i am trying to write a program for my university project the program is supposed to do first come first serve scheduling i have thought a lot about this function but i don't know how to make it work, i always get Segmentation fault: 11, i also tried to use temp.at(j) but it gave me Segmentation fault: 6, and i tried to minimise the vector so it would be in-bound by declaring the vectors outside the function, then use temp.size() instead of Processes but it also did't work.
void FCFS(Process ProcessDetails[], int Processes)
{
vector<int> temp;
vector<int> temp1;
int first = 0; //first value to compare with.
for(int j = 0; j < Processes; j++){ // to make sure that it passes through all elements.
for(int i = 0; i < Processes; i++){ // pass each PID and Burst time to vector temp and temp1.
if(ProcessDetails[i].ArrivalTime == first){
temp.operator[](j) = ProcessDetails[i].PID;
temp1.operator[](j) = ProcessDetails[i].BurstTime;
}
}
first++;// increase first value to declare the round is finished and start a new one.
}
for(int i = 0; i < Processes; i++){ // pass the sorted vector values back to the arrays.
ProcessDetails[i].PID = temp.operator[](i);
ProcessDetails[i].BurstTime = temp1.operator[](i);
}
}
the program works fine until it reaches this function, please help.
The behaviour of a vector's operator[]() is undefined if it is used to access elements that do not exist.
Since you have used default-constructed vectors, their size is zero - so they have no elements to access.
If you use the .at() member function, it will check the index and throw an exception (of type std::out_of_range, which is declared in the standard header <stdexcept>) when indices are invalid. You can confirm that by wrapping the code in an appropriate try/catch block.
To eliminate the problem, you need to reize the vector (e.g. add elements to it using push_back(), resize it using resize(), etc) before using operator[](). And ensure the index is valid, since operator[]() does not resize a std::vector.
Also, temp[j] is equivalent to temp.operator[](j). For types that supply an operator[]() function, the compiler handles turning expressions like temp[j] into a call of temp.operator[](j).
Your vectors have no elements.
Using the vector operator[] will therefore fail.
Use push_back, emplace, resize, or some other function to add elements to the vectors.
you will have to change your vector assignment to
if(ProcessDetails[i].ArrivalTime == first){
temp.push_back(ProcessDetails[i].PID);
temp1.push_back(ProcessDetails[i].BurstTime);
}
Related
I'm seeing strange behaviour in the output of a 'simple' function that generates the binary representation of an integer. I aim to use the function in another code, but first want to see it working by printing to the console. I first define the function dec2bin to take a number and the length of the required bitstring. After defining a vector that will be returned by the function, the vector is populated with the required binary values. In the main function, the vector is accessed and printed element-wise to the console. However, the output is incorrect unless I include an arbitrary cout statement within the vector populating for loop.
The function is defined as follows:
vector<int> dec2bin(int N, int j){
vector<int> myvec;
myvec.push_back(N);
for (int i = N-1; i >= 0; i--) {
int r = j >> i;
if (r & 1)
myvec[N-i-1] = 1;
else
myvec[N-i-1] = 0;
//cout << " ";
}
return myvec;
}
My main function is:
int main(){
int length = 8;
int num = 15;
vector<int> myvec = dec2bin(length,num);
for (int j=0; j<length; j++){cout << myvec[j];}
return 0;
}
When the commented out line (cout << " ") is included within the for loop the output is correct, with 00001111. Without that line (or with the line outside of the for loop) the output is incorrectly 00001141130.
I have read various things about pointers and memory allocation, so expect this is where I am going wrong; however, I can't quite see how to incorporate those ideas into my own code. Any advice would be greatly appreciated, many thanks.
On this line:
myvec[N-i-1] = 1;
You are invoking undefined behavior for any index other than 0.
myvec.push_back(N); pushes a single element to the vector, and you never add more elements to the vector. When you access a vector out of bounds, you will not get a compiler error or runtime exception, but your code has undefined behavior, which means the output could be literally anything.
Actually, it seems like you just confused push_back() with resize(). And instead of resize(), you can pass the size to the constructor. If you replace:
vector<int> myvec;
myvec.push_back(N);
with:
std::vector<int> myvec(N);
or:
std::vector<int> myvec;
myvec.resize(N);
You get a vector with N elements, and accessing myvec[i] is fine for any index 0 <= i < N. Alternatively, you could call myvec.reserve(10) to let the vector allocate space for N elements. The vector will then still be empty, and you have to push_back() any element you want to add.
You are accessing the unallocated memory because you only pushed back(push_back()) once and due to it you are getting undefined behaviour.
You should use .at() function instead of [] operator for accessing the vector element, it'll throw the exception out of bounds.
