I am working on learning vectors in my C++ object oriented 1 class and we have been introduced the concept of range based for loops. I decided to practice the range based for-loops separately so that I could get used to the syntax but I came across a weird issue.
#include<iostream>
using namespace std;
int main()
{
int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
for ( auto i: a)
{
cout << a[i] << " ";
}
return 0;
}
When I run the above code my output is the following.
2 3 4 5 6 7 8 9 0 1 Press any key to continue...
My output should read
1 2 3 4 5 6 7 8 9 0 Press any key to continue...
Can anyone tell me why my first index is skipped? I have visual studio 2013 professional.
You get the weird output because i in the range loop is the value from the array, not an index. That is,
for (auto i : a)
loops through the values of a. In your code you're effectively printing the sequence a[a[0]], a[a[1]], etc.
The code you probably want is
for (auto i : a) {
std::cout << i << std::endl;
}
Related
vector<vector<double>> weights
{
{1},
{1}
};
Above is my code to make a 2x1 vector each holding 1.
I would like to make a matrix of 2xN that I could use to multiply with that vector.
I have seen other stackoverflow questions that talk about creating matrices, and most of the ones I've seen are with fixed values, or user input.
But what I would like to do, is initialize the entire first column of N length with 1s, and the initialize the entire second column with a second vector I already have.
I am unsure how in C++ I could accomplish this. I'm way more familiar with R, and in R this is a pretty simple task. Any thoughts or guidance?
You mean like this?
std::vector<int> vinner {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
};
std::vector<std::vector<int>> v {
std::vector<int>(10, 1),
vinner
};
int main(int argc, char **argv)
{
for (auto i : v) {
for (auto j : i) {
std::cout << j << " ";
}
std::cout << "\n";
}
return 0;
}
Output:
$ clang++ -o vect vect.cpp -std=c++17
$ ./vect
1 1 1 1 1 1 1 1 1 1
1 2 3 4 5 6 7 8 9 10
an int vector is given, and I need to find the maximum sum of number sequences that are more than 3 positions away from each other. So an example:
Input: 10, 5, 6, 7, 16, 18, 12
Output: 29
Explanation:
candidate "routes":
10-7-12 = 29
10-16 = 26
10-18 = 28
5-16 = 21
5-18 = 23
5-12 = 17
...
16
18
12
I want to solve this problem using recursion and I have gotten to the point where the program tests all paths with numbers that are more than 3 positions away. However, it fails to test all the paths starting from the same digit.
Here my code:
#include <iostream>
using namespace std;
const int n = 7;
int sum = 0;
int temp=0;
void rec(int (*arr), int ind){
if (ind > n-3){
if(temp > sum)
sum = temp;
temp=0;
return;
}
temp += arr[ind];
cout<<arr[ind]<<" ";
for(int i=3;i<n;i++)
rec(arr, ind+i);
cout<<endl;
}
int main(){
int arr[n] = {10, 9, 9, 9, 8, 8, 7};
for(int i=0;i<n;i++)
rec(arr, i);
cout<<endl<<endl<<"Max: "<<sum;
return 0;}
Output:
10 9 7
8
8
7
9 8
8
7
9 8
7
9 7
8
8
7
Max: 26
Even though in this case the output is correct, you can see that it tests only 10-9-7 and not the other 10-8, 10-8 and 10-7. It seems it doesn´t even go in the first for-loop I called the function recursively.
Am I understanding something wrong or how do I accomplish that it tests all pairs that are more than 3 positions away?
There are 3 problems here:
Based on your output:
10 9 7
8
8
⋮
Your code actually loops through all the possible routes. Instead of thinking your second line 8 and third line 8 as two complete routes, they were actually derived from your first rout 10 9 7. So really think of it like this:
10 --- 9 --- 7
`- 8
`- 8
`- 7
9 --- 8
`- 8
`- 7
⋮
That's actually not your output. Plugged your code in, I got my output:
10 9
8
9 8
9
9
8
Max: 19
The reason that it didn't go through all routes is because your line:
if (ind > n-3)
{
⋮
return;
}
You are terminating your loop way too early. Instead, you should only end it right before it goes out of index. So it should be:
if (ind > n - 1)
{
⋮
}
Your logic actually has flaws.
