Buddy Memory Allocation Program | Set 1 (Allocation) (original) (raw)
Last Updated : 11 Jul, 2025
Prerequisite - Buddy System
**Question: Write a program to implement the buddy system of memory allocation in Operating Systems.
**Explanation -
The buddy system is implemented as follows- A list of free nodes, of all the different possible powers of 2, is maintained at all times (So if total memory size is 1 MB, we'd have 20 free lists to track-one for blocks of size 1 byte, 1 for 2 bytes, next for 4 bytes and so on).
When a request for allocation comes, we look for the smallest block bigger than it. If such a block is found on the free list, the allocation is done (say, the request is of 27 KB and the free list tracking 32 KB blocks has at least one element in it), else we traverse the free list _upwards till we find a big enough block. Then we keep splitting it in two blocks-one for adding to the next free list (of smaller size), one to traverse _down the tree till we reach the target and return the requested memory block to the user. If no such allocation is possible, we simply return null.
**Example:
Let us see how the algorithm proceeds by tracking a memory block of size 128 KB. Initially, the free list is: {}, {}, {}, {}, {}, {}, {}, { (0, 127) }
- **Request: 32 bytes
No such block found, so we traverse up and split the 0-127 block into 0-63, 64-127; we add 64-127 to list tracking 64 byte blocks and pass 0-63 downwards; again it is split into 0-31 and 32-63; since we have found the required block size, we add 32-63 to list tracking 32 byte blocks and return 0-31 to user.
List is: {}, {}, {}, {}, {}, { (32, 63) }, { (64, 127) }, {} - **Request: 7 bytes
No such block found-split block 32-63 into two blocks, namely 32-47 and 48-63; then split 32-47 into 32-39 and 40-47; finally, return 32-39 to user (internal fragmentation of 1 byte occurs)
List is: {}, {}, {}, { (40, 47) }, { (48, 63) }, {}, { (64, 127) }, {} - **Request: 64 bytes
Straight up memory segment 64-127 will be allocated as it already exists.
List is: {}, {}, {}, { (40, 47) }, { (48, 63) }, {}, {}, {} - Request: 56 bytes
Result: Not allocated
The result will be as follows:
**Figure - Buddy Allocation-128 shows the starting address of next possible block (if main memory size ever increases)
**Implementation -
C++ `
#include<bits/stdc++.h> using namespace std;
// Size of vector of pairs int size;
// Global vector of pairs to store // address ranges available in free list vector<pair<int, int>> free_list[100000];
// Map used as hash map to store the starting // address as key and size of allocated segment // key as value map<int, int> mp;
void initialize(int sz) {
// Maximum number of powers of 2 possible
int n = ceil(log(sz) / log(2));
size = n + 1;
for(int i = 0; i <= n; i++)
free_list[i].clear();
// Initially whole block of specified
// size is available
free_list[n].push_back(make_pair(0, sz - 1)); }
void allocate(int sz) {
// Calculate index in free list
// to search for block if available
int n = ceil(log(sz) / log(2));
// Block available
if (free_list[n].size() > 0)
{
pair<int, int> temp = free_list[n][0];
// Remove block from free list
free_list[n].erase(free_list[n].begin());
cout << "Memory from " << temp.first
<< " to " << temp.second << " allocated"
<< "\n";
// map starting address with
// size to make deallocating easy
mp[temp.first] = temp.second -
temp.first + 1;
}
else
{
int i;
for(i = n + 1; i < size; i++)
{
// Find block size greater than request
if(free_list[i].size() != 0)
break;
}
// If no such block is found
// i.e., no memory block available
if (i == size)
{
cout << "Sorry, failed to allocate memory \n";
}
// If found
else
{
pair<int, int> temp;
temp = free_list[i][0];
// Remove first block to split it into halves
free_list[i].erase(free_list[i].begin());
i--;
for(; i >= n; i--)
{
// Divide block into two halves
pair<int, int> pair1, pair2;
pair1 = make_pair(temp.first,
temp.first +
(temp.second -
temp.first) / 2);
pair2 = make_pair(temp.first +
(temp.second -
temp.first + 1) / 2,
temp.second);
free_list[i].push_back(pair1);
// Push them in free list
free_list[i].push_back(pair2);
temp = free_list[i][0];
