I made a program for school's little contest, I have:
#include <iostream>
#include <vector>
int main(){
int n;
std::cin >> n;
int c[n];
std::vector<int> r(n);
std::cout << "I made it through";
//rest of program
}
When i type in 1000000 and press enter program crashes without a word, exit code (0xC00000FD). I presumed it happens when i initialize vector, but i though it can handle a lot more. Am i doing something wrong? I know i can use pointer but i'd prefer not to touch things that are already working.
The stack is a very restricted resource.
Even though your compiler seems to implement C-style VLA's in C++ (int c[n];), it does not magically gain more memory.
Test 1) Success of reading n, and 2) n not being out-of-bounds for your use before executing the statement allocating that stack-array.
Default stack-size for Windows: 1MB sizeof(int): 4 Thus, about 250000 fit.
Default stack size for Linux: 8MB sizeof(int): 4 Thus, about 2000000 fit.
Workaround: Use dynamic allocation for the ints, like with a std::vector:
std::vector<int> c(n);
Alternatively, at least use a smart-pointer:
std::unique_ptr<int[]> c(new int[n]);
The problem isn't vector (which uses the heap for its underlying expanding storage in most STL implementations), but int c[n], which will allocate 1,000,000 4-byte integers in the stack, which is almost 4MB. On Win32 the stack is by default around 1MB, hence the overflow.
If you really need to use an array then change your c array to be allocated on the heap by using new, but don't forget to delete[], otherwise use of vector is preferred for most expanding storage scenarios. If you need a fixed-length array then consider array<class T, size_t N> (new in C++11) which adds bounds-checking.
You probably just need to allocate the array dynamically. In C++
#include <iostream>
#include <vector>
int main()
{
int n;
std::cin >> n;
int *c = new int[n];
if(nullptr == c) {
std::cerr << "Runtime error" << std::endl;
return 1;
}
std::vector<int> r(begin(n), end(n));
std::cout << "I made it through";
delete[] c;
return 0;
}
Also to make the vector after you have allocated c dynamically you can use begin, end
Related
The following code gives me a segmentation fault when run on a 2Gb machine, but works on a 4GB machine.
int main()
{
int c[1000000];
cout << "done\n";
return 0;
}
The size of the array is just 4Mb. Is there a limit on the size of an array that can be used in c++?
You're probably just getting a stack overflow here. The array is too big to fit in your program's stack region; the stack growth limit is usually 8 MiB or 1 MiB for user-space code on most mainstream desktop / server OSes. (Normal C++ implementations use the asm stack for automatic storage, i.e. non-static local variables arrays. This makes deallocating them happen for free when functions return or an exception propagates through them.)
If you dynamically allocate the array you should be fine, assuming your machine has enough memory.
int* array = new int[1000000]; // may throw std::bad_alloc
But remember that this will require you to delete[] the array manually to avoid memory leaks, even if your function exits via an exception. Manual new/delete is strongly discouraged in modern C++, prefer RAII.
A better solution would be to use std::vector<int> array (cppreference). You can reserve space for 1000000 elements, if you know how large it will grow. Or even resize it to default-construct them (i.e. zero-initialize the memory, unlike when you declare a plain C-style array with no initializer), like std::vector<int> array(1000000)
When the std::vector object goes out of scope, its destructor will deallocate the storage for you, even if that happens via an exception in a child function that's caught by a parent function.
In C or C++ local objects are usually allocated on the stack. You are allocating a large array on the stack, more than the stack can handle, so you are getting a stackoverflow.
Don't allocate it local on stack, use some other place instead. This can be achieved by either making the object global or allocating it on the global heap. Global variables are fine, if you don't use the from any other compilation unit. To make sure this doesn't happen by accident, add a static storage specifier, otherwise just use the heap.
