Intersection of two sorted vectors in C++
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
Intersection of two sorted vectors in C++ - can this be written any better?
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
int l = 0, r = 0;
while(l < nums1.size() && r < nums2.size()){
int left = nums1[l], right = nums2[r];
if(left == right){
result.push_back(right);
while(l < nums1.size() && nums1[l] == left )l++;
while(r < nums2.size() && nums2[r] == right )r++;
continue;
}
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
}
return result;
}
c++ reinventing-the-wheel
$endgroup$
add a comment |
$begingroup$
Intersection of two sorted vectors in C++ - can this be written any better?
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
int l = 0, r = 0;
while(l < nums1.size() && r < nums2.size()){
int left = nums1[l], right = nums2[r];
if(left == right){
result.push_back(right);
while(l < nums1.size() && nums1[l] == left )l++;
while(r < nums2.size() && nums2[r] == right )r++;
continue;
}
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
}
return result;
}
c++ reinventing-the-wheel
$endgroup$
4
$begingroup$
Do you know aboutstd::set_intersection()
? Reference and example implementations: en.cppreference.com/w/cpp/algorithm/set_intersection
$endgroup$
– user673679
13 hours ago
2
$begingroup$
@user673679 yes I did, and didn't want to use it;
$endgroup$
– Rick
13 hours ago
add a comment |
$begingroup$
Intersection of two sorted vectors in C++ - can this be written any better?
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
int l = 0, r = 0;
while(l < nums1.size() && r < nums2.size()){
int left = nums1[l], right = nums2[r];
if(left == right){
result.push_back(right);
while(l < nums1.size() && nums1[l] == left )l++;
while(r < nums2.size() && nums2[r] == right )r++;
continue;
}
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
}
return result;
}
c++ reinventing-the-wheel
$endgroup$
Intersection of two sorted vectors in C++ - can this be written any better?
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
int l = 0, r = 0;
while(l < nums1.size() && r < nums2.size()){
int left = nums1[l], right = nums2[r];
if(left == right){
result.push_back(right);
while(l < nums1.size() && nums1[l] == left )l++;
while(r < nums2.size() && nums2[r] == right )r++;
continue;
}
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
}
return result;
}
c++ reinventing-the-wheel
c++ reinventing-the-wheel
edited 2 hours ago
Peter Mortensen
25417
25417
asked 15 hours ago
RickRick
308112
308112
4
$begingroup$
Do you know aboutstd::set_intersection()
? Reference and example implementations: en.cppreference.com/w/cpp/algorithm/set_intersection
$endgroup$
– user673679
13 hours ago
2
$begingroup$
@user673679 yes I did, and didn't want to use it;
$endgroup$
– Rick
13 hours ago
add a comment |
4
$begingroup$
Do you know aboutstd::set_intersection()
? Reference and example implementations: en.cppreference.com/w/cpp/algorithm/set_intersection
$endgroup$
– user673679
13 hours ago
2
$begingroup$
@user673679 yes I did, and didn't want to use it;
$endgroup$
– Rick
13 hours ago
4
4
$begingroup$
Do you know about
std::set_intersection()
? Reference and example implementations: en.cppreference.com/w/cpp/algorithm/set_intersection$endgroup$
– user673679
13 hours ago
$begingroup$
Do you know about
std::set_intersection()
? Reference and example implementations: en.cppreference.com/w/cpp/algorithm/set_intersection$endgroup$
– user673679
13 hours ago
2
2
$begingroup$
@user673679 yes I did, and didn't want to use it;
$endgroup$
– Rick
13 hours ago
$begingroup$
@user673679 yes I did, and didn't want to use it;
$endgroup$
– Rick
13 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Indentation
Your indentation is not consistent. This makes the code hard to read and maintain. It should be fixed so you don't give other people headaches.
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
That is basically unreadable giberish (opinion of Martin).
