This question already has answers here:
Getting a stack overflow exception when declaring a large array
(8 answers)
Closed 8 years ago.
Something like this throws an error:
using namespace std;
int main()
{
int test[1000000] = {};
}
Something like this doesn't:
using namespace std;
int test[1000000] = {};
int main()
{
}
Why is that? A million ints isn't even too memory-demanding.
The first one allocates space on the stack. The second one allocates space in the data segment at compile/link time. The stack is of limited size.
Stack is not dynamic, but you can also do this
int* arr = new int[1000000];
but don't forget to delete it because this declares array in the heap which is dynamic memory and by deleting it from heap you prevent the memory leak.
Example :
delete arr;
This is just alternative how to use memory
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As the title suggests,
int [] a = new int[size];
constantly throws me an error.
I am not yet familiar with C++ so please help me tweak the code above to as closely similar to the syntax above (above was a given pseudo(?) code in a class) so I can create an array.
Thank you.
#include <iostream>
using namespace std;
// Test program
int main( )
{
int [] a = new int[10];
cout << a[0];
return 0;
}
Tried running the above code and it failed to compile.
new is for dynamic allocation (of an array on the heap in this case), and returns a pointer to the new array.
[] in the declaration are for declaring it to be a stack array (and the brackets belong the right side of the variable name).
So the two legal approaches your code is mixing would simplify to either:
int a[10]; // Declares a stack array of size 10
or:
int *a = new int[10]; // Dynamically allocates a heap array of size 10 and stores the pointer to it in `a`
The former is the better solution here (there's no benefit to a heap allocation for such a small array that is only used within the scope of this function, and it would require explicit delete[] a; later to properly clean it up).
That said, neither version initializes the values, so the contents of a[0] are uninitialized, and reading from them will get random garbage. Both versions can be changed to initialize the data with zeroes though, e.g.:
int a[10] = {0}; // Declares a stack array of size 10 and initializes to zero
or:
int *a = new int[10](); // Dynamically allocates a heap array of size 10 and stores the pointer to it in `a`, initializing to zero
Final note: Using raw new is generally frowned upon in modern C++. If you needed heap-allocated contiguous data storage, you're probably better off using a std::vector.
This question already has answers here:
Segmentation fault on large array sizes
(7 answers)
Closed 3 years ago.
Program with large global array:
int ar[2000000];
int main()
{
}
Program with large local array:
int main()
{
int ar[2000000];
}
When I declare an array with large size in the main function, the program crashes with "SIGSEGV (Segmentation fault)".
However, when I declare it as global, everything works fine. Why is that?
Declaring the array globally causes the compiler to include the space for the array in the data section of the compiled binary. In this case you have increased the binary size by 8 MB (2000000 * 4 bytes per int). However, this does mean that the memory is available at all times and does not need to be allocated on the stack or heap.
EDIT: #Blue Moon rightly points out that an uninitialized array will most likely be allocated in the bss data segment and may, in fact, take up no additional disk space. An initialized array will be allocated statically.
When you declare an array that large in your program you have probably exceeded the stack size of the program (and ironically caused a stack overflow).
A better way to allocate a large array dynamically is to use a pointer and allocate the memory on the heap like this:
using namespace std;
int main() {
int *ar;
ar = malloc(2000000 * sizeof(int));
if (ar != null) {
// Do something
free(ar);
}
return 0;
}
A good tutorial on the Memory Layout of C Programs can be found here.
This question already has answers here:
Segmentation fault on large array sizes
(7 answers)
Closed 3 years ago.
Program with large global array:
int ar[2000000];
int main()
{
}
Program with large local array:
int main()
{
int ar[2000000];
}
When I declare an array with large size in the main function, the program crashes with "SIGSEGV (Segmentation fault)".
However, when I declare it as global, everything works fine. Why is that?
Declaring the array globally causes the compiler to include the space for the array in the data section of the compiled binary. In this case you have increased the binary size by 8 MB (2000000 * 4 bytes per int). However, this does mean that the memory is available at all times and does not need to be allocated on the stack or heap.
EDIT: #Blue Moon rightly points out that an uninitialized array will most likely be allocated in the bss data segment and may, in fact, take up no additional disk space. An initialized array will be allocated statically.
When you declare an array that large in your program you have probably exceeded the stack size of the program (and ironically caused a stack overflow).
A better way to allocate a large array dynamically is to use a pointer and allocate the memory on the heap like this:
using namespace std;
int main() {
int *ar;
ar = malloc(2000000 * sizeof(int));
if (ar != null) {
// Do something
free(ar);
}
return 0;
}
A good tutorial on the Memory Layout of C Programs can be found here.
This question already has answers here:
Segmentation fault on large array sizes
(7 answers)
Closed 3 years ago.
Program with large global array:
int ar[2000000];
int main()
{
}
Program with large local array:
int main()
{
int ar[2000000];
}
When I declare an array with large size in the main function, the program crashes with "SIGSEGV (Segmentation fault)".
However, when I declare it as global, everything works fine. Why is that?
Declaring the array globally causes the compiler to include the space for the array in the data section of the compiled binary. In this case you have increased the binary size by 8 MB (2000000 * 4 bytes per int). However, this does mean that the memory is available at all times and does not need to be allocated on the stack or heap.
EDIT: #Blue Moon rightly points out that an uninitialized array will most likely be allocated in the bss data segment and may, in fact, take up no additional disk space. An initialized array will be allocated statically.
When you declare an array that large in your program you have probably exceeded the stack size of the program (and ironically caused a stack overflow).
A better way to allocate a large array dynamically is to use a pointer and allocate the memory on the heap like this:
using namespace std;
int main() {
int *ar;
ar = malloc(2000000 * sizeof(int));
if (ar != null) {
// Do something
free(ar);
}
return 0;
}
A good tutorial on the Memory Layout of C Programs can be found here.
This question already has answers here:
Stack overflow visual C++, potentially array size?
(2 answers)
Closed 9 years ago.
Why is this code giving segmentation fault? I am using code::blocks.
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a[555555];
}
This is what called stack overflow.
Generally stack size are small and you can't allocate memory in such a large amount on stack.
For this purpose programmers allocate it on heap (using dynamic allocation). In C, you can use malloc family function
int *a = malloc(sizeof(int)*55555); // Use free(a) to deallocate
In C++ you can use new operator
int *b = new int[555555]; // Use delete [] to deallocate
Because you are trying to allocate a bit over 2MB (Previously I failed at math and thought it was 2GB) of memory on stack, which then blows the stack.
Note: for windows the default stack size for a particular thread is 1MB and on Gnu/linux you can find out stack size value using ulimit -s command.
You've come to the right place to ask the question. ;)
The array is large and lives on the stack. The code crashes because it runs out of the limited stack space.
If you allocate a on the heap, the problem will likely disappear.
As other already had told you you are trying to allocate a large amount of memory in the stack which space is usually very limited.
See for instance:
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a[555555];
int* b = new int[555555];
delete [] b;
}
In that snipped, you have two arrays of integers, one allocated in the heap and the other allocated on the stack.
Here you can find some explanations about which are the differences between the heap and the stack:
What and where are the stack and heap?
http://gribblelab.org/CBootcamp/7_Memory_Stack_vs_Heap.html
I've got some considerations on your code.
First, modern compilers will recognize that a is unused and then it'll be thrown away.
But if you put a certain value into a certain position, a will be allocated, even though it's bigger than the stack size. The kernel will not allow you do that: that's why you get a SIGSEGV.
Finally, you should rely on either std::array or std::vector rather than pure C arrays.