I have a map of a vector of char's and a vector of strings. Every so often, if I've seen the vector of characters before, I'd like to add a string to my vector of strings. Below is my code to do that.
map<vector<char>, vector<string>>::iterator myIter = mMyMap.find(vChars);
if(myIter != mMyMap.end()) {
vector<string> vStrings = myIter->second;
mMyMap.erase(myIter);
vStrings.push_back(some_other_string);
mMyMap.insert(pair<vector<char>, vector<string>>(vChars, vStrings));
return TRUE;
}
The call to mMyMap.erase() seems to get stuck an in infinite loop though. I'm guessing it's because vStrings isn't getting a deep-copy of myIter->second.
Do I need to initalize vStrings like:
vector<string> vStrings(myIter->second);
Or what's the proper fix?
I don't see an error in the posted code fragment (other than a missing )). But may I suggest simplifying lines 2-8 to:
if(myIter != mMyMap.end()) {
myIter->second.push_back(some_other_string);
}
vector vStrings = myIter->second;
and
vector vStrings(myIter->second);
are same things. They both call copy constructor. And the copy is deep copy only. My guess is that the vector that is getting copied is too big(or long). Each element of the vector will be copied one by one. And hence the time.
Related
My apologies for the lengthy explanation.
I am working on a C++ application that loads two files into two 2D string vectors, rearranges those vectors, builds another 2D string vector, and outputs it all in a report. The first element of the two vectors is a code that identifies the owner of the item and the item in the vector. I pass the owner's identification to the program on start and loop through the two vectors in a nested while loop to find those that have matching first elements. When I do, I build a third vector with components of the first two, and I then need to capture any that don't match.
I was using the syntax "vector.erase(vector.begin() + i)" to remove elements from the two original arrays when they matched. When the loop completed, I had my new third vector, and I was left with two vectors that only had elements, which didn't match and that is what I needed. This was working fine as I tried the various owners in the files (the program accepts one owner at a time). Then I tried one that generated an out of range error.
I could not figure out how to do the erase inside of the loop without throwing the error (it didn't seem that swap and pop or erase-remove were feasible solutions). I solved my problem for the program with two extra nested while loops after building my third vector in this one.
I'd like to know how to make the erase method work here (as it seems a simpler solution) or at least how to check for my out of range error (and avoid it). There were a lot of "rows" for this particular owner; so debugging was tedious. Before giving up and going on to the nested while solution, I determined that the second erase was throwing the error. How can I make this work, or are my nested whiles after the fact, the best I can do? Here is the code:
i = 0;
while (i < AIvector.size())
{
CHECK:
j = 0;
while (j < TRvector.size())
{
if (AIvector[i][0] == TRvector[j][0])
{
linevector.clear();
// Add the necessary data from both vectors to Combo_outputvector
for (x = 0; x < AIvector[i].size(); x++)
{
linevector.push_back(AIvector[i][x]); // add AI info
}
for (x = 3; x < TRvector[j].size(); x++) // Don't need the the first three elements; so start with x=3.
{
linevector.push_back(TRvector[j][x]); // add TR info
}
Combo_outputvector.push_back(linevector); // build the combo vector
// then erase these two current rows/elements from their respective vectors, this revises the AI and TR vectors
AIvector.erase(AIvector.begin() + i);
TRvector.erase(TRvector.begin() + j);
goto CHECK; // jump from here because the erase will have changed the two increments
}
j++;
}
i++;
}
As already discussed, your goto jumps to the wrong position. Simply moving it out of the first while loop should solve your problems. But can we do better?
Erasing from a vector can be done cleanly with std::remove and std::erase for cheap-to-move objects, which vector and string both are. After some thought, however, I believe this isn't the best solution for you because you need a function that does more than just check if a certain row exists in both containers and that is not easily expressed with the erase-remove idiom.
Retaining the current structure, then, we can use iterators for the loop condition. We have a lot to gain from this, because std::vector::erase returns an iterator to the next valid element after the erased one. Not to mention that it takes an iterator anyway. Conditionally erasing elements in a vector becomes as simple as
auto it = vec.begin()
while (it != vec.end()) {
if (...)
it = vec.erase(it);
else
++it;
}
Because we assign erase's return value to it we don't have to worry about iterator invalidation. If we erase the last element, it returns vec.end() so that doesn't need special handling.
