value. So, using a has undefined behavior, and that extends to the entire behavior of the program on the code branch where a is used, which is the entire program execution.
I am not able to understand why garbage value is not coming
This is a very strange statement. I wonder: what kind of answer or explanation do you expect you might get? Something like:
Everyone who said that "uninitialized local variables start out containing random values" lied to you. WSL was wrong for giving you random values. You should have gotten 0, like you did with onlinegdb.com.
onlinegdb.com is buggy. It should have given truly random values.
The rules for const variables are special. When you say const int b = a;, it magically makes a's uninitialized value more predictable.
Are you expecting to get an answer like any of those? Because, no, none of those is true, none of those can possibly be true.
I'm sorry if it sounds like I'm teasing you here. I agree, it's surprising at first if an uninitialized local variable always starts out containing 0, because that's not very random, is it?
But the point is, the value of an uninitialized local variable is not defined. It is unspecified, indeterminate, and/or undefined. You cannot know what it is going to be. But that means that no value — no possible value — that it contains can ever be "wrong". In particular, onlinegdb.com is not wrong for not giving you random values: remember, it's not obligated to give you anything!
Think about it like this. Suppose you buy a carton of milk. Suppose it's printed on the label, "Contents may spoil — keep refrigerated." Suppose you leave the carton of milk on the counter overnight. That is, suppose you fail to properly refrigerate it. Suppose that, a day later, you realize your mistake. Horrified, you carefully open the milk carton and take a small taste, to see if it has spoiled. But you got lucky! It's still okay! It didn't spoil!
Now, at this point, what do you do?
Hastily put the milk in the refrigerator, and vow to be more careful next time.
March back to the store where you bought the milk, and accuse the shopkeeper of false advertising: "The label says 'contents may spoil', but it didn't!!" What do you think the shopkeeper is going to say?
This may seem like a silly analogy, but really, it's just like your C/C++ coding situation. The rules say you're supposed to initialize your local variables before you use them. You failed. Yet, somehow, you got predictable values anyway, at least under one compiler. But you can't complain about this, because it's not causing you a problem. And you can't depend on it, because as your experience with the other compiler showed you, it's not a guaranteed result.
Typically, local variables are stored on the stack. The stack gets used for all sorts of stuff. When you call a function, it receives a new stack frame where it can store its local variables, and where other stuff pertaining to that function call is stored, too. When a function returns, its stack frame is popped, but the memory is typically not cleared. That means that, when the next function is called, its newly-allocated stack frame may end up containing random bits of data left over from the previous function whose stack frame happened to occupy that part of stack memory.
So the question of what value an uninitialized local variable contains ends up depending on what the previous function might have been and what it might have left lying around on the stack.
In the case of main, it's quite possible that since it's the first function to be called, and the stack might start out empty, that main's stack frame always ends up being built on top of virgin, untouched, all-0 memory. That would mean that uninitialized variables in main might always seem to start out containing 0.
But this is not, not, not, not, not guaranteed!!!
Nobody said the stack was guaranteed to start out containing 0. Nobody said that there wasn't some startup code that ran before main that might have left some random garbage lying around on the stack.
If you want to enumerate possibilities, I can think of 3:
The function you're wondering about is one that always gets called first, or always gets called at a "deep leaf" of the call stack, meaning that it always gets a brand-new stack frame, and on a machine where the stack always starts out containing 0. Under these circumstances, uninitialized variables might always seem to start out containing 0.
The function you're wondering about does not always get a brand-new stack frame. It always gets a "dirty" stack frame with some previous function's random data lying around, and the program is such that, during every run, that previous function was doing something different and left something different on the stack, such that the next function's uninitialized local variables always seem to start out containing different, seemingly random values.
The function you're wondering about is always called right after a previous function that always does the same thing, meaning that it always leaves the same values lying around, meaning that the next function's uninitialized local variables always seem to start out with the same values every time, which aren't zero but aren't random.
