Mixing
2D iterator for loop
I'm having issues implementing an iterator class in regards to .end(). The problem I'm running into is when I try to use a for loop with my iterator class; it's currently stopping just before the last element in a 2D vector. However, I want it to go one element past the last element without causing a compiler error, and to include the last character in the print statement.
main.cpp
// file contents
// aaa
// bbb
// cce
// factory method (adds file contents to 2D <char> vector)
Base *a = Base::create("file");
cout << *a; // overloaded '<<'
Prints
a a a b b b c c e
Now, when I use a for loop with my iterator class, it doesn't include the last character.
for(auto it = a->begin(); it != a->end(); it++)
cout << *it << ' ';
Prints
a a a b b b c c
The .end prints the following
Base::iterator it = aa->end();
cout << *it << 'n';
// prints e
When I try a while loop, it includes the last character.
// Note: my iterator class is a nested class inside Base class.
const Base::iterator et = a->begin();
int i = 0;
while(i < 13) {
cout << *et << ' ';
et++;
}
Prints
a a a b b b c c e e e e e
I understand that a->end() is supposed to point just past the last character, but I don't understand how to implement that. When I increment past the last value in my operator++(int), it displays a segmentation fault. Currently, my overloaded increment method stops at the last character and doesn't go past it. With that said, how do I implement my ++(int) method to include the last element when printing from a for loop? Am I supposed to add a null element to the vector or something like that?
Inside Base.cpp
// This function is whats causing the issue. I know it looks ugly.
Base::iterator Base::iterator::operator++(int) {
// base is a Base *, x and y are coordinates for 2D vector
Base::iterator temp(base, x, y); // save value
// if iterator has not reached end
if( !((x == base->vec.size()-1) && (y == base->vec[0].size()-1)) )
{
// if y < row size
if(y < base->vec[0].size()-1)
y++;
// if y has reached end of row, increment x and start y on a new row
else if(x < base->vec.size() && y == base->vec[0].size()-1) {
y=0;
x++;
}
}
return temp;
}
Base::iterator Base::begin() {
return Base::iterator(this, 0, 0);
}
Base::iterator Base::end() {
return Base::iterator(this, vec.size()-1, vec[0].size()-1);
}
Rest of Base.cpp
#include "Base.h"
using namespace std;
// Factory method (instantiates 2D vector with file contents)
Base *Base::create(string filename) {/*removed irrelevant code */}
Base::~Base(){}
// prints 2D vector
ostream &operator<<(ostream &os, const Base &val){/*removed irrelevant code*/}
Base::iterator::iterator(Base *b, int m, int n): base(b), x(m), y(n) {}
Base::iterator::~iterator(){}
// returns a character inside 2D vector
char &Base::iterator::operator*() const {
return base->vec[x][y];
}
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base->vec[x][y] == *rhs;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return base->vec[x][y] != *rhs;
}
// Bunch of other functions
Any help would be appreciated.
c++ iterator
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
add a comment |
I'm having issues implementing an iterator class in regards to .end(). The problem I'm running into is when I try to use a for loop with my iterator class; it's currently stopping just before the last element in a 2D vector. However, I want it to go one element past the last element without causing a compiler error, and to include the last character in the print statement.
main.cpp
// file contents
// aaa
// bbb
// cce
// factory method (adds file contents to 2D <char> vector)
Base *a = Base::create("file");
cout << *a; // overloaded '<<'
Prints
a a a b b b c c e
Now, when I use a for loop with my iterator class, it doesn't include the last character.
for(auto it = a->begin(); it != a->end(); it++)
cout << *it << ' ';
Prints
a a a b b b c c
The .end prints the following
Base::iterator it = aa->end();
cout << *it << 'n';
// prints e
When I try a while loop, it includes the last character.
// Note: my iterator class is a nested class inside Base class.
const Base::iterator et = a->begin();
int i = 0;
while(i < 13) {
cout << *et << ' ';
et++;
}
Prints
a a a b b b c c e e e e e
I understand that a->end() is supposed to point just past the last character, but I don't understand how to implement that. When I increment past the last value in my operator++(int), it displays a segmentation fault. Currently, my overloaded increment method stops at the last character and doesn't go past it. With that said, how do I implement my ++(int) method to include the last element when printing from a for loop? Am I supposed to add a null element to the vector or something like that?