Because you always set your temp back to 0 once reached the last possible number inside the array, you actually lost some part of your previous sums. You can easily create a sequence that makes it fails, such as: 10, 9, 9, 7, 8, 20, 7.
Instead of having temp as a global, it would be better to set it local, and being passed to each recursions.
To do that, you would need to add the temp as a parameter of your rec function, and call it like:
void rec(int (*arr), int ind, int temp)
{
⋮
rec(arr, i, temp);
⋮
}
int main()
{
⋮
rec(arr, i, 0);
⋮
}
Or if you want to keep the signature the same in your main, you could also add a default argument in your rec:
void rec(int (*arr), int ind, int temp = 0)
Side note, I would considering using std::array or std::vector instead of c-style array, to avoid having a global size n or passing int (*arr) in your function. Plus you can easily make your arr as a longer sequence.
Suppose I want to remove the unique elements from an std::vector (not get rid of the duplicates, but retain only the elements that occur at least 2 times) and I want to achieve that in a pretty inefficient way - by calling std::count while std::remove_ifing. Consider the following code:
#include <algorithm>
#include <iostream>
#include <vector>
int main() {
std::vector<int> vec = {1, 2, 6, 3, 6, 2, 7, 4, 4, 5, 6};
auto to_remove = std::remove_if(vec.begin(), vec.end(), [&vec](int n) {
return std::count(vec.begin(), vec.end(), n) == 1;
});
vec.erase(to_remove, vec.end());
for (int i : vec) std::cout << i << ' ';
}
From reference on std::remove_if we know that the elements beginning from to_remove have unspecified values, but I wonder how unspecified they can really be.
To explain my concern a little further - we can see that the elements that should be removed are 1, 3, 5 and 7 - the only unique values. std::remove_if will move the 1 to the end but there is no guarantee that there will be a value 1 at the end after said operation. Can this be (due to that value being unspecified) that it will turn into 3 and make the std::count call return a count of (for example) 2 for the later encountered value 3?
Essentially my question is - is this guaranteed to work, and by work I mean to inefficiently erase unique elements from an std::vector?
I am interested in both language-lawyer answer (which could be "the standard says that this situation is possible, you should avoid it") and in-practice answer (which could be "the standard says that this situation is possible, but realistically there is no way of this value ending up as a completely differeny one, for example 3").
After the predicate returns true the first time, there will be one unspecified value in the range. That means any subsequent calls of the predicate will count an unspecified value. The count is therefore potentially incorrect, and you may either leave values unaffected that you intend to be discarded, or discard values that should be retained.
You could modify the predicate so it keeps a count of how many times it has returned true, and reduce the range accordingly. For example;
std::size_t count = 0;
auto to_remove = std::remove_if(vec.begin(), vec.end(), [&vec, &count](int n)
{
bool once = (std::count(vec.begin(), vec.end() - count, n) == 1);
if (once) ++count;
return once;
});
Subtracting an integral value from a vector's end iterator is safe, but that isn't necessarily true for other containers.
You misunderstood how std::remove_if works. The to-be-removed values are not necessarily shifted to the end. See:
Removing is done by shifting (by means of move assignment) the elements in the range in such a way that the elements that are not to be removed appear in the beginning of the range. cppreference
This is the only guarantee for the state of the range. According to my knowledge, it's not forbidden to shift all values around and it would still satisfy the complexity. So it might be possible that some compilers shift the unwanted values to the end but that would be just extra unnecessary work.
An example of possible implementation of removing odd numbers from 1 2 3 4 8 5:
v - read position
1 2 3 4 8 5 - X will denotes shifted from value = unspecified
^ - write position
v
1 2 3 4 8 5 1 is odd, ++read
^
v
2 X 3 4 8 5 2 is even, *write=move(*read), ++both
^
v
2 X 3 4 8 5 3 is odd, ++read
^
v
2 4 3 X 8 5 4 is even, *write=move(*read), ++both
^
v
2 4 8 X X 5 8 is even, *write=move(*read), ++both
^
2 4 8 X X 5 5 is odd, ++read
^ - this points to the new end.