But I hope it's obvious that you absolutely can't depend on any of this! And of course there's no reason to depend on any of this. If you want your local variables to have predictable values, you can simply initialize them. But if you're curious what happens when you don't, hopefully this explanation has helped you understand that.
Also, be aware that the explanation I've given here is somewhat of a simplification, and is not complete. There are systems that don't use a conventional stack at all, meaning that none of those possibilities 1, 2, or 3 could apply. There are systems that deliberately randomize the stack every time, either to help new programs not to accidentally become dependent on uninitialized variables, or to make sure that attackers can't exploit certain predictable results of a badly written program's undefined behavior.
When your operating system gives your program memory to work with, it will likely be zero to start (though not guaranteed). As your program calls functions it creates stack frames, and your program will effectively go from the .start assembly function to the int main() c function, so when main is called, no stack frame has written the memory that local variables are placed at. Therefore, a and b are both likely to be 0 (and b is guaranteed to be the same as a). However, it's not guaranteed to be 0, especially if you call some functions that have local variables or lots of parameters. For instance, if your code was instead
#include <stdio.h>
void foo()
{
int x = 42;
}
int main()
{
foo();
int a;
const int b = a;
printf("%d %d\n", a, b);
return 0;
}
then a would PROBABLY have the value 42 (in unoptimized builds), but that would depend on the ABI (https://en.wikipedia.org/wiki/Application_binary_interface) that your compiler uses and probably a few other things.
Basically, don't do that.
I came across a piece of code which is working fine as of now,but in my opinion its undefined behavior and might introduce a bug in future.
Pseudo code :
void OpertateLoad(int load_id)
{
int value = 0;
/* code to calculate value */
SetLoadRequest(load_id,&value);
/*some processing not involving value**/
}
void SetLoadRequest(int load_id, int* value)
{
/**some processing**/
LoadsArray[load_id] = *value;
/**some processing**/
}
In my understanding C compiler will not guarantee where Auto variables will be stored. It could be stack/register(if available and suitable for processing).
I am suspecting that if compiler decides to store value onto general purpose register then, SetLoadRequest function might refer to wrong data.
Am I getting it right?or I am overthinking it?
I am using IARARM compiler for ARM CORTEX M-4 processor.
----------:EDIT:----------
Answers Summarize that " Compiler will ensure that data is persisted between the calls, no matter where the variable is stored ".
Just want to confirm : Is this behavior also true 'if a function is returning the address of local auto variable and caller is de-referencing it?'.
If NO then is there anything in C standard which guarantees the behavior in both cases? Or As I stated earlier its undefined behavior?
You are overthinking it. The compiler knows that, if value is in a register, that it must be stored to memory before passing a pointer to that memory to SetLoadRequest.
More generally, don't think about stack and registers at all. The language says there's a variable (without saying how it's implemented), and that you can take its address and use that in another function to refer to the variable. So you can!
The language also says that local variables cease to exist when leaving a block, so this permission does not extend to returning pointers to local variables (which causes undefined behavior if the caller does anything at all with the pointer).
I am overthinking it?
Yes!
The compiler will take care of this. If it stores it in a register, it will know how to handle it (with a memory load).
C11 draft standard n1570:
6.2.4 Storage durations of objects
6 For such an object that does not have a variable length array type, its lifetime extends from entry into the block with which it is associated until execution of that block ends in any way. (Entering an enclosed block or calling a function suspends, but does not end, execution of the current block.) If the block is entered recursively, a new instance of the
object is created each time. The initial value of the object is indeterminate. If an
initialization is specified for the object, it is performed each time the declaration or
compound literal is reached in the execution of the block; otherwise, the value becomes
indeterminate each time the declaration is reached.
To my understanding, the scope of value is the body of OpertateLoad. However, SetLoadRequest assigns the value pointed to, so the actual value of value is copied. No undefined behaviour is involved.