Inside Base.cpp
// This function is whats causing the issue. I know it looks ugly.
Base::iterator Base::iterator::operator++(int) {
// base is a Base *, x and y are coordinates for 2D vector
Base::iterator temp(base, x, y); // save value
// if iterator has not reached end
if( !((x == base->vec.size()-1) && (y == base->vec[0].size()-1)) )
{
// if y < row size
if(y < base->vec[0].size()-1)
y++;
// if y has reached end of row, increment x and start y on a new row
else if(x < base->vec.size() && y == base->vec[0].size()-1) {
y=0;
x++;
}
}
return temp;
}
Base::iterator Base::begin() {
return Base::iterator(this, 0, 0);
}
Base::iterator Base::end() {
return Base::iterator(this, vec.size()-1, vec[0].size()-1);
}
Rest of Base.cpp
#include "Base.h"
using namespace std;
// Factory method (instantiates 2D vector with file contents)
Base *Base::create(string filename) {/*removed irrelevant code */}
Base::~Base(){}
// prints 2D vector
ostream &operator<<(ostream &os, const Base &val){/*removed irrelevant code*/}
Base::iterator::iterator(Base *b, int m, int n): base(b), x(m), y(n) {}
Base::iterator::~iterator(){}
// returns a character inside 2D vector
char &Base::iterator::operator*() const {
return base->vec[x][y];
}
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base->vec[x][y] == *rhs;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return base->vec[x][y] != *rhs;
}
// Bunch of other functions
Any help would be appreciated.
c++ iterator
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
If you're gonna wrap a 2D-vector it's probably better to use a 1D vector in the underlying implementation.
– super
Nov 20 '18 at 22:36
Ya, I thought about that, but that would require completely redoing all my methods to work with a 1D vector and adding additional class variables. Worst case, I may end up doing it. Just looking to see if there is any solution for 2D iterators.
– N. Colostate
Nov 20 '18 at 22:44
add a comment |
I'm having issues implementing an iterator class in regards to .end(). The problem I'm running into is when I try to use a for loop with my iterator class; it's currently stopping just before the last element in a 2D vector. However, I want it to go one element past the last element without causing a compiler error, and to include the last character in the print statement.
main.cpp
// file contents
// aaa
// bbb
// cce
// factory method (adds file contents to 2D <char> vector)
Base *a = Base::create("file");
cout << *a; // overloaded '<<'
Prints
a a a b b b c c e
Now, when I use a for loop with my iterator class, it doesn't include the last character.
for(auto it = a->begin(); it != a->end(); it++)
cout << *it << ' ';
Prints
a a a b b b c c
The .end prints the following
Base::iterator it = aa->end();
cout << *it << 'n';
// prints e
When I try a while loop, it includes the last character.
// Note: my iterator class is a nested class inside Base class.
const Base::iterator et = a->begin();
int i = 0;
while(i < 13) {
cout << *et << ' ';
et++;
}
Prints
a a a b b b c c e e e e e
I understand that a->end() is supposed to point just past the last character, but I don't understand how to implement that. When I increment past the last value in my operator++(int), it displays a segmentation fault. Currently, my overloaded increment method stops at the last character and doesn't go past it. With that said, how do I implement my ++(int) method to include the last element when printing from a for loop? Am I supposed to add a null element to the vector or something like that?
Inside Base.cpp
// This function is whats causing the issue. I know it looks ugly.