This will allocate in the BSS segment, which is a part of the heap. Since it's in static storage, it's zero initialized if you don't specify otherwise, unlike local variables (automatic storage) including arrays.
static int c[1000000];
int main()
{
cout << "done\n";
return 0;
}
A non-zero initializer will make a compiler allocate in the DATA segment, which is a part of the heap too. (And all the data for the array initializer will take space in the executable, including all the implicit trailing zeros, instead of just a size to zero-init in the BSS)
int c[1000000] = {1, 2, 3};
int main()
{
cout << "done\n";
return 0;
}
This will allocate at some unspecified location in the heap:
int main()
{
int* c = new int[1000000]; // size can be a variable, unlike with static storage
cout << "done\n";
delete[] c; // dynamic storage needs manual freeing
return 0;
}
Also, if you are running in most UNIX & Linux systems you can temporarily increase the stack size by the following command:
ulimit -s unlimited
But be careful, memory is a limited resource and with great power come great responsibilities :)
You array is being allocated on the stack in this case attempt to allocate an array of the same size using alloc.
Because you store the array in the stack. You should store it in the heap. See this link to understand the concept of the heap and the stack.
Your plain array is allocated in stack, and stack is limited to few magabytes, hence your program gets stack overflow and crashes.
Probably best is to use heap-allocated std::vector-based array which can grow almost to size of whole memory, instead of your plain array.
Try it online!
#include <vector>
#include <iostream>
int main() {
std::vector<int> c(1000000);
std::cout << "done\n";
return 0;
}
Then you can access array's elements as usual c[i] and/or get its size c.size() (number of int elements).
If you want multi-dimensional array with fixed dimensions then use mix of both std::vector and std::array, as following:
Try it online!
#include <vector>
#include <array>
#include <iostream>
int main() {
std::vector<std::array<std::array<int, 123>, 456>> c(100);
std::cout << "done\n";
return 0;
}
In example above you get almost same behavior as if you allocated plain array int c[100][456][123]; (except that vector allocates on heap instead of stack), you can access elements as c[10][20][30] same as in plain array. This example above also allocates array on heap meaning that you can have array sizes up to whole memory size and not limited by stack size.
To get pointer to the first element in vector you use &c[0] or just c.data().
there can be one more way that worked for me!
you can reduce the size of array by changing its data type:
int main()
{
short c[1000000];
cout << "done\n";
return 0;
}
or
int main()
{
unsigned short c[1000000];
cout << "done\n";
return 0;
}
The following code gives me a segmentation fault when run on a 2Gb machine, but works on a 4GB machine.
int main()
{
int c[1000000];
cout << "done\n";
return 0;
}
The size of the array is just 4Mb. Is there a limit on the size of an array that can be used in c++?
You're probably just getting a stack overflow here. The array is too big to fit in your program's stack region; the stack growth limit is usually 8 MiB or 1 MiB for user-space code on most mainstream desktop / server OSes. (Normal C++ implementations use the asm stack for automatic storage, i.e. non-static local variables arrays. This makes deallocating them happen for free when functions return or an exception propagates through them.)
If you dynamically allocate the array you should be fine, assuming your machine has enough memory.
int* array = new int[1000000]; // may throw std::bad_alloc
But remember that this will require you to delete[] the array manually to avoid memory leaks, even if your function exits via an exception. Manual new/delete is strongly discouraged in modern C++, prefer RAII.
A better solution would be to use std::vector<int> array (cppreference). You can reserve space for 1000000 elements, if you know how large it will grow. Or even resize it to default-construct them (i.e. zero-initialize the memory, unlike when you declare a plain C-style array with no initializer), like std::vector<int> array(1000000)
When the std::vector object goes out of scope, its destructor will deallocate the storage for you, even if that happens via an exception in a child function that's caught by a parent function.
In C or C++ local objects are usually allocated on the stack. You are allocating a large array on the stack, more than the stack can handle, so you are getting a stackoverflow.
Don't allocate it local on stack, use some other place instead. This can be achieved by either making the object global or allocating it on the global heap. Global variables are fine, if you don't use the from any other compilation unit. To make sure this doesn't happen by accident, add a static storage specifier, otherwise just use the heap.
