I wanted to access deleted array to see how the memory was changed it works till I deleted really big array then I get access violation exception. Please do not care about cout I know they are slow but I will get rid of them.
When I do it for 1000 elements array it is ok, when I do it for 1000000 i get an exception. I know that this is weird task but my teacher is stubborn and I can't find out how to deal with that.
EDIT: I know that I never should access that memory, but I also know that there is probably trick he will show then and tell that I am not right.
long max = 1000000;// for 10000 i do not get any exception.
int* t = new int[max];
cout<<max<<endl;
uninitialized_fill_n(t, max, 1);
delete[] t;
cout<<"deleted t"<<endl;
int x;
cin>>x;//wait little bit
int one = 1;
long counter = 0;
for(long i = 0; i < max; i++){
cout<<i<<endl;
if(t[i] != 1){
cout<<t[i]<<endl;
counter++;
}
}
That state of "deleted" memory is undefined, to access memory after delete is UNDEFINED BEHAVIOUR (meaning, the C++ specification allows "anything" to happen when you access such memory - including the appearance of it "working" sometimes, and "not working" sometimes).
You should NEVER access memory that has been deleted, and as shown in your larger array case, it may not work to do so, because the memory may no longer actually be available to your process.
You are not allowed to access to a released buffer
Accessing memory that is no longer in use results in undefined behaviour. You will not get any consistent patterns. If the original memory has not been overwritten after it became invalid, the values will be exactly what they used to be.
I find the answer to this similar question to be very clear in explaining the concept with a simple analogy.
A simple way to mimic this behaviour is to create a function which returns a pointer to a local variable, for example:
int *foo(){
int a=1;
return &a;
}
Related
Everything seems to run okay up until the return part of shuffle_array(), but I'm not sure what.
int * shuffle_array(int initialArray[], int userSize)
{
// Variables
int shuffledArray[userSize]; // Create new array for shuffled
srand(time(0));
for (int i = 0; i < userSize; i++) // Copy initial array into new array
{
shuffledArray[i] = initialArray[i];
}
for(int i = userSize - 1; i > 0; i--)
{
int randomPosition = (rand() % userSize;
temp = shuffledArray[i];
shuffledArray[i] = shuffledArray[randomPosition];
shuffledArray[randomPosition] = temp;
}
cout << "The numbers in the initial array are: ";
for (int i = 0; i < userSize; i++)
{
cout << initialArray[i] << " ";
}
cout << endl;
cout << "The numbers in the shuffled array are: ";
for (int i = 0; i < userSize; i++)
{
cout << shuffledArray[i] << " ";
}
cout << endl;
return shuffledArray;
}
Sorry if spacing is off here, not sure how to copy and past code into here, so I had to do it by hand.
EDIT: Should also mention that this is just a fraction of code, not the whole project I'm working on.
There are several issues of varying severity, and here's my best attempt at flagging them:
int shuffledArray[userSize];
This array has a variable length. I don't think that it's as bad as other users point out, but you should know that this isn't allowed by the C++ standard, so you can't expect it to work on every compiler that you try (GCC and Clang will let you do it, but MSVC won't, for instance).
srand(time(0));
This is most likely outside the scope of your assignment (you've probably been told "use rand/srand" as a simplification), but rand is actually a terrible random number generator compared to what else the C++ language offers. It is rather slow, it repeats quickly (calling rand() in sequence will eventually start returning the same sequence that it did before), it is easy to predict based on just a few samples, and it is not uniform (some values have a much higher probability of being returned than others). If you pursue C++, you should look into the <random> header (and, realistically, how to use it, because it's unfortunately not a shining example of simplicity).
Additionally, seeding with time(0) will give you sequences that change only once per second. This means that if you call shuffle_array twice quickly in succession, you're likely to get the same "random" order. (This is one reason that often people will call srand once, in main, instead.)
for(int i = userSize - 1; i > 0; i--)
By iterating to i > 0, you will never enter the loop with i == 0. This means that there's a chance that you'll never swap the zeroth element. (It could still be swapped by another iteration, depending on your luck, but this is clearly a bug.)
int randomPosition = (rand() % userSize);
You should know that this is biased: because the maximum value of rand() is likely not divisible by userSize, you are marginally more likely to get small values than large values. You can probably just read up the explanation and move on for the purposes of your assignment.
return shuffledArray;
This is a hard error: it is never legal to return storage that was allocated for a function. In this case, the memory for shuffledArray is allocated automatically at the beginning at the function, and importantly, it is deallocated automatically at the end: this means that your program will reuse it for other purposes. Reading from it is likely to return values that have been overwritten by some code, and writing to it is likely to overwrite memory that is currently used by other code, which can have catastrophic consequences.
Of course, I'm writing all of this assuming that you use the result of shuffle_array. If you don't use it, you should just not return it (although in this case, it's unlikely to be the reason that your program crashes).