Your second loop can be removed altogether. The C++ standard defines functions for searching inside STL containers. std::find_if searches for a value in a container that satisfies a condition and returns an iterator to it, or end() if it doesn't exist. You haven't declared your types anywhere so I'm just going to assume the rows are std::vector<std::string>>.
using row_t = std::vector<std::string>;
auto AI_it = AIVector.begin();
while (AI_it != AIVector.end()) {
// Find a row in TRVector with the same first element as *AI_it
auto TR_it = std::find_if (TRVector.begin(), TRVector.end(), [&AI_it](const row_t& row) {
return row[0] == (*AI_it)[0];
});
// If a matching row was found
if (TR_it != TRVector.end()) {
// Copy the line from AIVector
auto linevector = *AI_it;
// Do NOT do this if you don't guarantee size > 3
assert(TR_it->size() >= 3);
std::copy(TR_it->begin() + 3, TR_it->end(),
std::back_inserter(linevector));
Combo_outputvector.emplace_back(std::move(linevector));
AI_it = AIVector.erase(AI_it);
TRVector.erase(TR_it);
}
else
++AI_it;
}
As you can see, switching to iterators completely sidesteps your initial problem of figuring out how not to access invalid indices. If you don't understand the syntax of the arguments for find_if search for the term lambda. It is beyond the scope if this answer to explain what they are.
A few notable changes:
linevector is now encapsulated properly. There is no reason for it to be declared outside this scope and reused.
linevector simply copies the desired row from AIVector rather than push_back every element in it, as long as Combo_outputvector (and therefore linevector) contains the same type than AIVector and TRVector.
std::copy is used instead of a for loop. Apart from being slightly shorter, it is also more generic, meaning you could change your container type to anything that supports random access iterators and inserting at the back, and the copy would still work.
linevector is moved into Combo_outputvector. This can be a huge performance optimization if your vectors are large!
It is possible that you used an non-encapsulated linevector because you wanted to keep a copy of the last inserted row outside of the loop. That would prohibit moving it, however. For this reason it is faster and more descriptive to do it as I showed above and then simply do the following after the loop.
auto linevector = Combo_outputvector.back();
It started with converting a vector<string> to vector<double>. I use gcc without C++11 so I could not use this approach. After having a look at the algorithm template of transpose I tried to use the template instead, but there is a bug for std::stod used inside the template (for gcc 4.x which results in ‘stod’ is not a member of ‘std’) So why not write it yourself and I use while-loop with stringstream (expensive?):
int i = 0;
while (svec.begin() != svec.end()) {
stringstream(*svec.begin())>> dvec[i] ;//svec: strings, dvec: doubles
//dvec[i] = std::stod(*svec.begin());
++svec.begin(); i++;
}
Unfortunately I get: double free or corruption (out): 0x00007f4788000ba0 ***
I think the issue here is that
++svec.begin();
doesn't work the way you think it does. This doesn't advance forward the beginning of the vector. Instead, it gets a (temporary) iterator to the beginning of the vector, then increments that. As a result, you'll be stuck in an infinite loop. To fix this, use a more traditional "loop over a container" loop either by using a range-based for loop or just counting up the indices.
I also noticed that you're writing into dvec by index. Without seeing the code to initialize dvec, I can't be sure of whether this is safe, since if you didn't already resize the vector this will write off the end and lead to undefined behavior. Even if you did set it up properly, because of the broken loop, my guess is that this eventually writes off the end of the vector and is what directly triggers the issue.
First of all, let's clear up some confusion:
std::stod is a C++11 function, so if you do not use C++11, then it is not a "bug" if it cannot be found.
You mean std::transform and not transpose, don't you?
You may very well use std::transform without C++11 and without std::stod. Unfortunately you don't show the code with which you tried.
So why not write it yourself and I use while-loop with stringstream
Yes, why not?
(expensive?):
Measure it :)
But it's unlikely you'll notice a difference.
int i = 0;
while (svec.begin() != svec.end()) {
The loop condition does not make sense. begin() and end() do not change. This line effectively reads as "do this stuff as long as the vector does not become empty".
stringstream(*svec.begin())>> dvec[i] ;//svec: strings, dvec: doubles
//dvec[i] = std::stod(*svec.begin());
++svec.begin(); i++;
}
I'd say you are overthinking this. The crash you get may come from the fact that you access dvec[i] while dvec is still empty, and you never actually add elements to it. That's undefined behaviour.
It's really as simple as "loop through string vector, use a stringstream on each element to get a double value, and add that value to the result vector", expressed in C++03 as:
// loop through string vector
for (std::vector<std::string>::const_iterator iter = svec.begin(); iter != svec.end(); ++iter)
{
std::string const& element = *iter;
// use a stringstream to get a double value:
std::istringstream is(element);
double result;
is >> result;
// add the double value to the result vector:
dvec.push_back(result);
}
If you don't have stod (which was added in C++11) you can use strtod and apply it to a C-string that corresponds to the C++ string that you have. Copying from the approach that you linked to and making that change (and cleaning up the formatting):
std::vector<double> convertStringVectortoDoubleVector(
const std::vector<std::string>& stringVector){
std::vector<double> doubleVector(stringVector.size());
std::transform(stringVector.begin(), stringVector.end(),
doubleVector.begin(), [](const std::string& val)
{
return strtod(val.c_str(), 0);
});
return doubleVector;
}
Note that if the code hasn't already ensured that each string holds a valid representation of a floating-point value you'll have to add code to check whether strtod succeeded.