This will allocate in the BSS segment, which is a part of the heap. Since it's in static storage, it's zero initialized if you don't specify otherwise, unlike local variables (automatic storage) including arrays.
static int c[1000000];
int main()
{
cout << "done\n";
return 0;
}
A non-zero initializer will make a compiler allocate in the DATA segment, which is a part of the heap too. (And all the data for the array initializer will take space in the executable, including all the implicit trailing zeros, instead of just a size to zero-init in the BSS)
int c[1000000] = {1, 2, 3};
int main()
{
cout << "done\n";
return 0;
}
This will allocate at some unspecified location in the heap:
int main()
{
int* c = new int[1000000]; // size can be a variable, unlike with static storage
cout << "done\n";
delete[] c; // dynamic storage needs manual freeing
return 0;
}
Also, if you are running in most UNIX & Linux systems you can temporarily increase the stack size by the following command:
ulimit -s unlimited
But be careful, memory is a limited resource and with great power come great responsibilities :)
You array is being allocated on the stack in this case attempt to allocate an array of the same size using alloc.
Because you store the array in the stack. You should store it in the heap. See this link to understand the concept of the heap and the stack.
Your plain array is allocated in stack, and stack is limited to few magabytes, hence your program gets stack overflow and crashes.
Probably best is to use heap-allocated std::vector-based array which can grow almost to size of whole memory, instead of your plain array.
Try it online!
#include <vector>
#include <iostream>
int main() {
std::vector<int> c(1000000);
std::cout << "done\n";
return 0;
}
Then you can access array's elements as usual c[i] and/or get its size c.size() (number of int elements).
If you want multi-dimensional array with fixed dimensions then use mix of both std::vector and std::array, as following:
Try it online!
#include <vector>
#include <array>
#include <iostream>
int main() {
std::vector<std::array<std::array<int, 123>, 456>> c(100);
std::cout << "done\n";
return 0;
}
In example above you get almost same behavior as if you allocated plain array int c[100][456][123]; (except that vector allocates on heap instead of stack), you can access elements as c[10][20][30] same as in plain array. This example above also allocates array on heap meaning that you can have array sizes up to whole memory size and not limited by stack size.
To get pointer to the first element in vector you use &c[0] or just c.data().
there can be one more way that worked for me!
you can reduce the size of array by changing its data type:
int main()
{
short c[1000000];
cout << "done\n";
return 0;
}
or
int main()
{
unsigned short c[1000000];
cout << "done\n";
return 0;
}
The following code gives me a segmentation fault when run on a 2Gb machine, but works on a 4GB machine.
int main()
{
int c[1000000];
cout << "done\n";
return 0;
}
The size of the array is just 4Mb. Is there a limit on the size of an array that can be used in c++?
You're probably just getting a stack overflow here. The array is too big to fit in your program's stack region; the stack growth limit is usually 8 MiB or 1 MiB for user-space code on most mainstream desktop / server OSes. (Normal C++ implementations use the asm stack for automatic storage, i.e. non-static local variables arrays. This makes deallocating them happen for free when functions return or an exception propagates through them.)
If you dynamically allocate the array you should be fine, assuming your machine has enough memory.
int* array = new int[1000000]; // may throw std::bad_alloc
But remember that this will require you to delete[] the array manually to avoid memory leaks, even if your function exits via an exception. Manual new/delete is strongly discouraged in modern C++, prefer RAII.
A better solution would be to use std::vector<int> array (cppreference). You can reserve space for 1000000 elements, if you know how large it will grow. Or even resize it to default-construct them (i.e. zero-initialize the memory, unlike when you declare a plain C-style array with no initializer), like std::vector<int> array(1000000)
When the std::vector object goes out of scope, its destructor will deallocate the storage for you, even if that happens via an exception in a child function that's caught by a parent function.
In C or C++ local objects are usually allocated on the stack. You are allocating a large array on the stack, more than the stack can handle, so you are getting a stackoverflow.
Don't allocate it local on stack, use some other place instead. This can be achieved by either making the object global or allocating it on the global heap. Global variables are fine, if you don't use the from any other compilation unit. To make sure this doesn't happen by accident, add a static storage specifier, otherwise just use the heap.
