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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 seen weird for loop syntax in C++. Please see following program.
#include <iostream>
using namespace std;
int main()
{
int num, count = 0;
int array[]= {1,1,2,3,4,1,3,2,9,8,7};
cout<<"Which number would you like to know about how many times it appeared?"<<endl;
cin>>num;
for (int i:array)
{
if (i == num)
++count;
}
cout<<"Number "<<num<<" appeared "<<count<<" times in the array"<<endl;
return 0;
}
It is successfully run on GCC Linux platform.
Reference link Here.
So, My question is, Is it the correct way to use for loop in C++?
Also, Is for (int i:array) equivalent to for ( int i:array ; ; )?
Sorry for my poor english. Thanks in advance.
There are now (since C++11) two distinct syntaxes for for-loops, the old C-style:
for (initialization; test; increment)
and the new
for (declaration: container)
In the new syntax, the declaration declares a variable which is successively given the value of each element of the container. Common values of "declaration" are auto val, const auto val, auto& val, and const auto& val, depending on whether you want a reference to the value in the container or a copy, and whether you want the value constant or not.
Both syntaxes are correct. It rather depends what you want to do in the loop. My preference is to use the range based for unless I am going to need the loop counter or iterator ... in which case I fall back on the old style for.
See http://en.cppreference.com/w/cpp/language/range-for for the gory details of the specification (and what is meant by "container").
The syntax for (int i:array) iterates through each element in the array, compared to for (int i = 0; i<sizeof(array); i++) which creates a counter that automatically increments on each iteration of the loop. The counter can then be used to access elements of the array with array[i]
As for which one you'd use, it depends on what you want to do. In your example there isn't a need to keep track of which iteration of the loop you are on, so the former will work fine. If you wanted to, say, print the iteration number each time then you would use the latter.
P.S. your English is perfect :)
I had an algorithm that started out like
int sumLargest2 ( int * arr, size_t n )
{
int largest(max(arr[0], arr[1])), secondLargest(min(arr[0],arr[1]));
// ...
and I realized that the first is probably not optimal because calling max and then min is repetitious when you consider that the information required to know the minimum is already there once you've found the maximum. So I figured out that I could do
int largest = max(arr[0], arr[1]);
int secondLargest = arr[0] == largest ? arr[1] : arr[0];
to shave off the useless invocation of min, but I'm not sure that actually saves any number of operations. Are there any fancy bit-shifting algorithms that can do the equivalent of
int largest(max(arr[0], arr[1])), secondLargest(min(arr[0],arr[1]));
?????
In C++, you can use std::minmax to produce a std::pair of the minimum and the maximum. This is particularly easy in combination with std::tie:
#include <algorithm>
#include <utility>
int largest, secondLargest;
std::tie(secondLargest, largest) = std::minmax(arr[0], arr[1]);
GCC, at least, is capable of optimizing the call to minmax into a single comparison, identical to the result of the C code below.
In C, you could write the test out yourself:
int largest, secondLargest;
if (arr[0] < arr[1]) {
largest = arr[1];
secondLargest = arr[0];
} else {
largest = arr[0];
secondLargest = arr[1];
}
How about:
int largestIndex = arr[1] > arr[0];
int largest = arr[largestIndex];
int secondLargest = arr[1 - largestIndex];
The first line relies on an implicit cast of a boolean result to 1 in the case of true and 0 in the case of false.
I'm going to assume that you'd rather solve the larger problem... That is, getting the sum of the largest two numbers in an array.
What you are trying to do is a std::partial_sort().
Let's implement it.
int sumLargest2(int * arr, size_t n) {
int * first = arr;
int * middle = arr + 2;
int * last = arr + n;
std::partial_sort(first, middle, last, std::greater<int>());
return arr[0] + arr[1];
}
And if you're unable to modify arr, then I'd recommend looking into std::partial_sort_copy().
x = max(a, b);
y = a + b - x;
It won't necessarily be faster, but it will be different.
Also beware of overflows.
If your intention is to reduce the function call to find min mad max you can try std::minmax_element. This is available since C++11.
auto result = std::minmax_element(arr, arr+n);
std::cout<< "min:"<< *result.first<<"\n";
std::cout<< "max :" <<*result.second << "\n";
If you just want to find the bigger of two values go:
if(a > b)
{
largest = a;
second = b;
}
else
{
largest = b;
second = a;
}
No function calls, one comparison, two assignments.
I'm assuming C++...
Short answer, use std::minmax and compile with the right optimizations and the right instruction set parameters.
Long ugly answer, The compiler cannot make all the assumptions necessary to make it really, really fast. You can. In this case, you can change the algorithm to process all data first and you can force alignment on the data. Doing all this, you can use intrinsics to make it faster.
Although I haven't tested it in this particular case, I've seen enormous performance improvements using these guidelines.
