So I am trying to code for this question:
Yes, I have to use arrays since it is a requirement.
Consider the problem of adding two n-bit binary integers, stored in two n-element arrays A and B. The sum of the two integers should be stored in binary form in an (n+1) element array C . State the problem formally and write pseudocode for adding the two integers.
I know that the ans array contains the correct output at the end of the addd function. However, I am not able to output that answer.
Below is my code. Please help me figure where in the code I'm going wrong, and what I can do to change it so it works. I will be very grateful.
#include <iostream>
using namespace std;
int * addd(int a[], int n1, int b[], int n2)
{
int s;
if(n1<n2) {s=n2+1;}
else {s=n1+1;}
int ans[s];
int i=n1-1, j=n2-1, k=s-1;
int carry=0;
while(i>=0 && j>=0 && k>0)
{
ans[k]=(a[i]+b[j]+carry)%2;
//cout<<k<<" "<<ans[k]<<endl;
carry=(a[i]+b[j]+carry)/2;
i--; j--; k--;
}
//cout<<"Carry "<<carry<<endl;
ans[0]=carry;
return ans;
}
int main(int argc, const char * argv[]) {
// insert code here...
int a[]={0,0,0,1,1,1};
int n1=sizeof(a)/sizeof(a[0]);
int b[]={1,0,1,1,0,1};
int n2=sizeof(b)/sizeof(b[0]);
int *p=addd(a,6,b,6);
// cout<<p[1]<<endl;
// cout<<p[0]<<" "<<p[1]<<" "<<p[2]<<" "<<p[3]<<" "<<p[4]<<" "<<p[5]<<" "<<p[6]<<endl;
return 0;
}
using namespace std;
Don't write using namespace std;. I have a summary I paste in from a file of common issues when I'm active in the Code Review Stack Exchange, but I don't have that here. Instead, you should just declare the symbols you need, like using std::cout;
int * addd(int a[], int n1, int b[], int n2)
The parameters of the form int a[] are very odd. This comes from C and is actually transformed into int* a and is not passing the array per-se.
The inputs should be const.
The names are not clear, but I'm guessing that n1 is the size of the array? In the Standard Guidelines, you'll see that passing a pointer plus length is strongly discouraged. The Standard Guidelines Library supplies a simple span type to use for this instead.
And the length should be size_t not int.
Based on the description, I think each element is only one bit, right? So why are the arrays of type int? I'd use bool or perhaps int8_t as being easier to work with.
What are you returning? If a and b and their lengths are the input, where is the output that you are returning a pointer to the beginning of? This is not giving value semantics, as you are returning a pointer to something that must exist elsewhere so what is its lifetime?
int s;
int ans[s];
return ans;
Well, there's your problem. First of all, declaring an array of a size that's not a constant is not even legal. (This is a gnu extension that implements C's VLA feature but not without issues as it breaks the C++ type system)
Regardless of that, you are returning a pointer to the first element of the local array, so what happens to the memory when the function returns? Boom.
int s;
No. Initialize values when they are created.
if(n1<n2) {s=n2+1;}
else {s=n1+1;}
Learn the library.
How about:
const size_t s = 1+std::max(n1,n2);
and then the portable way to get your memory is:
std::vector<int> ans(s);
Your main logic will not work if one array is shorter than the other. The shorter input should behave as if it had leading zeros to match. Consider abstracting the problem of "getting the next bit" so you don't duplicate the code for handling each input and make an unreadable mess. You really should have learned to use collections and iterators first.
now:
return ans;
would work as intended since it is a value. You just need to declare the function to be the right type. So just use auto for the return type and it knows.
int n1=sizeof(a)/sizeof(a[0]);
Noooooooo.
There is a standard function to give the size of a built-in primitive array. But really, this should be done automatically as part of the passing, not as a separate thing, as noted earlier.
int *p=addd(a,6,b,6);
You wrote 6 instead of n1 etc.