// Remove first free block to
// further split
free_list[i].erase(free_list[i].begin());
}
cout << "Memory from " << temp.first
<< " to " << temp.second
<< " allocated" << "\n";
mp[temp.first] = temp.second -
temp.first + 1;
}
}}
// Driver code int main() {
// Uncomment following code for interactive IO
/*
int total,c,req;
cin>>total;
initialize(total);
while(true)
{
cin>>req;
if(req < 0)
break;
allocate(req);
}*/
initialize(128);
allocate(32);
allocate(7);
allocate(64);
allocate(56);
return 0;}
// This code is contributed by sarthak_eddy
Java
import java.io.; import java.util.;
class Buddy {
// Inner class to store lower
// and upper bounds of the allocated memory
class Pair
{
int lb, ub;
Pair(int a, int b)
{
lb = a;
ub = b;
}
}
// Size of main memory
int size;
// Array to track all
// the free nodes of various sizes
ArrayList<Pair> arr[];
// Else compiler will give warning
// about generic array creation
@SuppressWarnings("unchecked")
Buddy(int s)
{
size = s;
// Gives us all possible powers of 2
int x = (int)Math.ceil(Math.log(s) / Math.log(2));
// One extra element is added
// to simplify arithmetic calculations
arr = new ArrayList[x + 1];
for (int i = 0; i <= x; i++)
arr[i] = new ArrayList<>();
// Initially, only the largest block is free
// and hence is on the free list
arr[x].add(new Pair(0, size - 1));
}
void allocate(int s)
{
// Calculate which free list to search to get the
// smallest block large enough to fit the request
int x = (int)Math.ceil(Math.log(s) / Math.log(2));
int i;
Pair temp = null;
// We already have such a block
if (arr[x].size() > 0)
{
// Remove from free list
// as it will be allocated now
temp = (Pair)arr[x].remove(0);
System.out.println("Memory from " + temp.lb
+ " to " + temp.ub + " allocated");
return;
}
// If not, search for a larger block
for (i = x + 1; i < arr.length; i++) {
if (arr[i].size() == 0)
continue;
// Found a larger block, so break
break;
}
// This would be true if no such block was found
// and array was exhausted
if (i == arr.length)
{
System.out.println("Sorry, failed to allocate memory");
return;
}
// Remove the first block
temp = (Pair)arr[i].remove(0);
i--;
// Traverse down the list
for (; i >= x; i--) {
// Divide the block in two halves
// lower index to half-1
Pair newPair = new Pair(temp.lb, temp.lb
+ (temp.ub - temp.lb) / 2);
// half to upper index
Pair newPair2 = new Pair(temp.lb
+ (temp.ub - temp.lb + 1) / 2, temp.ub);
// Add them to next list
// which is tracking blocks of smaller size
arr[i].add(newPair);
arr[i].add(newPair2);
// Remove a block to continue the downward pass
temp = (Pair)arr[i].remove(0);
}
// Finally inform the user
// of the allocated location in memory
System.out.println("Memory from " + temp.lb
+ " to " + temp.ub + " allocated");
}
public static void main(String args[]) throws IOException
{
int initialMemory = 0, val = 0;
// Uncomment the below section for interactive I/O
/*Scanner sc=new Scanner(System.in);
initialMemory = sc.nextInt();
Buddy obj = new Buddy(initialMemory);
while(true)
{
val = sc.nextInt();// Accept the request
if(val <= 0)
break;
obj.allocate(val);// Proceed to allocate
}*/
initialMemory = 128;
// Initialize the object with main memory size
Buddy obj = new Buddy(initialMemory);
obj.allocate(32);
obj.allocate(7);
obj.allocate(64);
obj.allocate(56);
}}
Python3
import math
Size of vector of pairs
size = 0
Global list of lists to store
address ranges available in free list
free_list = [[] for _ in range(100000)]
Dictionary used as a hash map to store the starting
address as key and size of allocated segment
as the value
mp = {}
def initialize(sz): global size
# Maximum number of powers of 2 possible
n = math.ceil(math.log(sz) / math.log(2))
size = n + 1
for i in range(size):
free_list[i].clear()
# Initially the whole block of specified
# size is available
free_list[n].append((0, sz - 1))def allocate(sz): global size
# Calculate index in free list
# to search for a block if available
n = math.ceil(math.log(sz) / math.log(2))
# Block available
if len(free_list[n]) > 0:
temp = free_list[n][0]
# Remove block from free list
free_list[n].pop(0)
print("Memory from", temp[0], "to", temp[1], "allocated")