Since you're not passing 2 integers to the function, I'm assuming your using an array and want to iterate it somehow. You now have a choice to make: make 2 arrays and use min/max or use 1 array with both a and b. This decision alone can already influence the performance.
If you have 2 arrays, these can be allocated on 32-byte boundaries with aligned malloc's and then processed using intrinsics. If you are going for real, raw performance - this is the way to go.
F.ex, let's assume you have AVX2. (NOTE: I'm not sure if you do and you SHOULD check this using CPU id's!). Go to the cheat sheet here: https://software.intel.com/sites/landingpage/IntrinsicsGuide/ and pick your poison.
The intrinsics you're looking for are in this case probably:
_mm256_min_epi32
_mm256_max_epi32
_mm256_stream_load_si256
If you have to do this for the entire array, you probably want to keep all the stuff in a single __mm256 register before merging the individual items. E.g.: do a min/max per 256-bit vector, and when the loop is done, extract the 32-bit items and do a min/max on that.
Long nicer answer: So ... as for the compiler. Compilers do attempt to optimize these kinds of things, but run into problems.
If you have 2 different arrays that you process, the compiler has to know that they are different in order to be able to optimize it. This is the reason why stuff like restrict exists, which tells the compiler exactly this little thing you probably already knew while writing the code.
Also, the compiler doesn't know your memory is aligned, so it has to check this and branch... for each call. We don't want this; which means we want it to inline its stuff. So, add inline, put it in a header file and that's that. You can also use aligned to give him a hint.
Your compiler also didn't get the hint that the int* won't change over time. If it cannot change, it's a good idea to tell him that using the const keyword.
A compiler uses an instruction set to do the compilation. Normally, they already use SSE, but AVX2 can help a lot (as I've shown with the intrinsics above). If you can compile it with those flags, make sure to use them - they help a lot.
Run in release mode, compile with optimizations on 'fast' and see what happens under the hood. If you do all this, you should see vpmax... instructions appearing in the inner loops, which means that the compiler uses the intrinsics just fine.
I don't know what else you want to do in the loop... if you use all these instructions you should hit the memory speed on big arrays.
How about a time-space trade-off?
#include <utility>
template<typename T>
std::pair<T, T>
minmax(T const& a, T const& b)
{ return b < a ? std::make_pair(b, a) : std::make_pair(a, b); }
//main
std::pair<int, int> values = minmax(a[0], a[1]);
int largest = values.second;
int secondLargest = values.first;
I think this is a very basic question but I couldn't just figure it out.
I was used to using arrays in C++ but I'm now starting to learn vectors.
I was making a test code, and I came across a question.
First of all, here's the code I made:
#include <iostream>
#include <vector>
#include <numeric>
using namespace std;
int main(){
vector<double> score(10);
for(vector<double>::size_type i=0;i<20;i++) {
cout<<"Enter marks for student #"<<i+1<<":"<<flush;
cin>>score[i];
}
double total = accumulate(score.begin(), score.end(),0);
cout<<"Total score:"<<total<<endl<<"Average score:"<<total/score.size()<<flush;
return 0;
}
In the for sentence in line #9, I am declaring i as a vector<double>::size_type type (because I was told to do so).
I tested the code with the type said above replaced with an int, and it worked perfectly fine.
Why is vector<double>::size_type preferred compared to int?
size_type is guaranteed to be large enough for the largest supported vector size, vector::max_size(). int is not: on many common platforms, int has 32 bits, while max_size() is considerably larger than 231.
If you know the size is (and will always be) a small number like 20, then you can get away with using int or any other integer type instead of size_type. If you were to change the program, for example to read the size from the input, then it would go horribly wrong if that value were larger than INT_MAX; while using size_type, it would continue working for any value up to max_size(), which you can easily test for.
The embedded type vector<double>::size_type relates to the return value for various vector methods, such as .size(), hence the preference in that case to use matching types (the int = 0 would generally result in a sign mismatch warning). E.g.
for (vector<double>::size_type i = 0; i < score.size(); ++i) { // types match
// ...
}
for (int i = 0; i < score.size(); ++i) { // types mismatch
// ...
}
std::vector::size_type is required to be large enough to represent the maximum number of elements that could be contained in a container, in this case vector<double>::max_size(). In general, it maps to size_t.
In your case, there is no explicit reason to use the vector<double>::size_type (although it would technically be better to use size_type) since the loop runs from 0 to 20 and both are int. Hence the following would be ok.
for (int i = 0; i < 20; ++i) { // loops 20 times
// ...
}
Additional notes:
Favour iterator based loops over index based ones:
for (vector<double>::iterator i = score.begin(); i != score.end(); ++i) {
// ...
}
It wasn't tagged as C++11, but if that is possible, the range based for loops deals with a lot of this.
for (double& i : score) {
// ...