Using namespace
std;
is super bad
This is mention in nearly every C++ review. There is a large article on the subject here: Why is “using namespace std” considered bad practice?. The second answer is the best in my opinion (Martin) see
Multiple declarations in one is bad (thanks to terrible syntax binding rules)
The one declaration per line has been written about adnausium in best practice guides. Please for the sake of your reader declare one variable per line with its own exact type.
The syntax binding rules alluded to above is:
int* x, y; // Here x is int* and y in int
// confusing to a reader. Did you really mean to make y an int?
// Avoid this problem be declaring one variable per line
Typically, functions like this would be based on iterators to work on any container
Here your code is limited to only using vectors. But the algorithm you are using could be used by any container type with only small modifications. As a result your function could provide much more utility being written to use iterators.
The standard library was written such that iterators are the glue between algorithms and container.
It would be a lot simpler, if not necessarily more efficient at runtime, to just use some hash sets.
- This function could be generic in T rather than assuming
int
. - The repeated conditions make me feel like there's simplification waiting here, although exactly what that is eludes me in the two minutes I'm spending on this.
- Should take by
const
ref, not ref, so that you can operate on temporaries.
$endgroup$
1
$begingroup$
You caught the problems and I voted you up, but you could improve your answer by explaining what the issue for the first 4 bullet items.
$endgroup$
– pacmaninbw
14 hours ago
2
$begingroup$
@pacmaninbw: Added some context.
$endgroup$
– Martin York
11 hours ago
add a comment |
$begingroup$
I invite you to review @DeadMG's answer.
Rewriting following (most of) his advice, you'd get something like:
#include <cassert>
#include <algorithm>
#include <vector>
std::vector<T> intersection(std::vector<T> const& left_vector, std::vector<T> const& right_vector) {
auto left = left_vector.begin();
auto left_end = left_vector.end();
auto right = right_vector.begin();
auto right_end = right_vector.end();
assert(std::is_sorted(left, left_end));
assert(std::is_sorted(right, right_end));
std::vector<T> result;
while (left != left_end && right != right_end) {
if (*left == *right) {
result.push_back(*left);
++left;
++right;
continue;
}
if (*left < *right) {
++left;
continue;
}
assert(*left > *right);
++right;
}
return result;
}
I've always found taking pairs of iterators awkward, so I would not recommend such an interface. Instead, you could take simply take any "iterable", they need not even have the same value type, so long as they are comparable:
template <typename Left, typename Right>
std::vector<typename Left::value_type> intersection(Left const& left_c, Right const& right_c);
Also, note that I've included some assert
to validate the pre-conditions of the methods (the collections must be sorted) as well as internal invariants (if *left
is neither equal nor strictly less than *right
then it must be strictly greater).
I encourage you to use assert
liberally:
- They document intentions: pre-conditions, invariants, etc...
- They check that those intentions hold.
Documentation & Bug detection rolled in one, with no run-time (Release) cost.
$endgroup$
1
$begingroup$
Why don't you post that as its own questions. There are some improvements even if we don't move to iterators like using template template types.
$endgroup$
– Martin York
11 hours ago
1
$begingroup$
@MartinYork: I generally don't find "template template" to be an improvement, they're quite awkward to use, and tend to constrain the inputs more than intended.
$endgroup$
– Matthieu M.
10 hours ago
$begingroup$
@MatthieuM. You should reconsider the solution you posted.
$endgroup$
– Rick
5 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Indentation
Your indentation is not consistent. This makes the code hard to read and maintain. It should be fixed so you don't give other people headaches.
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
That is basically unreadable giberish (opinion of Martin).
Using namespace
std;
is super bad
This is mention in nearly every C++ review. There is a large article on the subject here: Why is “using namespace std” considered bad practice?. The second answer is the best in my opinion (Martin) see
Multiple declarations in one is bad (thanks to terrible syntax binding rules)
The one declaration per line has been written about adnausium in best practice guides. Please for the sake of your reader declare one variable per line with its own exact type.
The syntax binding rules alluded to above is:
int* x, y; // Here x is int* and y in int
// confusing to a reader. Did you really mean to make y an int?