So, in general, you cannot rely on count returning any meaningful values. Since in the case that move==copy (as is for ints) the resulting array is 2 4 8|4 8 5. Which has incorrect count both for the odd and even numbers. In case of std::unique_ptr the X==nullptr and thus the count for nullptr and removed values might be wrong. Other remaining values should not be left in the end part of the array as there were no copies done.
Note that the values are not unspecified as in you cannot know them. They are exactly the results of move assignments which might leave the value in unspecified state. If it specified the state of the moved-from variables ( asstd::unique_ptr does) then they would be known. E.g. if move==swap then the range will be permuted only.
I added some outputs:
#include <algorithm>
#include <iostream>
#include <vector>
#include <mutex>
int main() {
std::vector<int> vec = {1, 2, 6, 3, 6, 2, 7, 4, 4, 5, 6};
auto to_remove = std::remove_if(vec.begin(), vec.end(), [&vec](int n) {
std::cout << "number " << n << ": ";
for (auto i : vec) std::cout << i << ' ';
auto c = std::count(vec.begin(), vec.end(), n);
std::cout << ", count: " << c << std::endl;
return c == 1;
});
vec.erase(to_remove, vec.end());
for (int i : vec) std::cout << i << ' ';
}
and got
number 1: 1 2 6 3 6 2 7 4 4 5 6 , count: 1
number 2: 1 2 6 3 6 2 7 4 4 5 6 , count: 2
number 6: 2 2 6 3 6 2 7 4 4 5 6 , count: 3
number 3: 2 6 6 3 6 2 7 4 4 5 6 , count: 1
number 6: 2 6 6 3 6 2 7 4 4 5 6 , count: 4
number 2: 2 6 6 3 6 2 7 4 4 5 6 , count: 2
number 7: 2 6 6 2 6 2 7 4 4 5 6 , count: 1
number 4: 2 6 6 2 6 2 7 4 4 5 6 , count: 2
number 4: 2 6 6 2 4 2 7 4 4 5 6 , count: 3
number 5: 2 6 6 2 4 4 7 4 4 5 6 , count: 1
number 6: 2 6 6 2 4 4 7 4 4 5 6 , count: 3
2 6 6 2 4 4 6
As you can see the counts can be wrong. I'm not able to create an example for your special case but as a rule you have to worry about wrong results.
First the number 4 is counted twice and in the next step the number 4 is counted thrice. The counts are wrong and you can't rely on them.
Can anyone explain this code working?It is to find index of 2 elements in vector that add to produce the given target.I don't understand how STL works in this.
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
map<int,int>m;
vector<int>v;
if(nums.size()==0)
{
return v;
}
for(int i=0;i<nums.size();i++)
{
if(m.find(nums[i])==m.end())
{
m[target-nums[i]]=i+1;
}
else
{
v.push_back(m[nums[i]]);
v.push_back(i+1);
}
}
return v;
}
};
It's pretty easy. Say the vector is { 8, 4, 3, 2, 5 } and the target is 10. First number you find is 8, so now you know that you are looking for a 2 (because 8 + 2 is 10). So you add the new target 2 and the index of 8 (which is 1 because the indexes are 1 based) to the map. Next number is 4 so now you are looking for 6, so 6 and the index 2 get added to the map. Now the map looks like this
2 ==> 1
6 ==> 2
Eventually you'll find one of the targets in the map (in this case we'll find a two), the map gives us the index of the original 8, and we know the index of the 2 because we've just found it, so we can output both indexes.
The code could be improved, but I think it works.