You need to use push_back() for insertion into the vector or you can use vector.resize(size) during vector declaration then it'll save you from undefined behavior.
Thanx
I was writing a program in C++ where I need to have a 2d grid of pointers which point to objects which are stored in a vector. I tested out some part of the program and saw strange results in the output.
I changed the objects to integers and removed everything non-essential to cut it down to the code snippet below, but I still get a weird output.
vector<vector<int*>> lattice(10, vector<int*>(10));//grid of pointers
vector<int> relevant;//vector carrying actual values onto which pointers will point
for(int i = 0; i<10; i++){
int new_integer = i;
relevant.push_back(new_integer);//insert integer into vector
lattice[0][i] = &relevant[i];//let pointer point onto this value
}
//OUTPUT
for(int j = 0; j<10; j++){
cout<<*lattice[0][j]<<" ";
cout<<relevant[j]<<endl;
}
I get strange outputs like this:
19349144 0
19374040 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
Also, my output changes from run to run and depending on how big/small I make my grid.
I would expect all the values on the left and right to be equal, I guess there is something fundamental about pointers I haven't understood, so sorry if this is a very basic question.
Can someone please explain why I get strange outputs for some values of the grid?
I need to have a 2d grid of pointers which point to objects which are stored in a vector
That's impossible. Or rather, that's a guarantee for dereferencing invalid addresses. Why?
At any given time, an std::vector has enough space allocated for some limited number of elements. If you keep adding elements to it, it will eventually max out its storage. At some insertion, it will decide allocate a new stretch of memory to use for storing its data; move (or copy) existing data to the new storage area; free the old storage area; and then be able to add more elements.
When this happens, all existing pointers into objects in the vector become invalid. The memory they point to may continue to hold the previous values, but may also be used to store other data - there are no guarantees on that! In fact, dereferencing invalid pointers officially results in undefined behavior.
... and I do see that what I've described is exactly what you're doing with your code. Your older pointers become invalid.
Instead, consider keeping indices into the vector rather than pointers. Indices don't get invalidated by adding elements to the vector, and you can keep using them.
PS - I see you're using a vector-of-vectors. That's technically valid, but it is often unadvisable. Consider using a matrix class (e.g. from the Eigen library) or allocating a certain amount of memory using std::make_unique(), then using it to initialize a gsl::multi_span.
relevant.push_back invalidates all pointers/references/iterators to its elements (if the new size exceeds its current capacity).
Therefore you are dereferencing potentially invalid pointers when you do
*lattice[0][j]
later.
You can use a container that doesn't invalidate on insertion at the end, such as std::list or std::deque, instead of std::vector, for relevant.
(Or you can reserve a sufficient capacity with a call to .reserve first, so that the size on the .push_back operations will never exceed the capacity and will therefore never invalidate pointers, but that carries the risk of easily accidentally ignoring this requirement on later code changes, again causing UB.)
When a std::vector is inserted into, if its new size() will exceed its current capacity(), the vector has to reallocate its internal array to make room, which will invalidate any existing iterators and pointers to the old memory.
In your example, you can avoid that by reserve()'ing the capacity() ahead of time, eg:
vector<vector<int*>> lattice(10, vector<int*>(10));//grid of pointers
vector<int> relevant;//vector carrying actual values onto which pointers will point
//ADD THIS!
relevant.reserve(10);//pre-allocate the capacity
for(int i = 0; i<10; i++){
int new_integer = i;
relevant.push_back(new_integer);//insert integer into vector
lattice[0][i] = &relevant[i];//let pointer point onto this value
}
//OUTPUT
for(int j = 0; j<10; j++){
cout<<*lattice[0][j]<<" ";
cout<<relevant[j]<<endl;
}
Alternatively, you can pre-allocate the size() and then use operator[] instead of push_back(), eg:
vector<vector<int*>> lattice(10, vector<int*>(10));//grid of pointers
vector<int> relevant(10);//vector carrying actual values onto which pointers will point
for(int i = 0; i<10; i++){
int new_integer = i;
relevant[i] = new_integer;//insert integer into vector
lattice[0][i] = &relevant[i];//let pointer point onto this value
}
//OUTPUT
for(int j = 0; j<10; j++){
cout<<*lattice[0][j]<<" ";
cout<<relevant[j]<<endl;
}
I have to copy the first size element from a set of Solution (a class) named population to an array of solution named parents. I have some problems with iterators because i should do an hybrid solution between a normal for loop
and a for with iterators. The idea is this: when I'm at the ith iteration of the for I declare a new iterator that's pointing the beginning
of population, then I advance this iterator to the ith position, I take this solution element and I copy into parents[i]
Solution* parents; //it is filled somewhere else
std::set<Solution> population; //it is filled somewhere else
for (int i = 0; i < size; i++) {
auto it = population.begin();
advance(it, i);
parents[i] = *it;
}
Two error messages popup with this sentence: 'Expression: cannot dereference end map/set iterator'
and 'Expression: cannot advance end map/set iterator'
Any idea on how to this trick? I know it's kinda bad mixing array and set, i should use vector instead of array?