Base::iterator Base::iterator::operator++(int) {
// base is a Base *, x and y are coordinates for 2D vector
Base::iterator temp(base, x, y); // save value
// if iterator has not reached end
if( !((x == base->vec.size()-1) && (y == base->vec[0].size()-1)) )
{
// if y < row size
if(y < base->vec[0].size()-1)
y++;
// if y has reached end of row, increment x and start y on a new row
else if(x < base->vec.size() && y == base->vec[0].size()-1) {
y=0;
x++;
}
}
return temp;
}
Base::iterator Base::begin() {
return Base::iterator(this, 0, 0);
}
Base::iterator Base::end() {
return Base::iterator(this, vec.size()-1, vec[0].size()-1);
}
Rest of Base.cpp
#include "Base.h"
using namespace std;
// Factory method (instantiates 2D vector with file contents)
Base *Base::create(string filename) {/*removed irrelevant code */}
Base::~Base(){}
// prints 2D vector
ostream &operator<<(ostream &os, const Base &val){/*removed irrelevant code*/}
Base::iterator::iterator(Base *b, int m, int n): base(b), x(m), y(n) {}
Base::iterator::~iterator(){}
// returns a character inside 2D vector
char &Base::iterator::operator*() const {
return base->vec[x][y];
}
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base->vec[x][y] == *rhs;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return base->vec[x][y] != *rhs;
}
// Bunch of other functions
Any help would be appreciated.
c++ iterator
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
I'm having issues implementing an iterator class in regards to .end(). The problem I'm running into is when I try to use a for loop with my iterator class; it's currently stopping just before the last element in a 2D vector. However, I want it to go one element past the last element without causing a compiler error, and to include the last character in the print statement.
main.cpp
// file contents
// aaa
// bbb
// cce
// factory method (adds file contents to 2D <char> vector)
Base *a = Base::create("file");
cout << *a; // overloaded '<<'
Prints
a a a b b b c c e
Now, when I use a for loop with my iterator class, it doesn't include the last character.
for(auto it = a->begin(); it != a->end(); it++)
cout << *it << ' ';
Prints
a a a b b b c c
The .end prints the following
Base::iterator it = aa->end();
cout << *it << 'n';
// prints e
When I try a while loop, it includes the last character.
// Note: my iterator class is a nested class inside Base class.
const Base::iterator et = a->begin();
int i = 0;
while(i < 13) {
cout << *et << ' ';
et++;
}
Prints
a a a b b b c c e e e e e
I understand that a->end() is supposed to point just past the last character, but I don't understand how to implement that. When I increment past the last value in my operator++(int), it displays a segmentation fault. Currently, my overloaded increment method stops at the last character and doesn't go past it. With that said, how do I implement my ++(int) method to include the last element when printing from a for loop? Am I supposed to add a null element to the vector or something like that?
Inside Base.cpp
// This function is whats causing the issue. I know it looks ugly.
Base::iterator Base::iterator::operator++(int) {
// base is a Base *, x and y are coordinates for 2D vector
Base::iterator temp(base, x, y); // save value
// if iterator has not reached end
if( !((x == base->vec.size()-1) && (y == base->vec[0].size()-1)) )
{
// if y < row size
if(y < base->vec[0].size()-1)
y++;
// if y has reached end of row, increment x and start y on a new row
else if(x < base->vec.size() && y == base->vec[0].size()-1) {
y=0;
x++;
}
}
return temp;
}
Base::iterator Base::begin() {
return Base::iterator(this, 0, 0);
}
Base::iterator Base::end() {
return Base::iterator(this, vec.size()-1, vec[0].size()-1);
}
Rest of Base.cpp
#include "Base.h"
using namespace std;
// Factory method (instantiates 2D vector with file contents)
Base *Base::create(string filename) {/*removed irrelevant code */}
Base::~Base(){}
// prints 2D vector
ostream &operator<<(ostream &os, const Base &val){/*removed irrelevant code*/}
Base::iterator::iterator(Base *b, int m, int n): base(b), x(m), y(n) {}
Base::iterator::~iterator(){}
// returns a character inside 2D vector
char &Base::iterator::operator*() const {
return base->vec[x][y];
}
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base->vec[x][y] == *rhs;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return base->vec[x][y] != *rhs;
}
// Bunch of other functions
Any help would be appreciated.
c++ iterator
c++ iterator
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
edited Nov 20 '18 at 23:12
N. Colostate
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
asked Nov 20 '18 at 22:25
N. ColostateN. Colostate
356
356
If you're gonna wrap a 2D-vector it's probably better to use a 1D vector in the underlying implementation.
– super
Nov 20 '18 at 22:36
Ya, I thought about that, but that would require completely redoing all my methods to work with a 1D vector and adding additional class variables. Worst case, I may end up doing it. Just looking to see if there is any solution for 2D iterators.