Anyway, with the previous edits, it becomes:
using std::size;
const auto p = addd (a, size(a), b, size(b));
Finally, concerning:
cout<<p[0]<<" "<<p[1]<<" "<<p[2]<<" "<<p[3]<<" "<<p[4]<<" "<<p[5]<<" "<<p[6]<<endl;
How about using loops?
for (auto val : p) cout << val;
cout << '\n';
oh, don't use endl. It's not needed for cout which auto-flushes anyway, and it's slow. Modern best practice is to use '\n' and then flush explicitly if/when needed (like, never).
Let's look at:
int ans[s];
Apart that this is not even part of the standard and probably the compiler is giving you some warnings (see link), that command allocate temporary memory in the stack which gets deallocated on function exit: that's why you are getting every time different results, you are reading garbage, i.e. memory that in the meantime might have been overwritten.
You can replace it for example with
int* ans = new int[s];
Don't forget though to deallocate the memory when you have finished using the buffer (outside the function), to avoid memory leakage.
Some other notes:
int s;
if(n1<n2) {s=n2+1;}
else {s=n1+1;}
This can be more elegantly written as:
const int s = (n1 < n2) ? n2 + 1 : n1 + 1;
Also, the actual computation code is imprecise as it leads to wrong results if n1 is not equal to n2: You need further code to finish processing the remaining bits of the longest array. By the way you don't need to check on k > 0 because of the way you have defined s.
The following should work:
int i=n1-1, j=n2-1, k=s-1;
int carry=0;
while(i>=0 && j>=0)
{
ans[k]=(a[i]+b[j]+carry)%2;
carry=(a[i]+b[j]+carry)/2;
i--; j--; k--;
}
while(i>=0) {
ans[k]=(a[i]+carry)%2;
carry=(a[i]+carry)/2;
i--; k--;
}
while(j>=0) {
ans[k]=(b[j]+carry)%2;
carry=(b[j]+carry)/2;
j--; k--;
}
ans[0]=carry;
return ans;
}
If You Must Only Use C Arrays
Returning ans is returning the pointer to a local variable. The object the pointer refers to is no longer valid after then function has returned, so trying to read it would lead to undefined behavior.
One way to fix this is to pass in the address to an array to hold your answer, and populate that, instead of using a VLA (which is a non-standard C++ extension).
A VLA (variable length array) is an array which takes its size from a run-time computed value. In your case:
int s;
//... code that initializes s
int ans[s];
ans is a VLA because you are not using a constant to determine the array size. However, that is not a standard feature of the C++ language (it is an optional one in the C language).
You can modify your function so that ans is actually provided by the caller.
int * addd(int a[], int n1, int b[], int n2, int ans[])
{
//...
And then the caller would be responsible for passing in a large enough array to hold the answer.
Your function also appears to be incomplete.
while(i>=0 && j>=0 && k>0)
{
ans[k]=(a[i]+b[j]+carry)%2;
//cout<<k<<" "<<ans[k]<<endl;
carry=(a[i]+b[j]+carry)/2;
i--; j--; k--;
}
If one array is shorter than the other, then the index for the shorter array will reach 0 first. Then, when that corresponding index goes negative, the loop will stop, without handling the remaining terms in the longer array. This essentially makes the corresponding entries in ans be uninitialized. Reading those values results in undefined behavior.
To address this, you should populate the remaining entries in ans with the correct calculation based on carry and the remaining entries in the longer array.
A More C++ Approach
The original answer above was provided assuming you were constrained to only using C style arrays for both input and output, and that you wanted an answer that would allow you to stay close to your original implementation.
Below is a more C++ oriented solution, assuming you still need to provide C arrays as input, but otherwise no other constraint.