// Avoid this problem be declaring one variable per line
Typically, functions like this would be based on iterators to work on any container
Here your code is limited to only using vectors. But the algorithm you are using could be used by any container type with only small modifications. As a result your function could provide much more utility being written to use iterators.
The standard library was written such that iterators are the glue between algorithms and container.
It would be a lot simpler, if not necessarily more efficient at runtime, to just use some hash sets.
- This function could be generic in T rather than assuming
int
. - The repeated conditions make me feel like there's simplification waiting here, although exactly what that is eludes me in the two minutes I'm spending on this.
- Should take by
const
ref, not ref, so that you can operate on temporaries.
$endgroup$
1
$begingroup$
You caught the problems and I voted you up, but you could improve your answer by explaining what the issue for the first 4 bullet items.
$endgroup$
– pacmaninbw
14 hours ago
2
$begingroup$
@pacmaninbw: Added some context.
$endgroup$
– Martin York
11 hours ago
add a comment |
$begingroup$
Indentation
Your indentation is not consistent. This makes the code hard to read and maintain. It should be fixed so you don't give other people headaches.
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
That is basically unreadable giberish (opinion of Martin).
Using namespace
std;
is super bad
This is mention in nearly every C++ review. There is a large article on the subject here: Why is “using namespace std” considered bad practice?. The second answer is the best in my opinion (Martin) see
Multiple declarations in one is bad (thanks to terrible syntax binding rules)
The one declaration per line has been written about adnausium in best practice guides. Please for the sake of your reader declare one variable per line with its own exact type.
The syntax binding rules alluded to above is:
int* x, y; // Here x is int* and y in int
// confusing to a reader. Did you really mean to make y an int?
// Avoid this problem be declaring one variable per line
Typically, functions like this would be based on iterators to work on any container
Here your code is limited to only using vectors. But the algorithm you are using could be used by any container type with only small modifications. As a result your function could provide much more utility being written to use iterators.
The standard library was written such that iterators are the glue between algorithms and container.
It would be a lot simpler, if not necessarily more efficient at runtime, to just use some hash sets.
- This function could be generic in T rather than assuming
int
. - The repeated conditions make me feel like there's simplification waiting here, although exactly what that is eludes me in the two minutes I'm spending on this.
- Should take by
const
ref, not ref, so that you can operate on temporaries.
$endgroup$
1
$begingroup$
You caught the problems and I voted you up, but you could improve your answer by explaining what the issue for the first 4 bullet items.
$endgroup$
– pacmaninbw
14 hours ago
2
$begingroup$
@pacmaninbw: Added some context.
$endgroup$
– Martin York
11 hours ago
add a comment |
$begingroup$
Indentation
Your indentation is not consistent. This makes the code hard to read and maintain. It should be fixed so you don't give other people headaches.
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
That is basically unreadable giberish (opinion of Martin).
Using namespace
std;
is super bad
This is mention in nearly every C++ review. There is a large article on the subject here: Why is “using namespace std” considered bad practice?. The second answer is the best in my opinion (Martin) see
Multiple declarations in one is bad (thanks to terrible syntax binding rules)
The one declaration per line has been written about adnausium in best practice guides. Please for the sake of your reader declare one variable per line with its own exact type.
The syntax binding rules alluded to above is:
int* x, y; // Here x is int* and y in int
// confusing to a reader. Did you really mean to make y an int?
// Avoid this problem be declaring one variable per line
Typically, functions like this would be based on iterators to work on any container
Here your code is limited to only using vectors. But the algorithm you are using could be used by any container type with only small modifications. As a result your function could provide much more utility being written to use iterators.
The standard library was written such that iterators are the glue between algorithms and container.
It would be a lot simpler, if not necessarily more efficient at runtime, to just use some hash sets.
- This function could be generic in T rather than assuming
int
. - The repeated conditions make me feel like there's simplification waiting here, although exactly what that is eludes me in the two minutes I'm spending on this.
- Should take by
const
ref, not ref, so that you can operate on temporaries.