– N. Colostate
Nov 20 '18 at 22:44
add a comment |
If you're gonna wrap a 2D-vector it's probably better to use a 1D vector in the underlying implementation.
– super
Nov 20 '18 at 22:36
Ya, I thought about that, but that would require completely redoing all my methods to work with a 1D vector and adding additional class variables. Worst case, I may end up doing it. Just looking to see if there is any solution for 2D iterators.
– N. Colostate
Nov 20 '18 at 22:44
If you're gonna wrap a 2D-vector it's probably better to use a 1D vector in the underlying implementation.
– super
Nov 20 '18 at 22:36
If you're gonna wrap a 2D-vector it's probably better to use a 1D vector in the underlying implementation.
– super
Nov 20 '18 at 22:36
Ya, I thought about that, but that would require completely redoing all my methods to work with a 1D vector and adding additional class variables. Worst case, I may end up doing it. Just looking to see if there is any solution for 2D iterators.
– N. Colostate
Nov 20 '18 at 22:44
Ya, I thought about that, but that would require completely redoing all my methods to work with a 1D vector and adding additional class variables. Worst case, I may end up doing it. Just looking to see if there is any solution for 2D iterators.
– N. Colostate
Nov 20 '18 at 22:44
add a comment |
1 Answer
1
active
oldest
votes
Base::iterator(this, vec.size()-1, vec[0].size()-1); this will return a valid last element. So you need to change it to Base::iterator(this, vec.size(), 0);, and also update the conditions to switch to a new row in your loop.
Something like:
// if iterator has not reached end
if(x < base->vec.size())
{
++y;
// if y has reached end of row, increment x and start y on a new row
if(y >= base->vec[0].size() {
y=0;
++x;
}
}
Also the iterators are wrong:
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base == rhs.base && x == rhs.x && y == ris.y;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return !(base == rhs.base && x == rhs.x && y == ris.y);
}
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
How do I resolve the segmentation fault? return Base::iterator(this, vec.size(), vec[0].size()); displays a segmentation fault. (3, 3) is out of bounds. The 2D vector only has indices (0,0) -> (2,2).
– N. Colostate
Nov 20 '18 at 22:38
1
Oops, bad bound, x < size() and bad bound for the end iterator, should be size(), 0.
– Matthieu Brucher
Nov 20 '18 at 22:46
Still gives a segmentation fault. Maybe I have to do something with my iterator class constructor? Thats where its currently flagging it as a segmentation fault. It's saying it can't assign values (3,0) to base*[x][y] since its out of bounds.
– N. Colostate
Nov 20 '18 at 22:53
It has probably to do with your comparison operators. Just compare base, x, y. You can't dereference end, as it's always a non existing object.
– Matthieu Brucher
Nov 20 '18 at 22:54
1
Compare base, x and y, don't compare the values themselves, they could even be the same for different iterators. Iterators compare "pointers", not the pointed values.
– Matthieu Brucher
Nov 20 '18 at 23:37
|
show 3 more comments
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
Base::iterator(this, vec.size()-1, vec[0].size()-1); this will return a valid last element. So you need to change it to Base::iterator(this, vec.size(), 0);, and also update the conditions to switch to a new row in your loop.
Something like:
// if iterator has not reached end
if(x < base->vec.size())
{
++y;
// if y has reached end of row, increment x and start y on a new row
if(y >= base->vec[0].size() {
y=0;
++x;
}
}
Also the iterators are wrong:
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base == rhs.base && x == rhs.x && y == ris.y;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return !(base == rhs.base && x == rhs.x && y == ris.y);
}
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
How do I resolve the segmentation fault? return Base::iterator(this, vec.size(), vec[0].size()); displays a segmentation fault. (3, 3) is out of bounds. The 2D vector only has indices (0,0) -> (2,2).
– N. Colostate
Nov 20 '18 at 22:38
1
Oops, bad bound, x < size() and bad bound for the end iterator, should be size(), 0.
– Matthieu Brucher
Nov 20 '18 at 22:46
Still gives a segmentation fault. Maybe I have to do something with my iterator class constructor? Thats where its currently flagging it as a segmentation fault. It's saying it can't assign values (3,0) to base*[x][y] since its out of bounds.