C Array Wrapper
A C array does not provide the amenities that you may be accustomed to have when using C++ containers. To gain some of these nice to have features, you can write an adapter that allows a C array to behave like a C++ container.
template <typename T, std::size_t N>
struct c_array_ref {
typedef T ARR_TYPE[N];
ARR_TYPE &arr_;
typedef T * iterator;
typedef std::reverse_iterator<T *> reverse_iterator;
c_array_ref (T (&arr)[N]) : arr_(arr) {}
std::size_t size () { return N; }
T & operator [] (int i) { return arr_[i]; }
operator ARR_TYPE & () { return arr_; }
iterator begin () { return &arr_[0]; }
iterator end () { return begin() + N; }
reverse_iterator rbegin () { return reverse_iterator(end()); }
reverse_iterator rend () { return reverse_iterator(begin()); }
};
Use C Array References
Instead of passing in two arguments as information about the array, you can pass in the array by reference, and use template argument deduction to deduce the array size.
Return a std::array
Although you cannot return a local C array like you attempted in your question, you can return an array that is wrapped inside a struct or class. That is precisely what the convenience container std::array provides. When you use C array references and template argument deduction to obtain the array size, you can now compute at compile time the proper array size that std::array should have for the return value.
template <std::size_t N1, std::size_t N2>
std::array<int, ((N1 < N2) ? N2 : N1) + 1>
addd(int (&a)[N1], int (&b)[N2])
{
Normalize the Input
It is much easier to solve the problem if you assume the arguments have been arranged in a particular order. If you always want the second argument to be the larger array, you can do that with a simple recursive call. This is perfectly safe, since we know the recursion will happen at most once.
if (N2 < N1) return addd(b, a);
Use C++ Containers (or Look-Alike Adapters)
We can now convert our arguments to the adapter shown earlier, and also create a std::array to hold the output.
c_array_ref<int, N1> aa(a);
c_array_ref<int, N2> bb(b);
std::array<int, std::max(N1, N2)+1> ans;
Leverage Existing Algorithms if Possible
In order to deal with the short comings of your original program, you can adjust your implementation a bit in an attempt to remove special cases. One way to do that is to store the result of adding the longer array to 0 and storing it into the output. However, this can mostly be accomplished with a simple call to std::copy.
ans[0] = 0;
std::copy(bb.begin(), bb.end(), ans.begin() + 1);
Since we know the input consists of only 1s and 0s, we can compute straight addition from the shorter array into the longer array, without concern for carry (that will be addressed in the next step). To compute this addition, we apply std::transform with a lambda expression.
std::transform(aa.rbegin(), aa.rend(), ans.rbegin(),
ans.rbegin(),
[](int a, int b) -> int { return a + b; });
Lastly, we can make a pass over the output array to fix up the carry computation. After doing so, we are ready to return the result. The return is possible because we are using std::array to represent the answer.
for (auto i = ans.rbegin(); i != ans.rend()-1; ++i) {
*(i+1) += *i / 2;
*i %= 2;
}
return ans;
}
A Simpler main Function
We now only need to pass in the two arrays to the addd function, since template type deduction will discover the sizes of the arrays. In addition, the output generator can be handled more easily with an ostream_iterator.
int main(int, const char * []) {
int a[]={1,0,0,0,1,1,1};
int b[]={1,0,1,1,0,1};
auto p=addd(a,b);
std::copy(p.begin(), p.end(),
std::ostream_iterator<int>(std::cout, " "));
return 0;
}
Try it online!
If I may editorialize a bit... I think this is a deceptively difficult question for beginners, and as-stated should flag problems in the design review long before any attempt at coding. It's telling you to do things that are not good/typical/idiomatic/proper in C++, and distracting you with issues that get in the way of the actual logic to be developed.
Consider the core algorithm you wrote (and Antonio corrected): that can be understood and discussed without worrying about just how A and B are actually passed in for this code to use, or exactly what kind of collection it is. If they were std::vector, std::array, or primitive C array, the usage would be identical. Likewise, how does one return the result out of the code? You populate ans here, and how it is gotten into and/or out of the code and back to main is not relevant.
Primitive C arrays are not first-class objects in C++ and there are special rules (inherited from C) on how they are passed as arguments.