You use std::copy_n.
#include <algorithm>
extern Solution* parents; //it is filled somewhere else
extern std::set<Solution> population; //it is filled somewhere else
std::copy_n(population.begin(), size, parents);
It seems like size may be incorrectly set. To ensure that your code behaves as expected, you should just use the collection's size directly:
auto it = population.begin();
for (int i = 0; i < population.size(); i++) {
parents[i] = *it;
++it;
}
This can also be solved with a much simpler expression:
std::copy(population.begin(), population.end(), parents);
I have to copy the first size element from a set [..] to an array
You can use std::copy_n.
for (int i = 0; i < size; i++) {
auto it = population.begin();
advance(it, i);
The problem with this is that you're iterating over the linked list in every iteration. This turns the copy operation from normally linear complexity to quadratic.
Expression: cannot dereference end map/set iterator'
The problem here appears to be that your set doesn't contain at least size number of elements. You cannot copy size number of elements if there aren't that many. I suggest that you would copy less elements when the set is smaller.
i should use vector instead of array?
Probably. Is the array very large? Is the size of the vector not known at compile time? If so, use a vector.
I'm still quite inexperienced in C++ and i'm trying to write sum code to add numbers precisely. This is a dll plugin for some finite difference software and the code is called several million times during a run. I want to write a function where any number of arguments can be passed in and the sum will be returned. My code looks like:
#include <cstdarg>
double SumFunction(int numArgs, ...){ // this allows me to pass any number
// of arguments to my function.
va_list args;
va_start(args,numArgs); //necessary prerequisites for using cstdarg
double myarray[10];
for (int i = 0; i < numArgs; i++) {
myarray[i] = va_arg(args,double);
} // I imagine this is sloppy code; however i cannot create
// myarray{numArgs] because numArgs is not a const int.
sum(myarray); // The actual method of addition is not relevant here, but
//for more complicated methods, I need to put the summation
// terms in a list.
vector<double> vec(numArgs); // instead, place all values in a vector
for (int i = 0; i < numArgs; i++) {
vec.at(i) = va_arg(args,double);
}
sum(vec); //This would be passed by reference, of course. The function sum
// doesn't actually exist, it would all be contained within the
// current function. This is method is twice as slow as placing
//all the values in the static array.
double *vec;
vec = new double[numArgs];
for (int i = 0; i < (numArgs); i++) {
vec[i] = va_arg(args,double);
}
sum(vec); // Again half of the speed of using a standard array and
// increasing in magnitude for every extra dynamic array!
delete[] vec;
va_end(args);
}
So the problem I have is that using an oversized static array is sloppy programming, but using either a vector or a dynamic array slows the program down considerably. So I really don't know what to do. Can anyone help, please?
One way to speed the code up (at the cost of making it more complicated) is to reuse a dynamic array or vector between calls, then you will avoid incurring the overhead of memory allocation and deallocation each time you call the function.
For example declare these variables outside your function either as global variables or as member variables inside some class. I'll just make them globals for ease of explanation:
double* sumArray = NULL;
int sumArraySize = 0;
In your SumFunction, check if the array exists and if not allocate it, and resize if necessary:
double SumFunction(int numArgs, ...){ // this allows me to pass any number
// of arguments to my function.
va_list args;
va_start(args,numArgs); //necessary prerequisites for using cstdarg
// if the array has already been allocated, check if it is large enough and delete if not:
if((sumArray != NULL) && (numArgs > sumArraySize))
{
delete[] sumArray;
sumArray = NULL;
}
// allocate the array, but only if necessary:
if(sumArray == NULL)
{
sumArray = new double[numArgs];
sumArraySize = numArgs;
}
double *vec = sumArray; // set to your array, reusable between calls
for (int i = 0; i < (numArgs); i++) {
vec[i] = va_arg(args,double);
}
sum(vec, numArgs); // you will need to pass the array size
va_end(args);
// note no array deallocation
}
The catch is that you need to remember to deallocate the array at some point by calling a function similar to this (like I said, you pay for speed with extra complexity):
void freeSumArray()
{
if(sumArray != NULL)
{
delete[] sumArray;
sumArray = NULL;
sumArraySize = 0;
}
}
You can take a similar (and simpler/cleaner) approach with a vector, allocate it the first time if it doesn't already exist, or call resize() on it with numArgs if it does.