Inside a function, it's fine to pass a pointer to automatic storage to another function, but it's never okay to return that. If you can't use std::vector (which is the best option here, IMO), you have three other options:
have shuffle_array accept a shuffledArray[] that is the same size as initialArray already, and return nothing;
have shuffle_array modify initialArray instead (the shuffling algorithm that you are using is in-place, meaning that you'll get correct results even if you don't copy the original input)
dynamically allocate the memory for shuffledArray using new, which will prevent it from being automatically reclaimed at the end of the function.
Option 3 requires you to use manual memory management, which is generally frowned upon these days. I think that option 1 or 2 are best. Option 1 would look like this:
void shuffle_array(int initialArray[], int shuffledArray[], int userSize) { ... }
where userSize is the size of both initialArray and shuffledArray. In this scenario, the caller needs to own the storage for shuffledArray.
You should NOT return a pointer to local variable. After the function returns, shuffledArray gets deallocated and you're left with a dangling pointer.
You cannot return a local array. The local array's memory is released when you return (did the compiler warn you about that). If you do not want to use std::vector then create yr result array using new
int *shuffledArray = new int[userSize];
your caller will have to delete[] it (not true with std::vector)
When you define any non static variables inside a function, those variables will reside in function's stack. Once you return from function, the function's stack is gone. In your program, you are trying to return a local array which will be gone once control is outside of shuffle_array().
To solve this, either you need to define the array globally (which I won't prefer because using global variables are dangerous) or use dynamic memory allocation for the array which will create space for the array in heap rather than allocating the space on the function's stack. You can use std::vectors also, if you are familiar with vectors.
To allocate memory dynamically, you have to use new as mentioned below.
int *shuffledArray[] = new int[userSize];
and once you completed using shuffledArray, you need to free the memory as below.
delete [] shuffledArray;
otherwise your program will leak memory.
int main() {
int* i = new int(1);
i++;
*i=1;
delete i;
}
Here is my logic:
I increment I by 1, and then assign a value to it. Then I delete the I, so I free the memory location while leaking the original memory. Where is my problem?
I also tried different versions. Every time, as long as I do the arithmetics and delete the pointer, my program crashes.
What your program shows is several cases of undefined behaviour:
You write to memory that hasn't been allocated (*i = 1)
You free something that you didn't allocate, effectively delete i + 1.
You MUST call delete on exactly the same pointer-value that you got back from new - nothing else. Assuming the rest of your code was valid, it would be fine to do int *j = i; after int *i = new int(1);, and then delete j;. [For example int *i = new int[2]; would then make your i++; *i=1; valid code]
Who allocates is who deallocates. So you should not be able to delete something you did not new by yourself. Furthermore, i++;*i=1; is UB since you may access a restricted memory area or read-only memory...
The code made no sense . I think You have XY problem. If you could post your original problem there will be more chance to help you.
In this case you need to have a short understanding how the heap memory management works. in particular implementation of it, when you allocate an object you receive a pointer to the start of the memory available to you to work with. However, the 'really' allocated memory starts a bit 'earlier'. This means the allocated block is a bit more than you have requested to allocate. The start of the block is the address you have received minus some offset. Thus, when you pass the incremented pointer to the delete it tries to find the internal information at the left side of it. And because your address is now incremented this search fails what results in a crash. That's in short.
The problem lies here:
i++;
This line doesn't increment the value i points to, but the pointer itself by the number of bytes an int has (4 on 32-bit platform).
You meant to do this:
(*i)++;
Let's take it step by step:
int* i = new int(1); // 1. Allocate a memory.
i++; // 2. Increment a pointer. The pointer now points to
// another location.
*i=1; // 3. Dereference a pointer which points to unknown
// memory. This could cause segmentation fault.
delete i; // 4. Delete the unknown memory which is undefined
// behavior.
In short: If you don't own a piece of memory you can't do arithmetic with it neither delete it!
I had posted a question on this but I thought to use memcmp() function instead.
Writing data to memory in C++
int x = 1;
int fileptr = 0;
void *data = malloc(4096);
memcpy((int *)data+fileptr, &x, sizeof(int));
Then I read the values back
int y;
fileptr = 0;
memcpy(&y, (int *)data+fileptr, sizeof(int));
cout<<y;
In this way, I get a different output for variable y (some long integer values). Please need immediate help.
You declare a pointer data but never initialize it. So the behavior is undefined. You have to point data somewhere, i.e. allocate memory. See here.
Unless there's some code you're leaving out, data is never initialized, and hence points to some random location. Since your application isn't crashing entirely, you're probably getting (somewhat) lucky and ending up with data pointing into a location on the stack which is written to by other code; hence the change in value.
Allocate some memory for data before you write to it and this won't happen.
You havent allocated any storage for data - so when you copy the int data there you are causing no end of problems by corrupting the stack.
You might want to do something like
int x = 1;
void *data=new int();
memcpy(data, &x, sizeof(x)); // never use the type use the var - more resilient to change
That actually invokes undefined behaviour, as you have not allocated memory for data. Where should it store the bit-patterns of x then?