Returning is even worse, and returning dynamic-sized things was a major headache in C and memory management like this is a major source of bugs and security flaws. What we want is value semantics.
Second, using arrays and subscripts is not idiomatic in C++. You use iterators and abstract over the exact nature of the collection. If you were interested in writing super-efficent back-end code that doesn't itself deal with memory management (it's called by other code that deals with the actual collections involved) it would look like std::merge which is a venerable function that dates back to the early 90's.
template< class InputIt1, class InputIt2, class OutputIt >
OutputIt merge( InputIt1 first1, InputIt1 last1,
InputIt2 first2, InputIt2 last2,
OutputIt d_first );
You can find others with similar signatures, that take two different ranges for input and outputs to a third area. If you write addp exactly like this, you could call it with primitive C arrays of hardcoded size:
int8_t A[] {0,0,0,1,1,1};
int8_t B[] {1,0,1,1,0,1};
int8_t C[ ??? ];
using std::begin; std::end;
addp (begin(A),end(A), begin(B), end(B), begin(C));
Note that it's up to the caller to have prepared an output area large enough, and there's no error checking.
However, the same code can be used with vectors, or even any combination of different container types. This could populate a std::vector as the result by passing an insertion iterator. But in this particular algorithm that's difficult since you're computing it in reverse order.
std::array
Improving upon the situation with primitive C arrays, you could use the std::array class which is exactly the same array but without the strange passing/returning rules. It's actually just a primitive C array inside a wrapping struct. See this documentation: https://en.cppreference.com/w/cpp/container/array
So you could write it as:
using BBBNum1 = std::array<int8_t, 6>
BBBNum1 addp (const BBBNum1& A, const BBBNum1& B) { ... }
The code inside can use A[i] etc. in the same way you are, but it also can get the size via A.size(). The issue here is that the inputs are the same length, and the output is the same as well (not 1 larger). Using templates, it could be written to make the lengths flexible but still only specified at compile time.
std::vector
The vector is like an array but with a run-time length. It's dynamic, and the go-to collection you should reach for in C++.
using BBBNum2 = std::vector<int8_t>
BBBNum2 addp (const BBBNum2& A, const BBBNum2& B) { ... }
Again, the code inside this function can refer to B[j] etc. and use B.size() exactly the same as with the array collection. But now, the size is a run-time property, and can be different for each one.
You would create your result, as in my first post, by giving the size as a constructor argument, and then you can return the vector by-value. Note that the compiler will do this efficiently and not actually have to copy anything if you write:
auto C = addp (A, B);
now for the real work
OK, now that this distraction is at least out of the way, you can worry about actually writing the implementation. I hope you are convinced that using vector instead of a C primitive array does not affect your problem logic or even the (available) syntax of using subscripts. Especially since the problem referred to psudocode, I interpret its use of "array" as "suitable indexable collection" and not specifically the primitive C array type.
The issue of going through 2 sequences together and dealing with differing lengths is actually a general purpose idea. In C++20, the Range library has things that make quick work of this. Older 3rd party libraries exist as well, and you might find it called zip or something like that.
But, let's look at writing it from scratch.
You want to read an item at a time from two inputs, but neatly make it look like they're the same length. You don't want to write the same code three times, or elaborate on the cases where A is shorter or where B may be shorter... just abstract out the idea that they are read together, and if one runs out it provides zeros.
This is its own piece of code that can be applied twice, to A and to B.
class backwards_bit_reader {
const BBBnum2& x;
size_t index;
public:
backwards_bit_reader(const BBBnum2& x) : x{x}, index{x.size()} {}
bool done() const { return index == 0; }
int8_t next()
{
if (done()) return 0; // keep reading infinite leading zeros
--index;
return x[index];
}
};
Now you can write something like:
backwards_bit_reader A_in { A };
backwards_bit_reader B_in { B };
while (!A_in.done() && !B_in.done()) {
const a = A_in.next();
const b = B_in.next();
const c = a+b+carry;
carry = c/2; // update
C[--k]= c%2;
}
C[0]= carry; // the final bit, one longer than the input
It can be written far more compactly, but this is clear.
another approach
The problem is, is writing backwards_bit_reader beyond what you've learned thus far? How else might you apply the same logic to both A and B without duplicating the statements?