# map starting address with
# size to make deallocating easy
mp[temp[0]] = temp[1] - temp[0] + 1
else:
i = n + 1
while i < size and not free_list[i]:
i += 1
# If no such block is found
# i.e., no memory block available
if i == size:
print("Sorry, failed to allocate memory")
else:
temp = free_list[i][0]
# Remove first block to split it into halves
free_list[i].pop(0)
i -= 1
while i >= n:
# Divide block into two halves
pair1 = (temp[0], temp[0] + (temp[1] - temp[0]) // 2)
pair2 = (temp[0] + (temp[1] - temp[0] + 1) // 2, temp[1])
# Push them in free list
free_list[i].append(pair1)
free_list[i].append(pair2)
temp = free_list[i][0]
# Remove first free block to further split
free_list[i].pop(0)
i -= 1
print("Memory from", temp[0], "to", temp[1], "allocated")
mp[temp[0]] = temp[1] - temp[0] + 1Driver code
def main(): # Uncomment the following code for interactive IO ''' total = int(input()) initialize(total) while True: req = int(input()) if req < 0: break allocate(req) '''
initialize(128)
allocate(32)
allocate(7)
allocate(64)
allocate(56)if name == "main": main()
# This code is contributed by utkarshcode1.C#
using System; using System.Collections.Generic;
public class Buddy {
// Inner class to store lower
// and upper bounds of the
// allocated memory
class Pair
{
public int lb, ub;
public Pair(int a, int b)
{
lb = a;
ub = b;
}
}
// Size of main memory
int size;
// Array to track all
// the free nodes of various sizes
List<Pair> []arr;
// Else compiler will give warning
// about generic array creation
Buddy(int s)
{
size = s;
// Gives us all possible powers of 2
int x = (int)Math.Ceiling(Math.Log(s) /
Math.Log(2));
// One extra element is added
// to simplify arithmetic calculations
arr = new List<Pair>[x + 1];
for (int i = 0; i <= x; i++)
arr[i] = new List<Pair>();
// Initially, only the largest block is free
// and hence is on the free list
arr[x].Add(new Pair(0, size - 1));
}
void allocate(int s)
{
// Calculate which free list to search
// to get the smallest block
// large enough to fit the request
int x = (int)Math.Ceiling(Math.Log(s) /
Math.Log(2));
int i;
Pair temp = null;
// We already have such a block
if (arr[x].Count > 0)
{
// Remove from free list
// as it will be allocated now
temp = (Pair)arr[x][0];
arr[x].RemoveAt(0);
Console.WriteLine("Memory from " + temp.lb +
" to " + temp.ub + " allocated");
return;
}
// If not, search for a larger block
for (i = x + 1; i < arr.Length; i++)
{
if (arr[i].Count == 0)
continue;
// Found a larger block, so break
break;
}
// This would be true if no such block
// was found and array was exhausted
if (i == arr.Length)
{
Console.WriteLine("Sorry, failed to" +
" allocate memory");
return;
}
// Remove the first block
temp = (Pair)arr[i][0];
arr[i].RemoveAt(0);
i--;
// Traverse down the list
for (; i >= x; i--)
{
// Divide the block in two halves
// lower index to half-1
Pair newPair = new Pair(temp.lb, temp.lb +
(temp.ub - temp.lb) / 2);
// half to upper index
Pair newPair2 = new Pair(temp.lb + (temp.ub -
temp.lb + 1) / 2, temp.ub);
// Add them to next list which is
// tracking blocks of smaller size
arr[i].Add(newPair);
arr[i].Add(newPair2);
// Remove a block to continue
// the downward pass
temp = (Pair)arr[i][0];
arr[i].RemoveAt(0);
}
// Finally inform the user
// of the allocated location in memory
Console.WriteLine("Memory from " + temp.lb +
" to " + temp.ub + " allocated");
}
// Driver Code
public static void Main(String []args)
{
int initialMemory = 0;
initialMemory = 128;
// Initialize the object with main memory size
Buddy obj = new Buddy(initialMemory);
obj.allocate(32);
obj.allocate(7);
obj.allocate(64);
obj.allocate(56);
}}
// This code is contributed by 29AjayKumar
JavaScript
`
Output
Memory from 0 to 31 allocated Memory from 32 to 39 allocated Memory from 64 to 127 allocated Sorry, failed to allocate memory
**Time Complexity -
If the main memory size is **n, we have log(n) number of different powers of 2 and hence log(n) elements in the array (named **arr in the code) tracking free lists. To allocate a block, we only need to traverse the array once upwards and once downwards, hence time complexity is **O(2log(n)) or simply **O(logn)