$endgroup$
Indentation
Your indentation is not consistent. This makes the code hard to read and maintain. It should be fixed so you don't give other people headaches.
if(left < right){
while(l < nums1.size() && nums1[l] == left )l++;
}else while( r < nums2.size() && nums2[r] == right )r++;
That is basically unreadable giberish (opinion of Martin).
Using namespace
std;
is super bad
This is mention in nearly every C++ review. There is a large article on the subject here: Why is “using namespace std” considered bad practice?. The second answer is the best in my opinion (Martin) see
Multiple declarations in one is bad (thanks to terrible syntax binding rules)
The one declaration per line has been written about adnausium in best practice guides. Please for the sake of your reader declare one variable per line with its own exact type.
The syntax binding rules alluded to above is:
int* x, y; // Here x is int* and y in int
// confusing to a reader. Did you really mean to make y an int?
// Avoid this problem be declaring one variable per line
Typically, functions like this would be based on iterators to work on any container
Here your code is limited to only using vectors. But the algorithm you are using could be used by any container type with only small modifications. As a result your function could provide much more utility being written to use iterators.
The standard library was written such that iterators are the glue between algorithms and container.
It would be a lot simpler, if not necessarily more efficient at runtime, to just use some hash sets.
- This function could be generic in T rather than assuming
int
. - The repeated conditions make me feel like there's simplification waiting here, although exactly what that is eludes me in the two minutes I'm spending on this.
- Should take by
const
ref, not ref, so that you can operate on temporaries.
edited 1 hour ago
Peter Mortensen
25417
25417
answered 15 hours ago
DeadMGDeadMG
759612
759612
1
$begingroup$
You caught the problems and I voted you up, but you could improve your answer by explaining what the issue for the first 4 bullet items.
$endgroup$
– pacmaninbw
14 hours ago
2
$begingroup$
@pacmaninbw: Added some context.
$endgroup$
– Martin York
11 hours ago
add a comment |
1
$begingroup$
You caught the problems and I voted you up, but you could improve your answer by explaining what the issue for the first 4 bullet items.
$endgroup$
– pacmaninbw
14 hours ago
2
$begingroup$
@pacmaninbw: Added some context.
$endgroup$
– Martin York
11 hours ago
1
1
$begingroup$
You caught the problems and I voted you up, but you could improve your answer by explaining what the issue for the first 4 bullet items.
$endgroup$
– pacmaninbw
14 hours ago
$begingroup$
You caught the problems and I voted you up, but you could improve your answer by explaining what the issue for the first 4 bullet items.
$endgroup$
– pacmaninbw
14 hours ago
2
2
$begingroup$
@pacmaninbw: Added some context.
$endgroup$
– Martin York
11 hours ago
$begingroup$
@pacmaninbw: Added some context.
$endgroup$
– Martin York
11 hours ago
add a comment |
$begingroup$
I invite you to review @DeadMG's answer.
Rewriting following (most of) his advice, you'd get something like:
#include <cassert>
#include <algorithm>
#include <vector>
std::vector<T> intersection(std::vector<T> const& left_vector, std::vector<T> const& right_vector) {
auto left = left_vector.begin();
auto left_end = left_vector.end();
auto right = right_vector.begin();
auto right_end = right_vector.end();
assert(std::is_sorted(left, left_end));
assert(std::is_sorted(right, right_end));
std::vector<T> result;
while (left != left_end && right != right_end) {
if (*left == *right) {
result.push_back(*left);
++left;
++right;
continue;
}
if (*left < *right) {
++left;
continue;
}
assert(*left > *right);
++right;
}
return result;
}
I've always found taking pairs of iterators awkward, so I would not recommend such an interface. Instead, you could take simply take any "iterable", they need not even have the same value type, so long as they are comparable:
template <typename Left, typename Right>
std::vector<typename Left::value_type> intersection(Left const& left_c, Right const& right_c);
Also, note that I've included some assert
to validate the pre-conditions of the methods (the collections must be sorted) as well as internal invariants (if *left
is neither equal nor strictly less than *right
then it must be strictly greater).