You should be learning to recognize what's sometimes called "code smell". Repeating the same block of code multiple times, and repeating the same steps with nothing changed but which variable it's applying to, should be seen as ugly and unacceptable.
You can at least cut back the cases by ensuring that B is always the longer one, if they are of different length. Do this by swapping A and B if that's not the case, as a preliminary step. (Actually implementing that well is another digression)
But the logic is still nearly duplicated, since you have to deal with the possibility of the carry propagating all the way to the end. Just now you have 2 copies instead of 3.
Extending the shorter one, at least in façade, is the only way to write one loop.
how realistic is this problem?
It's simplified to the point of being silly, but if it's not done in base 2 but with larger values, this is actually implementing multi-precision arithmetic, which is a real thing people want to do. That's why I named the type above BBBNum for "Bad Binary Bignum".
Getting down to an actual range of memory and wanting the code to be fast and optimized is also something you want to do sometimes. The BigNum is one example; you often see this with string processing. But we'll want to make an efficient back-end that operates on memory without knowing how it was allocated, and higher-level wrappers that call it.
For example:
void addp (const int8_t* a_begin, const int8_t* a_end,
const int8_t* b_begin, const int8_t* b_end,
int8_t* result_begin, int8_t* result_end);
will use the provided range for output, not knowing or caring how it was allocated, and taking input that's any contiguous range without caring what type of container is used to manage it as long as it's contiguous. Note that as you saw with the std::merge example, it's more idiomatic to pass begin and end rather than begin and size.
But then you have helper functions like:
BBBNum2 addp (const BBBNum2& A, const BBBNum2& B)
{
BBBNum result (1+std::max(A.size(),B.size());
addp (A.data(), A.data()+A.size(), B.data(), B.data()+B.size(), C.data(), C.data()+C.size());
}
Now the casual user can call it using vectors and a dynamically-created result, but it's still available to call for arrays, pre-allocated result buffers, etc.
I thought that a cool way of using vectors could be to have one vector class template hold an two separate int variables for x/y-coordinates to graph.
example:
std::vector<int, int> *name*;
// First int. being the x-intercept on a graph
// Second int. being the y-intercept on a graph
(I also understand that I could just make every even/odd location or two separate vectors to classify each x/y-coordinate, but for me I would just like to see if this could work)
However, after making this vector type, I came across an issue with assigning which int within the vector will be written to or extracted from. Could anyone tell me how to best select and std::cout both x/y ints appropriately?
P.S. - My main goal, in using vectors this way, is to make a very basic graph output to Visual Studio terminal. While being able to change individual x/y-intercepts by 'selecting' and changing if needed. These coordinates will be outputted to the terminal via for/while loops.
Also, would anyone like to list out different ways to best make x/y-coordinates with different containers?
Your question rather broad, in other words it is asking for a bit too much. I will just try to give you some pointers from which you can work your way to what you like.
A) equidistant x
If your x values are equidistant, ie 0, 0.5, 1, 1.5 then there is no need to store them, simply use a
std::vector<int> y;
if the number of variables is not known at compile time, otherwise a
std::array<int,N> y;
B) arbitrary x
There are several options that depend on what you actually want to do. For simply storing (x,y)-pairs and printing them on the screen, they all work equally well.
map
std::map<int,int> map_x_to_y = { { 1,1}, {2,4}, {3,9}};
// print on screen
for (const auto& xy : map_x_to_y) {
std::cout << xy.first << ":" xy.second;
}
a vector of pairs
std::vector<std::pair<int,int>> vector_x_and_y = { { 1,1}, {2,4}, {3,9}};
Printing on screen is actually the same as with map. The advantage of the map is that it has its elements ordered, while this is not the case for the vector.