I know this might seem as a duplicate question, but it is not, I have a problem with a function and don''t know why it behaves like that.
I have a vector which holds elements of type MyMaterial** (std::vector). At a point in my program, I will know an element, "currentElement", and I will want to remove it.
I tried doing this:
myMaterials.erase(currentElement);
But here is the problem: Instead of only deleting "currentElement", it also deletes all elements added after it. Why does it do that and how can I solve it?
I must mention that I don''t know the position of "currentElement" in the vector, and i prefere not to search for it, I''m hoping there is another way.
If you are using std::vector, then:
Erase elements Removes from the vector either a single element
(position) or a range of elements ([first,last)).
Maybe you are looking for something like this:
How do I remove an item from a stl vector with a certain value?
To use vector::erase(iterator) to remove an element from a vector, you may either have to know its index OR iterate thru the list to hunt for it.
Luckily, There is the std::map, and this is how you would work it
std::map<std::string,myMaterial> myMaterials;
myMaterial mat;//assuming it doesnt take any args
myMaterials['myMaterialXYZ'] = mat; ///add it to the array
myMaterials.erase('myMaterialXYZ'); ///Erase by key..(or "name" in this case)
Now you can easily track string names instead of ever changing index positions...and memory locations, which by the way may be another ace up your sleeve.
I couldn''t really use the examples given by you above, because I got all kinds of errors, due to the type of elements the vector holds. But I made a function which managed to delete the specific element:
int i=0;
int found=0;
MyMaterial **material = gMaterials.begin();
while(material != gMaterials.end() && found == 0)
{
if(currentElement == material)
{
gMaterials.erase(gMaterials.begin() + i, gMaterials.begin() + i+1);
found = 1;
}
i++;
cloth++;
}
I don''t know how good/correct it is, but it does the job.
Thank you very much for the suggestions and the help.
I've implemented a merge function for vectors, which basically combines to sorted vectors in a one sorted vector. (yes, it is for a merge sort algorithm). I was trying to make my code faster and avoid overheads, so I decided not to use the push_back method on the vector, but try to use the array syntax instead which has lesser over head. However, something is going terribly wrong, and the output is messed up when i do this. Here's the code:
while(size1<left.size() && size2 < right.size()) //left and right are the input vectors
{
//it1 and it2 are iterators on the two sorted input vectors
if(*it1 <= *it2)
{
final.push_back(*it1); //final is the final vector to output
//final[count] = *it1; // this does not work for some reason
it1++;
size1++;
//cout<<"count ="<<count<<" size1 ="<<size1<<endl;
}
else
{
final.push_back(*it2);
//final[count] = left[size2];
it2++;
size2++;
}
count++;
//cout<<"count ="<<count<<" size1 ="<<size1<<"size2 = "<<size2<<endl;
}
It seems to me that the two methods should be functionally equivalent.
PS I have already reserved space for the final vector so that shouldnt be a problem.
You can't add new objects to vector using operator[]. .reserve() doesn't add them neither. You have to either use .resize() or .push_back().
Also, you are not avoiding overheads at all; call cost of operator[] isn't really much better that push_back() one, so until you profile your code thorougly, just use push_back. You can still use reserve to make sure unneccessary allocations won't be made.
In most of the cases, "optimizations" like this don't really help. If you want to make your code faster, profile it first and look for the hot paths.
There is a huge difference between
vector[i] = item;
and
vector.push_back(item);
Differences:
The first one modifies the element at index i and i must be valid index. That is,
0 <= i < vector.size() must be true
If i is an invalid index, the first one invokes undefined behavior, which means ANYTHING can happen. You could, however, use at() which throws exception if i is invalid:
vector.at(i) = item; //throws exception if i is invalid
The second one adds an element to the vector at the end, which means the size of the vector increases by one.
Since, sematically both of them do different thing, choose the one which you need.
void insert_string( std::vector<std::string> & strings, const std::string &s )
{
std::vector<std::string>::iterator it=lower_bound(strings.begin(),strings.end(),s);
if(strings.size()>0) std::cout<<*it<<" is found\n"; // ****
strings.insert(it,s);
}
When attempting to use this function, the first insertion goes fine. The second insertion will output "[firststring] is found" and then segfault. If I comment out the if/cout line, I can repeatedly call and no segfaults occur.
I've also tried doing something like std::string tmp=*it; which will then segfault on that line. While printing is not a huge deal, what I'm really trying to do is check if the string at the position found by lower_bound is the same as the string that is trying to be inserted (i.e., if(*it==s), which is segfaulting just like the above two examples).
What am I missing here?
Thanks!
Check for the condition if it == strings.end(), if it is don't print it. This could cause the issue. Are you sure the string you're trying to check is in the vector of strings?