I encourage you to use assert
liberally:
- They document intentions: pre-conditions, invariants, etc...
- They check that those intentions hold.
Documentation & Bug detection rolled in one, with no run-time (Release) cost.
$endgroup$
1
$begingroup$
Why don't you post that as its own questions. There are some improvements even if we don't move to iterators like using template template types.
$endgroup$
– Martin York
11 hours ago
1
$begingroup$
@MartinYork: I generally don't find "template template" to be an improvement, they're quite awkward to use, and tend to constrain the inputs more than intended.
$endgroup$
– Matthieu M.
10 hours ago
$begingroup$
@MatthieuM. You should reconsider the solution you posted.
$endgroup$
– Rick
5 hours ago
add a comment |
$begingroup$
I invite you to review @DeadMG's answer.
Rewriting following (most of) his advice, you'd get something like:
#include <cassert>
#include <algorithm>
#include <vector>
std::vector<T> intersection(std::vector<T> const& left_vector, std::vector<T> const& right_vector) {
auto left = left_vector.begin();
auto left_end = left_vector.end();
auto right = right_vector.begin();
auto right_end = right_vector.end();
assert(std::is_sorted(left, left_end));
assert(std::is_sorted(right, right_end));
std::vector<T> result;
while (left != left_end && right != right_end) {
if (*left == *right) {
result.push_back(*left);
++left;
++right;
continue;
}
if (*left < *right) {
++left;
continue;
}
assert(*left > *right);
++right;
}
return result;
}
I've always found taking pairs of iterators awkward, so I would not recommend such an interface. Instead, you could take simply take any "iterable", they need not even have the same value type, so long as they are comparable:
template <typename Left, typename Right>
std::vector<typename Left::value_type> intersection(Left const& left_c, Right const& right_c);
Also, note that I've included some assert
to validate the pre-conditions of the methods (the collections must be sorted) as well as internal invariants (if *left
is neither equal nor strictly less than *right
then it must be strictly greater).
I encourage you to use assert
liberally:
- They document intentions: pre-conditions, invariants, etc...
- They check that those intentions hold.
Documentation & Bug detection rolled in one, with no run-time (Release) cost.
$endgroup$
1
$begingroup$
Why don't you post that as its own questions. There are some improvements even if we don't move to iterators like using template template types.
$endgroup$
– Martin York
11 hours ago
1
$begingroup$
@MartinYork: I generally don't find "template template" to be an improvement, they're quite awkward to use, and tend to constrain the inputs more than intended.
$endgroup$
– Matthieu M.
10 hours ago
$begingroup$
@MatthieuM. You should reconsider the solution you posted.
$endgroup$
– Rick
5 hours ago
add a comment |
$begingroup$
I invite you to review @DeadMG's answer.
Rewriting following (most of) his advice, you'd get something like:
#include <cassert>
#include <algorithm>
#include <vector>
std::vector<T> intersection(std::vector<T> const& left_vector, std::vector<T> const& right_vector) {
auto left = left_vector.begin();
auto left_end = left_vector.end();
auto right = right_vector.begin();
auto right_end = right_vector.end();
assert(std::is_sorted(left, left_end));
assert(std::is_sorted(right, right_end));
std::vector<T> result;
while (left != left_end && right != right_end) {
if (*left == *right) {
result.push_back(*left);
++left;
++right;
continue;
}
if (*left < *right) {
++left;
continue;
}
assert(*left > *right);
++right;
}
return result;
}
I've always found taking pairs of iterators awkward, so I would not recommend such an interface. Instead, you could take simply take any "iterable", they need not even have the same value type, so long as they are comparable:
template <typename Left, typename Right>
std::vector<typename Left::value_type> intersection(Left const& left_c, Right const& right_c);
Also, note that I've included some assert
to validate the pre-conditions of the methods (the collections must be sorted) as well as internal invariants (if *left
is neither equal nor strictly less than *right
then it must be strictly greater).