C) not using any container
For leightweight calculations you can consider to not store the (xy) pairs at all, but simply use a function:
int fun(int x) { return x*x; }
TL;DR / more focussed
A vector stores one type. You cannot have a std::vector<int,int>. If you look at the documentation of std::vector you will find that the second template parameter is an allocator (something you probably dont have to care about for some time). If you want to store two values as one element in a vector you either have to use std::vector<std::pair<double,double>> or a different container.
PS
I used std::pair in the examples above. However, I do consider it as good practice to name things whenever I can and leave std::pair for cases when I simply cannot give names better than first and second. In this spirit you can replace std::pair in the above examples with a
struct data_point {
int x;
int y;
};
I'm working on a basic DFS algorithm for my oriented graph in c++.
What i got, is a class Vertex which have a char value and a list of adjacence.
Given that, i'm trying to sort a list of object (in my case vertex, but let's say something else) by its value.
My class is something like that:
class Foo {
private:
char x;
list<Foo*> listOfChildrenObject;
};
let's say we got 5 Foo's object.
A,B,C,D,E
now, A is the father of every object in my program.
let's say i've inserted them in this way:
A (the father), D, C, B, E
and then, i want to print them in order:
A,B,C,D,E
and for that, i want to use listOfChildrenObject.sort() to do that.
there's a way to sort my list of object by the char value?
Ok, i think i got it.
sorry for the dumb and repeated question, i resolved adding a lambda function to my sort, and changing the list to a vector (i realized that i had no reason to use a list)
in this way:
std::sort(v.begin(), v.end(), [](Foo* a, Foo* b) {return a->getValue() < b->getValue(); });
TL;DR: Help me pass vector array input into a function that originally took in int pointers and ints.
I am in a college class that is learning about well-known algorithms. In this class, we use C++ to code the given algorithms. There was no pre-req class to learn C++, so my knowledge is pretty low when it comes to some of the major stuff with programming.
My problem: I have to create a program that takes an input file, sorts it with the user's choice of sorting algorithm, and write the results to an output file. My original code that works perfectly uses an input file of 20 items, placed into an array of length 20, and sorts no problem with each individual sorting algorithm.
Since last night, the only thing I have changed is that my input goes to a vector array, since the teacher will give us files of varying length (10 items to 1,000,000 items). I have four sorting algorithms that need to sort these given input files. Only one of them works, and it does not pass any variables into the function.
The other 3 originally passed in array pointers and other variables, however they do not work with my new input now going to a vector array instead of an int array. I know that what I am passing in needs to be changed, but I have no idea how to do this correctly. I have tried many different ways from sources found on the internet (with pointers and references), but I have had no luck. Here is some snipets of the code I'm using.
vector<int> A;
void insertionSort() // This works no problem as is.
void split(int* A, int* B, int low, int high) //code for Merge-Sort
{
//recurisvely calls split() and splitMerge()
}
void splitMerge(int* A, int* B, int low, int mid, int high) // more code for Merge-Sort
{
// more code for Merge-Sort
}
//quickSort() and countSort() also pass ints and do not work either.
//small part of main()
for (i = 0; unsorted >> temp; i++)
{
A.push_back(temp);
cout << A[i] << "\n";
length++; //I use int length extensively in for-loops in my functions
}
Last thing. I do not get an error when trying to run the Merge-Sort. It just freezes my console window.
conversion from vector to array is done this way
vector vInts;
...fill the vector
int * intArr=vInts[0]
so you don't need to modify your code too much
I believe there is not enough code to make an accurate prediction on where the error may be, but I think the problem is that the sorting algorithms is doing something like A++ with you pointers to access the next member.
Because arrays store the pointer to the next array inside the object and not on adjacent cells of memory, your algorithms are cycling through stuff they shouldn't.
If this is the case, the solution to your problem is to use an iterator instead of a pointer.
void split(A.begin(), B.begin(), int low, int high)