This will allocate in the BSS segment, which is a part of the heap. Since it's in static storage, it's zero initialized if you don't specify otherwise, unlike local variables (automatic storage) including arrays.
static int c[1000000];
int main()
{
cout << "done\n";
return 0;
}
A non-zero initializer will make a compiler allocate in the DATA segment, which is a part of the heap too. (And all the data for the array initializer will take space in the executable, including all the implicit trailing zeros, instead of just a size to zero-init in the BSS)
int c[1000000] = {1, 2, 3};
int main()
{
cout << "done\n";
return 0;
}
This will allocate at some unspecified location in the heap:
int main()
{
int* c = new int[1000000]; // size can be a variable, unlike with static storage
cout << "done\n";
delete[] c; // dynamic storage needs manual freeing
return 0;
}
Also, if you are running in most UNIX & Linux systems you can temporarily increase the stack size by the following command:
ulimit -s unlimited
But be careful, memory is a limited resource and with great power come great responsibilities :)
You array is being allocated on the stack in this case attempt to allocate an array of the same size using alloc.
Because you store the array in the stack. You should store it in the heap. See this link to understand the concept of the heap and the stack.
Your plain array is allocated in stack, and stack is limited to few magabytes, hence your program gets stack overflow and crashes.
Probably best is to use heap-allocated std::vector-based array which can grow almost to size of whole memory, instead of your plain array.
Try it online!
#include <vector>
#include <iostream>
int main() {
std::vector<int> c(1000000);
std::cout << "done\n";
return 0;
}
Then you can access array's elements as usual c[i] and/or get its size c.size() (number of int elements).
If you want multi-dimensional array with fixed dimensions then use mix of both std::vector and std::array, as following:
Try it online!
#include <vector>
#include <array>
#include <iostream>
int main() {
std::vector<std::array<std::array<int, 123>, 456>> c(100);
std::cout << "done\n";
return 0;
}
In example above you get almost same behavior as if you allocated plain array int c[100][456][123]; (except that vector allocates on heap instead of stack), you can access elements as c[10][20][30] same as in plain array. This example above also allocates array on heap meaning that you can have array sizes up to whole memory size and not limited by stack size.
To get pointer to the first element in vector you use &c[0] or just c.data().
there can be one more way that worked for me!
you can reduce the size of array by changing its data type:
int main()
{
short c[1000000];
cout << "done\n";
return 0;
}
or
int main()
{
unsigned short c[1000000];
cout << "done\n";
return 0;
}
File extension: .cpp
I have the following code:
int main() {
int N; cin >> N;
int myArray[N];
return 0;
}
I get an error when I'm trying to run that program if I input N as 1,000,000. However, when I set myArray[N] to myArray[1000000], it doesn't. Why does this happen?
int myArray[N]; is not valid in C++. This behavior was introduced in C99 but never in C++, possibly because it causes a lot of ugly stuff to happen behind the scenes in order to make it work, and it would make the generated code less efficient as a result. In fact this feature was even reversed in C11 where its support is merely optional and not mandatory anymore. Use std::vector<int> instead, or any similar standard container of your choice.
First of all VLA (variable length arrays) is an extension to C++. Compilers are supporting that since usually they support also C which has this functionality in standard.
Second problem this array is allocated on stack.
Stack has very limited size. So when your N has very big value application may crash since stack will overflow.
In this case you should use std::vector which will allocate data on heap.
Question is why array with static array size do not crash?
There can be couple reasons.
Compiler notices that array is not used and based on "As if" rule removes array.
Compiler knows size of the array at compile time, so required stack size is know. This information may be propagated to linker and application is build with bigger stack size then default value (in case of one suorce code application it may be possible). Disclaimer: this is my guessing, I didn't verified this in any form (by testing, or compiler documentation), but I've found this SO answer which confirms my suspicions.
The size of static arrays array[N] must be known at compile time.
Use std::vector for dynamic arrays:
// Example program
#include <iostream>
#include <string>
#include <vector>
int main()
{
int N; std::cin >> N;
std::cout << N << std::endl;
std::vector<int> myArray(N);
std::cout << myArray.size() << std::endl;
return 0;
}
That happens because size of static arrays must be known at compile time.
It is strongly recommended to use std::vector instead of arrays for more flexibility, and safety (this is always the answer: Use a vector if possible). You may use std::vector::reserve to request capacity be at least the length you want it to be. Use std::vector::capacity to see the current capacity.