}
Or even using a for_each with a lambda.
The vector size_type is what's used by vector to do size comparisons. If you had a loop that used an int as a counter and compared against the vector's actual size, you'd get warnings from the compiler about signed vs. unsigned integer comparisons:
for( int i=0; i<score.size(); i++ ){ // <-- signed vs. unsigned comparisons
// do something...
}
Your problem is two fold.
First, you are writing beyond the end of the std::vector -- the std::vector has 10 elements, and you are writing to 20.
In order to fix this, and to follow the Don't Repeat Yourself principle, you'd change your code as follows:
int main(){
std::vector<double> score(20);
for(std::vector<double>::size_type i=0;i<score.size();i++) {
where I both made the vector larger and used its size to determine how much to write.
Now, when we try to replace that long clause with int:
int main(){
std::vector<double> score(20);
for(int i=0;i<score.size();i++) {
we get a singed/unsigned comparison (probably):
i<score.size()
where score.size() is an unsigned value of type std::vector<double>::size_type, and i is an int.
Compilers often give warnings in these cases, as it is really easy to get nonsensical results (if i < 0, the comparison will usually result in the negative number comparing larger than the positive one!) In addition, if the size of the vector is larger than the max value of int (on some systems, as small as 32767, usually at least 2147483647, and sometimes much larger -- this is a value the compiler is reasonably free to pick for itself, the C++ standard does not specify it fully), the loop will fail spectacularly.
Now, while the type of std::vector<double>::size() is std::vector<double>::size_type, this is (in every implementation I've experienced) just std::size_t. So that is a shorter way to say it:
for(std::size_t i=0;i<score.size();i++) {
(std::size_t is defined in <cstddef> as an aside).
If you are programming in C++11, you can do one better:
int i=0;
for(double& student:score) {
std::cout<<"Enter marks for student #"<<++i<<":"<<std::flush;
std::cin>>student;
}
which is a range-based for loop and avoid the problem of indexing entirely.
This might seem like a weird question, but how would I create a C++ function that tells whether a given C++ function that takes as a parameter a variable of type X and returns a variable of type X, is injective in the space of machine representation of those variables, i.e. never returns the same variable for two different variables passed to it?
(For those of you who weren't Math majors, maybe check out this page if you're still confused about the definition of injective: http://en.wikipedia.org/wiki/Injective_function)
For instance, the function
double square(double x) { return x*x};
is not injective since square(2.0) = square(-2.0),
but the function
double cube(double x) { return x*x*x};
is, obviously.
The goal is to create a function
template <typename T>
bool is_injective(T(*foo)(T))
{
/* Create a set std::set<T> retVals;
For each element x of type T:
if x is in retVals, return false;
if x is not in retVals, add it to retVals;
Return true if we made it through the above loop.
*/
}
I think I can implement that procedure except that I'm not sure how to iterate through every element of type T. How do I accomplish that?
Also, what problems might arise in trying to create such a function?
You need to test every possible bit pattern of length sizeof(T).
There was a widely circulated blog post about this topic recently: There are Only Four Billion Floats - So Test Them All!
In that post, the author was able to test all 32-bit floats in 90 seconds. Turns out that would take a few centuries for 64-bit values.
So this is only possible with small input types.
Multiple inputs, structs, or anything with pointers are going to get impossible fast.
BTW, even with 32-bit values you will probably exhaust system memory trying to store all the output values in a std::set, because std::set uses a lot of extra memory for pointers. Instead, you should use a bitmap that's big enough to hold all 2^sizeof(T) output values. The specialized std::vector<bool> should work. That will take 2^sizeof(T) / 8 bytes of memory.
Maybe what you need is std::numeric_limits. To store the results, you may use an unordered_map (from std if you're using C++11, or from boost if you're not).
You can check the limits of the data types, maybe something like this might work (it's a dumb solution, but it may get you started):
template <typename T>
bool is_injective(T(*foo)(T))
{
std::unordered_map<T, T> hash_table;
T min = std::numeric_limits<T>::min();
T max = std::numeric_limits<T>::max();
for(T it = min; i < max; ++i)
{
auto result = hash_table.emplace(it, foo(it));
if(result.second == false)
{
return false;
}
}
return true;
}
Of course, you may want to restrict a few of the possible data types. Otherwise, if you check for floats, doubles or long integers, it'll get very intensive.
but the function
double cube(double x) { return x*x*x};
is, obviously.
It is obviously not. There are 2^53 more double values representable in [0..0.5) than in [0..0.125).
As far as I know, you cannot iterate all possible values of a type in C++.
But, even if you could, that approach would get you nowhere. If your type is a 64 bit integer, you might have to iterate through 2^64 values and keep track of the result for all of them, which is not possible.
Like other people said, there is no solution for a generic type X.