I need to make a big array in one task (more than 10^7).
And what I found that if i do it int main the code wouldnt work (the program will exit before doing cout "Process returned -1073741571 (0xC00000FD)").
If I do it outside everything will work.
(I am using Code::Blocks 17.12)
// dont work
#include <bits/stdc++.h>
using namespace std;
const int N = 1e7;
int main() {
int a[N];
cout << 1;
return 0;
}
// will work
#include <bits/stdc++.h>
using namespace std;
const int N = 1e7;
int a[N];
int main() {
cout << 1;
return 0;
}
So I have questions:
-Why it happens?
-What can I do to define array in int main()? (actually if i do vector same size in int main() everything will work and it is strange)
There are four main types of memory which are interesting for C++ programmers: stack, heap, static memory, and the memory of registers.
In
const int N = 1e7;
int main(){int a[N];}
stack memory is deployed.
This type of memory is usually more limited than the heap and the static memory in size. For that reason, the error code is returned.
Operator new (or other function which allocates memory in heap) is needed so as to use heap:
const int N = 1e7;
int main(){int* a = new int[N]; delete a;}
Usually, the operator new is not used explicitly.
std::vector uses heap (i.e. it uses new or something of the lower level underneath) (as opposed to the std::array or the C-style array, e.g. int[N]). Because of that, std::vector is usually capable of holding bigger chunks of data than the std::array or the C-style array.
If you do
const int N = 1e7;
int a[N];
int main(){}
static memory is utilized. It's usually less limited in size than the stack memory.
To wrap up, you used stack in int main(){int a[N];}, static memory in int a[N]; int main(){}, and heap in int main(){std::vector<int> v(N);}, and, because of that, received different results.
Use heap for big arrays (via the std::vector or the operator new, examples are given above).
The problem is that your array is actually very big. Assuming that int is 4 bytes, 10 000 000 integers will be 40 000 000bytes which is about 40 Mb. In windows maximum stack size is 1Mb and on modern Linux 8Mb. As local variables are located in stack so youre allocating your 40mb array in 1mb or 8mb stack (if youre in windows or linux respectively). So your program runs out of stack space. In case of global array its ok, because global variables are located in bss(data) segment of program which has static size and is not changing. And in case of std::vector your array is allocated in dynamic memory e.g. in heap, thats why your program is not crashing. If you don't want to use std::vector you can dynamically allocate an array on heap like following
int* arrayPtr = new int[N]
Then you need to free unused dynamically allocated memory with delete operator:
delete arrayPtr;
But in this case you need to know how to work with pointers. Or if you want it to not be dynamic and be only in main, you can make your array in main static (I think 99.9% this will work 😅 and I think you need to try) like this
int main() {static int arr[N];return 0;}
Which will be located in data segment (like global variable)
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;
}
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;
}