I would like to fill a vector<int>
using std::fill
, but instead of one value, the vector should contain numbers in increasing order after.
I tried achieving this by iterating the third parameter of the function by one, but this would only give me either vectors filled with 1 or 2 (depending of the position of the ++
operator).
Example:
vector<int> ivec; int i = 0; std::fill(ivec.begin(), ivec.end(), i++); // elements are set to 1 std::fill(ivec.begin(), ivec.end(), ++i); // elements are set to 2
Read a value into x , use that value as index into the vector, and increase the value at that index. So if the input for x is 1 , then it's equivalent to vec[1]++ , that is the second (remember that indexes are zero based) will be increased by one.
Preferably use std::iota
like this:
std::vector<int> v(100) ; // vector with 100 ints. std::iota (std::begin(v), std::end(v), 0); // Fill with 0, 1, ..., 99.
That said, if you don't have any c++11
support (still a real problem where I work), use std::generate
like this:
struct IncGenerator { int current_; IncGenerator (int start) : current_(start) {} int operator() () { return current_++; } }; // ... std::vector<int> v(100) ; // vector with 100 ints. IncGenerator g (0); std::generate( v.begin(), v.end(), g); // Fill with the result of calling g() repeatedly.
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