#include <iostream>
#include <vector>
int main () {
std::vector<int> ivec;
ivec.reserve(100);
std::cout << ivec.capacity() << std::endl;
return 0;
}
Output:
100
Only if you have a very good reason to prefer arrays over vectors, you may dynamically allocate an array. Using std::shared_ptr makes this process much safer and convenient. Here's how it's done the way you want:
#include <iostream>
#include <memory>
int main () {
int N;
std::cin >> N;
std::shared_ptr<int> arr_ptr (new int[N], std::default_delete<int[]>());
for (int i = 0; i != N; ++i) {
arr_ptr.get()[i] = i * 2;
}
for (int i = 0; i != N; ++i) {
std::cout << arr_ptr.get()[i] << std::endl;
}
return 0;
}
Input:
10
Output:
0
2
4
6
8
10
12
14
16
18
That happens because, in C++, the size of static arrays declared with array[N] must be known at compile time and thus your error is propably your compiler which tells you that he must know the size inbeforehand. As stated use std::vector when you need dynamic arrays.
This question already has answers here:
What is the difference between Static and Dynamic arrays in C++?
(13 answers)
Closed 9 years ago.
I'm a beginer. I'm confused about the difference between them. I have read a few answers and realized one of the difference is that the dynamic array can be deleted while the normal array can't. But are there any other differences? Such as their functions, sizes or whatsoever?
Well I have read such an example, and I don't see any difference if I replace the dynamic array {p= new (nothrow) int[i];} with a normal array {int p [i];}.
#include <iostream>
#include <new>
using namespace std;
int main ()
{
int i,n;
int * p;
cout << "How many numbers would you like to type? ";
cin >> i;
p= new (nothrow) int[i];
if (p == 0)
cout << "Error: memory could not be allocated";
else
{
for (n=0; n<i; n++)
{
cout << "Enter number: ";
cin >> p[n];
}
cout << "You have entered: ";
for (n=0; n<i; n++)
cout << p[n] << ", ";
delete[] p;
}
return 0;
}
But are there any other differences?
Compile and run this, see what happens:
int main(int argc, char** argv){
char buffer[1024*1024*64];
buffer[0] = 0;
return 0;
}
Explanation:
Well, normal array is placed either on stack or within code segment (if it is global variable or static local variable). At least on windows/linux PCs. The stack has limited size (although you can change it using ulimit in linux and compiler settings on windows). So your array is too big for the stack, your program will instantly crash upon entering the function with that array ("segmentation fault" on linux, "stack overflow" or "access violation" on windows (forgoet which one)). Default limit for array size is 1Megabyte on windows (x86), and 8 megabytes on linux.
You cannot determine size of a block allocated with new. int *p = new int[146]; std::cout << sizeof(p) << std::endl. will print sizeof(int*), not size of allocated memory. However, sizeof works on arrays.
Theoretically, using dynamic memory allocation, you can allocate as much memory as you want (operating system may impose limits, though, and on 32bit system maximum allocated block size will be 2..3 GB). You can also control resource usage by freeing the memory, so your program won't be eating system ram/swap file for no reason.
Dynamic arrays are not automatically freed, you have delete them manually.
That's just a brief overview.
Dynamic memory allocation provides finer resource control and removes some limitations of local variables.
Speaking of which, although you CAN use new/delete, if you want to use arrays with variable size, you should use std::vector instead. Manual memory management is prone to errors, so you should make compiler do it for you when possible. As a result, it is advised to at least study STL(Standard Template Library) , smart pointers, RAII, Rule of Three.
Dynamic is just that. You can change the size of the resulting array at runtime rather than compile time. There are compiler extensions that allow you to use static arrays (int array[CONSTANT]) like a dynamic array in that you can specify the size at runtime but those are nonstandard.
Dynamic arrays are also allocated on the heap rather than the stack. This allows you to use all of the available memory if necessary to create the array. The stack has a limited size that depends on the OS.
Allocating memory on the heap allows you to for example do this:
int* createArray(int n) {
int* arr = new int[n];
return arr;
}
int main()
{
int* myArray = createArray(10);
// this array is now valid and can be used
delete[] myArray;
}
I'm sure there are numerous other intricacies between dynamic vs static arrays but those points are the big ones.