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Sorting a vector of custom objects
13 answers
I have a vector<data> info
where data
is defined as:
struct data{
string word;
int number;
};
I need to sort info
by the length of the word strings. Is there a quick and simple way to do it?
Use a comparison function:
bool compareByLength(const data &a, const data &b)
{
return a.word.size() < b.word.size();
}
and then use std::sort
in the header #include <algorithm>
:
std::sort(info.begin(), info.end(), compareByLength);
Just make a comparison function/functor:
bool my_cmp(const data& a, const data& b)
{
// smallest comes first
return a.word.size() < b.word.size();
}
std::sort(info.begin(), info.end(), my_cmp);
Or provide an bool operator<(const data& a) const
in your data
class:
struct data {
string word;
int number;
bool operator<(const data& a) const
{
return word.size() < a.word.size();
}
};
or non-member as Fred said:
struct data {
string word;
int number;
};
bool operator<(const data& a, const data& b)
{
return a.word.size() < b.word.size();
}
and just call std::sort()
:
std::sort(info.begin(), info.end());
Yes: you can sort using a custom comparison function:
std::sort(info.begin(), info.end(), my_custom_comparison);
my_custom_comparison
needs to be a function or a class with an operator()
overload (a functor) that takes two data
objects and returns a bool
indicating whether the first is ordered prior to the second (i.e., first < second
). Alternatively, you can overload operator<
for your class type data
; operator<
is the default ordering used by std::sort
.
Either way, the comparison function must yield a strict weak ordering of the elements.
As others have mentioned, you could use a comparison function, but you can also overload the < operator and the default less<T>
functor will work as well:
struct data {
string word;
int number;
bool operator < (const data& rhs) const {
return word.size() < rhs.word.size();
}
};
Then it's just:
std::sort(info.begin(), info.end());
Edit
As James McNellis pointed out, sort
does not actually use the less<T>
functor by default. However, the rest of the statement that the less<T>
functor will work as well is still correct, which means that if you wanted to put struct data
s into a std::map
or std::set
this would still work, but the other answers which provide a comparison function would need additional code to work with either.