I have the following code snippet.
int j;
printf("%d",j);
As expected, I get a garbage value.
32039491
But when I include a loop in the above snippet, like
int j;
print("%d",j);
while(j);
I get the following output on multiple trials of the program.
0
I always thought local variables are initialized to a garbage value by default, but it looks like variables get auto initialized when a loop is used.
It is having indeterminate value. It can be anything.
Quoting C11 §6.7.9
If an object that has automatic storage duration is not initialized explicitly, its value is
indeterminate. [...]
Automatic local variables, unless initialized explicitly, will contain indeterminate value. In case you try to use a variable while it holds indeterminate value and either
does not have the address taken
can have trap representation
the usage will lead to undefined behavior.
As expected, I get a garbage value.
Then your expectation is unjustifiably hopeful. When you use the indeterminate value of an uninitialized object, you generally get (and for your code snippets alone you do get) undefined behavior. Printing a garbage value is but one of infinitely many possible manifestations.
I always thought local variables are initialized to a garbage value by default, but it looks like variables get auto initialized when a loop is used.
You thought wrong, and you're also drawing the wrong conclusion. Both of your code snippets, when standing alone, exhibit undefined behavior. You cannot safely rely on any particular result.
I thought that once a function returns, all the local variables declared within (barring those with static keyword) are garbage collected. But when I am trying out the following code, it still prints the value after the function has returned. Can anybody explain why?
int *fun();
main() {
int *p;
p = fun();
printf("%d",*p); //shouldn't print 5, for the variable no longer exists at this address
}
int *fun() {
int q;
q = 5;
return(&q);
}
There's no garbage collection in C. Once the scope of a variable cease to exist, accessing it in any means is illegal. What you see is UB(Undefined behaviour).
It's undefined behavior, anything can happen, including appearing to work. The memory probably wasn't overwritten yet, but that doesn't mean you have the right to access it. Yet you did! I hope you're happy! :)
If you really want it to loose the value, perhaps call another function with at least a few lines of code in it, before doing the printf by accessing the location. Most probably your value would be over written by then.
But again as mentioned already this is undefined behavior. You can never predict when (or if at all) it crashes or changes. But you cannot rely upon it 'changing or remaining the same' and code an application with any of these assumptions.
What i am trying to illustrate is, when you make another function call after returning from previous one, another activation record is pushed on to the stack, most likely over writing the previous one including the variable whose value you were accessing via pointer.
No body is actually garbage collecting or doing a say memset 0 once a function and it's data goes out of scope.
C doesn't support garbage collection as supported by Java. Read more about garbage collection here
Logically, q ceases to exist when fun exits.
Physically (for suitably loose definitions of "physical"), the story is a bit more complicated, and depends on the underlying platform. C does not do garbage collection (not that garbage collection applies in this case). That memory cell (virtual or physical) that q occupied still exists and contains whatever value was last written to it. Depending on the architecture / operating system / whatever, that cell may still be accessible by your program, but that's not guaranteed:
6.2.4 Storage durations of objects
2 The lifetime of an object is the portion of program execution during which storage is
guaranteed to be reserved for it. An object exists, has a constant address,33)
and retains
its last-stored value throughout its lifetime.34)
If an object is referred to outside of its
lifetime, the behavior is undefined. The value of a pointer becomes indeterminate when
the object it points to (or just past) reaches the end of its lifetime.
33) The term ‘‘constant address’’ means that two pointers to the object constructed at possibly different
times will compare equal. The address may be different during two different executions of the same
program.
34) In the case of a volatile object, the last store need not be explicit in the program.
"Undefined behavior" is the C language's way of dealing with problems by not dealing with them. Basically, the implementation is free to handle the situation any way it chooses to, up to ignoring the problem completely and letting the underlying OS kill the program for doing something naughty.
In your specific case, accessing that memory cell after fun had exited didn't break anything, and it had not yet been overwritten. That behavior is not guaranteed to be repeatable.