– N. Colostate
Nov 20 '18 at 22:53
It has probably to do with your comparison operators. Just compare base, x, y. You can't dereference end, as it's always a non existing object.
– Matthieu Brucher
Nov 20 '18 at 22:54
1
Compare base, x and y, don't compare the values themselves, they could even be the same for different iterators. Iterators compare "pointers", not the pointed values.
– Matthieu Brucher
Nov 20 '18 at 23:37
|
show 3 more comments
Base::iterator(this, vec.size()-1, vec[0].size()-1); this will return a valid last element. So you need to change it to Base::iterator(this, vec.size(), 0);, and also update the conditions to switch to a new row in your loop.
Something like:
// if iterator has not reached end
if(x < base->vec.size())
{
++y;
// if y has reached end of row, increment x and start y on a new row
if(y >= base->vec[0].size() {
y=0;
++x;
}
}
Also the iterators are wrong:
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base == rhs.base && x == rhs.x && y == ris.y;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return !(base == rhs.base && x == rhs.x && y == ris.y);
}
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
How do I resolve the segmentation fault? return Base::iterator(this, vec.size(), vec[0].size()); displays a segmentation fault. (3, 3) is out of bounds. The 2D vector only has indices (0,0) -> (2,2).
– N. Colostate
Nov 20 '18 at 22:38
1
Oops, bad bound, x < size() and bad bound for the end iterator, should be size(), 0.
– Matthieu Brucher
Nov 20 '18 at 22:46
Still gives a segmentation fault. Maybe I have to do something with my iterator class constructor? Thats where its currently flagging it as a segmentation fault. It's saying it can't assign values (3,0) to base*[x][y] since its out of bounds.
– N. Colostate
Nov 20 '18 at 22:53
It has probably to do with your comparison operators. Just compare base, x, y. You can't dereference end, as it's always a non existing object.
– Matthieu Brucher
Nov 20 '18 at 22:54
1
Compare base, x and y, don't compare the values themselves, they could even be the same for different iterators. Iterators compare "pointers", not the pointed values.
– Matthieu Brucher
Nov 20 '18 at 23:37
|
show 3 more comments
Base::iterator(this, vec.size()-1, vec[0].size()-1); this will return a valid last element. So you need to change it to Base::iterator(this, vec.size(), 0);, and also update the conditions to switch to a new row in your loop.
Something like:
// if iterator has not reached end
if(x < base->vec.size())
{
++y;
// if y has reached end of row, increment x and start y on a new row
if(y >= base->vec[0].size() {
y=0;
++x;
}
}
Also the iterators are wrong:
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base == rhs.base && x == rhs.x && y == ris.y;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return !(base == rhs.base && x == rhs.x && y == ris.y);
}
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
Base::iterator(this, vec.size()-1, vec[0].size()-1); this will return a valid last element. So you need to change it to Base::iterator(this, vec.size(), 0);, and also update the conditions to switch to a new row in your loop.
Something like:
// if iterator has not reached end
if(x < base->vec.size())
{
++y;
// if y has reached end of row, increment x and start y on a new row
if(y >= base->vec[0].size() {
y=0;
++x;
}
}
Also the iterators are wrong:
bool Base::iterator::operator==(const Base::iterator& rhs) const {
return base == rhs.base && x == rhs.x && y == ris.y;
}
bool Base::iterator::operator!=(const Base::iterator& rhs) const {
return !(base == rhs.base && x == rhs.x && y == ris.y);
}
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
edited Nov 20 '18 at 23:39
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
answered Nov 20 '18 at 22:32
Matthieu BrucherMatthieu Brucher
15.2k32140
15.2k32140
How do I resolve the segmentation fault? return Base::iterator(this, vec.size(), vec[0].size()); displays a segmentation fault. (3, 3) is out of bounds. The 2D vector only has indices (0,0) -> (2,2).
– N. Colostate
Nov 20 '18 at 22:38
1
Oops, bad bound, x < size() and bad bound for the end iterator, should be size(), 0.
– Matthieu Brucher
Nov 20 '18 at 22:46
Still gives a segmentation fault. Maybe I have to do something with my iterator class constructor? Thats where its currently flagging it as a segmentation fault. It's saying it can't assign values (3,0) to base*[x][y] since its out of bounds.