When using a std::vector the optimizer must consider that relocation is possible and this introduces an extra indirection.
In other words the code for
v[index] += value;
where v is for example a std::vector<int> is expanded to
int *p = v._begin + index;
*p += value;
i.e. from vector you need first to get the field _begin (that contains where the content starts in memory), then apply the index, and then dereference to get the value and mutate it.
If the code performing the computation on the elements of the vector in a loop calls any unknown non-inlined code, the optimizer is forced to assume that unknown code may mutate the _begin field of the vector and this will require doing the two-steps indirection for each element.
(NOTE: that the vector is passed with a cost std::vector<T>& reference is totally irrelevant: a const reference doesn't mean that the vector is const but simply puts a limitation on what operations are permitted using that reference; external code could have a non-const reference to access the vector and constness can also be legally casted away... constness of references is basically ignored by the optimizer).
One way to remove this extra lookup (if you know that the vector is not being resized during the computation) is to cache this address in a local and use that instead of the vector operator [] to access the element:
int *p = &v[0];
for (int i=0,n=v.size(); i<n; i++) {
/// use p[i] instead of v[i]
}
This will generate code that is almost as efficient as a static array because, given that the address of p is not published, nothing in the body of the loop can change it and the value p can be assumed constant (something that cannot be done for v._begin as the optimizer cannot know if someone else knows the address of _begin).
I'm saying "almost" because a static array only requires indexing, while using a dynamically allocated area requires "base + indexing" access; most CPUs however provide this kind of memory access at no extra cost. Moreover if you're processing elements in sequence the indexing addressing becomes just a sequential memory access but only if you can assume the start address constant (i.e. not in the case of std::vector<T>::operator[]).
Assuming that the "max storage ever needed" is in the order of 10-50, I'd say using a local array is perfectly fine.
Using vector<T> will use 3 * sizeof(*T) (at least) to track the contents of the vector. So if we compare that to an array of double arr[10];, then that's 7 elements more on the stack of equal size (or 8.5 in 32-bit build). But you also need a call to new, which takes a size argument. So that takes up AT LEAST one, more likely 2-3 elements of stackspace, and the implementation of new is quite possibly not straightforward, so further calls are needed, which take up further stack-space.
If you "don't know" the number of elements, and need to cope with quite large numbers of elements, then using a hybrid solution, where you have a small stack-based local array, and if numargs > small_size use vector, and then pass vec.data() to the function sum.
I'm new to programming and I was wondering, how to get the size of an array, that is, get the size of how many elements are inside the array. For example if I declare an array of size 10, but only input 3 elements into the array, how would I determine the size of this array? If I don't know how many elements I placed in initially.
If you declare an array, e.g. int array[10], then its size is always 10 * sizeof(int). There is no way to know how many times you've accessed it; you'd need to keep track of that manually.
You should consider using container classes, e.g. std::vector:
std::vector<int> vec;
vec.push_back(5);
vec.push_back(10);
vec.push_back(42);
std::cout << vec.size() << "\n"; // Prints "3"
If you declare an old-style array of 10 elements, e.g. std::string words[10], the size of the array is always 10 strings. Even with the new style (std::array), it would be a fixed size.
You might be looking for a std::vector<>. This doesn't have a fixed size, but does have a .size() method. Therefore, if you add three elements to it, it will have .size()==3
to get the array size (in number of elements) assuming you do not know it in advance
use sizeof(a)/sizeof(a[0])
see the below example program. I used C but it should carry over to C++ just fine
#include <stdio.h>
int main(){
int a[10];
printf("%d elements\n",sizeof(a)/sizeof(a[0]));
return 0;
}
//output: 10 elements
There's several possible ways, but they depend on your definition.
If you know there is a value the user won't input (also known as a sentinel value), you can use a function like memset, to set the entire array to that unused value. You would then iterate through the list counting all the variables in the list that don't match that value.
The other way is to build your own array class, which counts whenever the array is modified (you'd have to overload the = and [] functions as appropriate).
You could also build a dynamically linked list, so as the user adds variables, the count can either be determined by walking the list or keeping count.
But, if you're taking the array as the basic array, with no idea as to it's actual starting state, and no idea what to expect from the user (given this is your program, this shouldn't occur), then generally speaking, no, there is known way to know this.
You maintain a counter variable count initialized to 0.
Whenever you are adding to array increment the count by 1.
Whenever you are removing from array decrement the count by 1.
anytime count will give you the size of the array.
Suggestion:
int[10] arr;
//init all to null
for (int i =0; i < 10; i++)
arr[i] = 0;
arr[0]=1;
arr[2]=5;
int sz = 0;
for (int j = 0; j < 10; j++)
if (arr[j] != 0) sz++;