If you have allocated memory (using malloc), and still get wrong output, then in that case, you are doing something terribly wrong. See your program again and then compile and run it again. Because the output cannot be anything other than 1. See this:
http://www.ideone.com/SBoHk
Compare your program with mine (in the above link) word by word, and see if there is anything you are missing.
I have the following program:
//simple array memory test.
#include <iostream>
using namespace std;
void someFunc(float*, int, int);
int main() {
int convert = 2;
float *arr = new float[17];
for(int i = 0; i < 17; i++) {
arr[i] = 1.0;
}
someFunc(arr, 17, convert);
for(int i = 0; i < 17; i++) {
cout << arr[i] << endl;
}
return 0;
}
void someFunc(float *arr, int num, int flag) {
if(flag) {
delete []arr;
}
}
When I put the following into gdb and insert a break point at float *arr ..., I step through the program and observe the following:
Printing the array arr after it has been initialized gives me 1 17 times.
Inside someFunc too, I print arr before delete to get the same print as above.
Upon going back into main, when I print arr, I get the first digit as 0 followed by 16 1.0s.
My questions:
1. Once the array has been deleted in someFunc, how am I still able to access arr without a segfault in someFunc or main?
2. The code snippet above is a test version of another piece of code that runs in a bigger program. I observe the same behaviour in both places (first digit is 0 but all others are the same. If this is some unexplained memory error, how am I observing the same thing in different areas?
3. Some explanations to fill the gaps in my understanding are most welcome.
A segfault occurs when you access a memory address that isn't mapped into the process. Calling delete [] releases memory back to the memory allocator, but usually not to the OS.
The contents of the memory after calling delete [] are an implementation detail that varies across compilers, libraries, OSes and especially debug-vs-release builds. Debug memory allocators, for instance, will often fill the memory with some tell-tale signature like 0xdeadbeef.
Dereferencing a pointer after it has been deleteed is undefined behavior, which means that anything can happen.
Once the array has been deleted, any access to it is undefined behavior.
There's no guarantee that you'll get a segment violation; in fact,
typically you won't. But there's no guarantee of what you will get; in
larger programs, modifying the contents of the array could easily result
in memory corruption elsewhere.
delete gives the memory back to the OS memory manager, but does not necessarily clears the contents in the memory(it should not, as it causes overhead for nothing). So the values are retained in the memory. And in your case, you are accessing the same memory -- so it will print what is in the memory -- it is not necessarily an undefined behaviour(depends on memory manager)
Re 1: You can't. If you want to access arr later, don't delete it.
C++ doesn't check for array boundaries. Only if you access a memory which you are not allowed to you will get segfault
Why the array is not overflowed (e.g. error alert) when the array is declared globally, in other why I'm able to fill it with unlimited amount of elements (through for) even it's limited by size in declaration and it does alert when I declare the array locally inside the main ?
char name[9];
int main(){
int i;
for( int i=0; i<18; ++i){
cin>>name[i];
}
cout<<"Inside the array: ";
for(i=0; i<20; i++)
cout<<name[i];
return 0;
}
C++ does not check bounds errors for arrays of any kind. Reading or writing outside of the array bounds causes what is known as "undefined behaviour", which means anything could happen. In your case, it seems that what happens is that it appears to work, but the program will still be in an invalid state.
If you want bounds checking, use a std::vector and its at() member function:
vector <int> a( 3 ); // vector of 3 ints
int n = a.at( 0 ); // ok
n = a.at( 42 ); // throws an exception
C++ does not have array bounds checking so the language never check to see if you have exceeded the end of your array but as others have mentioned bad things can be expected to happen.
Global variables exists in the static segment which is totally separate from your stack. It also does not contain important information like return addresses. When you exceed an array's boundaries you are effectively corrupting memory. It just so happens that corrupting the stack is more likely to cause more visible bad things than corrupting the data segment. All of this depends on the way your operating system organizes a process's memory.
its undefined behavior. Anything can happen.
You cannot assume too much about the memory layout of variables. It may run on your computer with these parameters, but totally fail when you increase your access bounds, or run that code on another machine. So if you seriously want to write code, don't let this become a habit.
I'd go one step further and state that C/C++ do not have arrays. What they have is array-like syntactic sugar that is immediately translated to pointer arithmetic, which cannot be checked, as pointers can be used to access potentially all of memory. Any checking that the compiler may manage to perform based on static sizes and constant bounds on an index is a happy accident, but you cannot rely on it.
Here's an oddity that stunned me when I first saw it:
int a[10], i;
i = 5;
a[i] = 42; // Looks normal.
5[a] = 37; // But what's this???
std::cout << "Array element = " << a[i] << std::endl;
But the odd-looking line is perfectly legal C++. This example emphasizes that arrays in C/C++ are a fiction.
Neil Butterworth already commented on the benefits of using std::vector and the at() access method for it, and I cannot second his recommendation strongly enough. (Unfortunately, the designers of STL blew a golden opportunity to make checked access the [] operators, with the at() methods the unchecked operators. This has probably cost the C++ programming community millions of hours and millions of dollars, and will continue to do so.)