I am trying to get union of 4 arrays using set_union. Here is the code I have so far:
int setA[5] = {2, 4, 5, 7, 8};
int setB[7] = {1, 2, 3, 4, 5, 6, 7};
int setC[5] = {2, 5, 8, 8, 15};
int setD[6] = {1, 4, 4, 6, 7, 12};
int AunionB[12];
int CunionD[11];
int finalUnion[23];
int *lastAunionB;
int *lastCunionD;
ostream_iterator<int> screen(cout, " ");
lastAunionB = set_union(setA, setA+5, setB, setB+7, AunionB);
cout << "AunionB = ";
copy(AunionB, lastAunionB, screen);
cout << endl;
lastCunionD = set_union(setC, setC+5, setD, setD+6, CunionD);
cout << "CunionD = ";
copy(CunionD, lastCunionD, screen);
cout << endl;
set_union(AunionB, AunionB+12, CunionD, CunionD+11, finalUnion);
cout << "Final Union = ";
copy(finalUnion, finalUnion+23, screen);
cout << endl;
When I ran the code, I got the following output:
AunionB = 1 2 3 4 5 6 7 8
CunionD = 1 2 4 4 5 6 7 8 8 12 15
Final Union = 1 2 3 4 5 6 7 2 4 4 5 6 7 8 8 12 15 52187240 1 1863041424 32767 0 0
Therefore, the unions of setA and setB works as intended as does the union of setC and setD. However, when I try to get the union of all for sets, it doesn't work! I'm guessing the last 5 values of finalUnion are the address fields but how do I remove them? Also, the union itself is incorrect and I can't understand why.
The size of the AunionB and Cuniond is not 12 and 11 because:
Elements from the second range that have an equivalent element in the first range are not copied to the resulting range.
Try this code:
int setA[5] = { 2, 4, 5, 7, 8 };
int setB[7] = { 1, 2, 3, 4, 5, 6, 7 };
int setC[5] = { 2, 5, 8, 8, 15 };
int setD[6] = { 1, 4, 4, 6, 7, 12 };
int AunionB[12];
int CunionD[11];
int finalUnion[23];
int *lastAunionB;
int *lastCunionD;
ostream_iterator<int> screen(cout, " ");
lastAunionB = set_union(setA, setA + 5, setB, setB + 7, AunionB);
cout << "AunionB = ";
copy(AunionB, lastAunionB, screen);
cout << endl;
lastCunionD = set_union(setC, setC + 5, setD, setD + 6, CunionD);
cout << "CunionD = ";
copy(CunionD, lastCunionD, screen);
cout << endl;
int *finalUnionEnd;
finalUnionEnd = set_union(AunionB, lastAunionB, CunionD, lastCunionD, finalUnion);
cout << "Final Union = ";
copy(finalUnion, finalUnionEnd, screen);
cout << endl;
And then you got the right result:
Final Union = 1 2 3 4 4 5 6 7 8 8 12 15
A Union operation removes values that the two sets have in common.
Note that AUnionB has 8 elements (not the 12 that your code predicts).
You need to adjust your union-of-union code to account for the actual size of the two initial unions. You have everything prepared to do it correctly:
int *lastFinalUnion = set_union(AunionB, lastAunionB, CunionD, lastCunionD, finalUnion);
Note that set C has two distinct occurrences of 8 and set D has two distinct occurrences of 4, which is why they appear duplicated in the intermediate result.
UPDATE
Also, I tried your code and i'm getting the answer as 1 2 3 4 5 6 7 2 4 4 5 6 7 8 8 12 15 . Shouldn't the answer be 1 2 3 4 4 5 6 7 8 8 12 15
I believe you are correct, but I'm not in front of a C++ compiler to step through and see what's doing on, or to verify your output. The actual code was edited in by another SO member, but it looks correct to me.
In the simplest case, set_union performs the "union" operation from set theory: the output range contains a copy of every element that is contained in [first1, last1), [first2, last2), or both. The general case is more complicated, because the input ranges may contain duplicate elements. The generalization is that if a value appears m times in [first1, last1) and n times in [first2, last2) (where m or n may be zero), then it appears max(m,n) times in the output range.
https://www.sgi.com/tech/stl/set_union.html