– N. Colostate
Nov 20 '18 at 22:53
It has probably to do with your comparison operators. Just compare base, x, y. You can't dereference end, as it's always a non existing object.
– Matthieu Brucher
Nov 20 '18 at 22:54
1
Compare base, x and y, don't compare the values themselves, they could even be the same for different iterators. Iterators compare "pointers", not the pointed values.
– Matthieu Brucher
Nov 20 '18 at 23:37
|
show 3 more comments
How do I resolve the segmentation fault? return Base::iterator(this, vec.size(), vec[0].size()); displays a segmentation fault. (3, 3) is out of bounds. The 2D vector only has indices (0,0) -> (2,2).
– N. Colostate
Nov 20 '18 at 22:38
1
Oops, bad bound, x < size() and bad bound for the end iterator, should be size(), 0.
– Matthieu Brucher
Nov 20 '18 at 22:46
Still gives a segmentation fault. Maybe I have to do something with my iterator class constructor? Thats where its currently flagging it as a segmentation fault. It's saying it can't assign values (3,0) to base*[x][y] since its out of bounds.
– N. Colostate
Nov 20 '18 at 22:53
It has probably to do with your comparison operators. Just compare base, x, y. You can't dereference end, as it's always a non existing object.
– Matthieu Brucher
Nov 20 '18 at 22:54
1
Compare base, x and y, don't compare the values themselves, they could even be the same for different iterators. Iterators compare "pointers", not the pointed values.
– Matthieu Brucher
Nov 20 '18 at 23:37
How do I resolve the segmentation fault? return Base::iterator(this, vec.size(), vec[0].size()); displays a segmentation fault. (3, 3) is out of bounds. The 2D vector only has indices (0,0) -> (2,2).
– N. Colostate
Nov 20 '18 at 22:38
How do I resolve the segmentation fault? return Base::iterator(this, vec.size(), vec[0].size()); displays a segmentation fault. (3, 3) is out of bounds. The 2D vector only has indices (0,0) -> (2,2).
– N. Colostate
Nov 20 '18 at 22:38
1
1
Oops, bad bound, x < size() and bad bound for the end iterator, should be size(), 0.
– Matthieu Brucher
Nov 20 '18 at 22:46
Oops, bad bound, x < size() and bad bound for the end iterator, should be size(), 0.
– Matthieu Brucher
Nov 20 '18 at 22:46
Still gives a segmentation fault. Maybe I have to do something with my iterator class constructor? Thats where its currently flagging it as a segmentation fault. It's saying it can't assign values (3,0) to base*[x][y] since its out of bounds.
– N. Colostate
Nov 20 '18 at 22:53
Still gives a segmentation fault. Maybe I have to do something with my iterator class constructor? Thats where its currently flagging it as a segmentation fault. It's saying it can't assign values (3,0) to base*[x][y] since its out of bounds.
– N. Colostate
Nov 20 '18 at 22:53
It has probably to do with your comparison operators. Just compare base, x, y. You can't dereference end, as it's always a non existing object.
– Matthieu Brucher
Nov 20 '18 at 22:54
It has probably to do with your comparison operators. Just compare base, x, y. You can't dereference end, as it's always a non existing object.
– Matthieu Brucher
Nov 20 '18 at 22:54
1
1
Compare base, x and y, don't compare the values themselves, they could even be the same for different iterators. Iterators compare "pointers", not the pointed values.
– Matthieu Brucher
Nov 20 '18 at 23:37
Compare base, x and y, don't compare the values themselves, they could even be the same for different iterators. Iterators compare "pointers", not the pointed values.
– Matthieu Brucher
Nov 20 '18 at 23:37
|
show 3 more comments
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If you're gonna wrap a 2D-vector it's probably better to use a 1D vector in the underlying implementation.
– super
Nov 20 '18 at 22:36
Ya, I thought about that, but that would require completely redoing all my methods to work with a 1D vector and adding additional class variables. Worst case, I may end up doing it. Just looking to see if there is any solution for 2D iterators.
– N. Colostate
Nov 20 '18 at 22:44