I encourage you to use assert
liberally:
- They document intentions: pre-conditions, invariants, etc...
- They check that those intentions hold.
Documentation & Bug detection rolled in one, with no run-time (Release) cost.
$endgroup$
I invite you to review @DeadMG's answer.
Rewriting following (most of) his advice, you'd get something like:
#include <cassert>
#include <algorithm>
#include <vector>
std::vector<T> intersection(std::vector<T> const& left_vector, std::vector<T> const& right_vector) {
auto left = left_vector.begin();
auto left_end = left_vector.end();
auto right = right_vector.begin();
auto right_end = right_vector.end();
assert(std::is_sorted(left, left_end));
assert(std::is_sorted(right, right_end));
std::vector<T> result;
while (left != left_end && right != right_end) {
if (*left == *right) {
result.push_back(*left);
++left;
++right;
continue;
}
if (*left < *right) {
++left;
continue;
}
assert(*left > *right);
++right;
}
return result;
}
I've always found taking pairs of iterators awkward, so I would not recommend such an interface. Instead, you could take simply take any "iterable", they need not even have the same value type, so long as they are comparable:
template <typename Left, typename Right>
std::vector<typename Left::value_type> intersection(Left const& left_c, Right const& right_c);
Also, note that I've included some assert
to validate the pre-conditions of the methods (the collections must be sorted) as well as internal invariants (if *left
is neither equal nor strictly less than *right
then it must be strictly greater).
I encourage you to use assert
liberally:
- They document intentions: pre-conditions, invariants, etc...
- They check that those intentions hold.
Documentation & Bug detection rolled in one, with no run-time (Release) cost.
answered 11 hours ago
Matthieu M.Matthieu M.
2,1871810
2,1871810
1
$begingroup$
Why don't you post that as its own questions. There are some improvements even if we don't move to iterators like using template template types.
$endgroup$
– Martin York
11 hours ago
1
$begingroup$
@MartinYork: I generally don't find "template template" to be an improvement, they're quite awkward to use, and tend to constrain the inputs more than intended.
$endgroup$
– Matthieu M.
10 hours ago
$begingroup$
@MatthieuM. You should reconsider the solution you posted.
$endgroup$
– Rick
5 hours ago
add a comment |
1
$begingroup$
Why don't you post that as its own questions. There are some improvements even if we don't move to iterators like using template template types.
$endgroup$
– Martin York
11 hours ago
1
$begingroup$
@MartinYork: I generally don't find "template template" to be an improvement, they're quite awkward to use, and tend to constrain the inputs more than intended.
$endgroup$
– Matthieu M.
10 hours ago
$begingroup$
@MatthieuM. You should reconsider the solution you posted.
$endgroup$
– Rick
5 hours ago
1
1
$begingroup$
Why don't you post that as its own questions. There are some improvements even if we don't move to iterators like using template template types.
$endgroup$
– Martin York
11 hours ago
$begingroup$
Why don't you post that as its own questions. There are some improvements even if we don't move to iterators like using template template types.
$endgroup$
– Martin York
11 hours ago
1
1
$begingroup$
@MartinYork: I generally don't find "template template" to be an improvement, they're quite awkward to use, and tend to constrain the inputs more than intended.
$endgroup$
– Matthieu M.
10 hours ago
$begingroup$
@MartinYork: I generally don't find "template template" to be an improvement, they're quite awkward to use, and tend to constrain the inputs more than intended.
$endgroup$
– Matthieu M.
10 hours ago
$begingroup$
@MatthieuM. You should reconsider the solution you posted.
$endgroup$
– Rick
5 hours ago
$begingroup$
@MatthieuM. You should reconsider the solution you posted.
$endgroup$
– Rick
5 hours ago
add a comment |
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4
$begingroup$
Do you know about
std::set_intersection()
? Reference and example implementations: en.cppreference.com/w/cpp/algorithm/set_intersection$endgroup$
– user673679
13 hours ago
2
$begingroup$
@user673679 yes I did, and didn't want to use it;
$endgroup$
– Rick
13 hours ago