use std fill to populate vector with increasing numbers

Question

I would like to fill a vector lt int gt  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 lt int gt  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

User · Answer

If you d rather not use C  11 features  you can use std  generate    include  lt algorithm gt   include  lt iostream gt   include  lt vector gt   struct Generator       Generator     m value  0           int operator       return m value          int m value      int main         std  vector lt int gt  ivec  10         std  generate  ivec begin    ivec end    Generator           std  vector lt int gt   const iterator it  end   ivec end        for   it   ivec begin    it    end    it             std  cout  lt  lt   it  lt  lt  std  endl            This program prints 0 to 9

User · Answer

There is another option - without using iota  For each   lambda expression can be used   vector lt int gt  ivec 10      the vector of size 10 int i   0                incrementor for each ivec begin    ivec end      amp   int amp  item      i  item    i       Two important things why it s working    Lambda captures outer scope   amp   which means that i will be available inside expression item passed as a reference  therefore  it is mutable inside the vector

User · Answer

You should use std  iota algorithm  defined in  lt numeric gt       std  vector lt int gt  ivec 100     std  iota ivec begin    ivec end    0      ivec will become   0  99    Because std  fill just assigns the given fixed value to the elements in the given range  n1 n2   And std  iota fills the given range  n1  n2  with sequentially increasing values  starting with the initial value and then using   value You can also use std  generate as an alternative  Don t forget that std  iota is C  11 STL algorithm  But a lot of modern compilers support it e g  GCC  Clang and VS2012   http   msdn microsoft com en-us library vstudio jj651033 aspx  P S  This function is named after the integer function   from the programming language APL  and signifies a Greek letter iota  I speculate that originally in APL this odd name was chosen because it resembles an    integer     even though in mathematics iota is widely used to denote the imaginary part of a complex number

User · Answer

std  iota is limited to a sequence n  n 1  n 2       But what if you want to fill an array with a generic sequence f 0   f 1   f 2   etc   Often  we can avoid a state tracking generator  For example   int a 7   auto f      int x    return x x     transform a  a 7  a   a  f  int  amp x   return f  amp x - a        will produce the sequence of squares  0 1 4 9 16 25 36   However  this trick will not work with other containers   If you re stuck with C  98  you can do horrible things like   int f int  amp x    int y    int   long   amp x   sizeof int   return y y      and then  int a 7   transform  int    0    int    0    7  a  f     But I would not recommend it

User · Answer

If you really want to use std  fill and are confined to C  98 you can use something like the following    include  lt algorithm gt   include  lt iterator gt   include  lt iostream gt   include  lt vector gt   struct increasing       increasing int start    x start         operator int    const   return x          mutable int x      int main int argc  char  argv          using namespace std       vector lt int gt  v 10       fill v begin    v end    increasing 0        copy v begin    v end    ostream iterator lt int gt  cout             cout  lt  lt  endl      return 0

User · Answer

Preferably use std  iota like this   std  vector lt int gt  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 lt int gt  v 100       vector with 100 ints  IncGenerator g  0   std  generate  v begin    v end    g      Fill with the result of calling g   repeatedly

User · Answer

In terms of performance you should initialize the vector with use of reserve   combined with push back   functions like in the example below   const int numberOfElements   10   std  vector lt int gt  data  data reserve numberOfElements    for int i   0  i  lt  numberOfElements  i        data push back i     All the std  fill  std  generate  etc  are operating on range of existing vector content  and  therefore the vector must be filled with some data earlier  Even doing the following  std  vector lt int gt  data 10   creates a vector with all elements set to its default value  i e  0 in case of int    The above code avoids to initialize vector content before filling it with the data you really want  Performance of this solution is well visible on large data sets

User · Answer

I know this is old question  but I am currently playing with library to handle exactly this problem  It requires c  14    include  htl hpp   htl  Token     std  vector lt int gt  vec     0     100      or for  auto const e    0     100               supports also custom steps      0    3  100     0  4  7  10

User · Answer

Speaking of boost   auto ivec   boost  copy range lt std  vector lt int gt  gt  boost  irange 5  10

User · Answer

I created a simple templated function  Sequence    for generating sequences of numbers  The functionality follows the seq   function in R  link   The nice thing about this function is that it works for generating a variety of number sequences and types    include  lt iostream gt   include  lt vector gt   template  lt typename T gt  std  vector lt T gt  Sequence T min  T max  T by      size t n elements     max - min    by    1    std  vector lt T gt  vec n elements     min -  by    for  size t i   0  i  lt  vec size      i        min    by      vec i    min        return vec      Example usage   int main         auto vec   Sequence 0   10   0 5       for auto  amp v   vec            std  cout  lt  lt  v  lt  lt  std  endl            The only caveat is that all of the numbers should be of the same inferred type  In other words  for doubles or floats  include decimals for all of the inputs  as shown   Updated  June 14  2018

User · Answer

this also works   j 0  for std  vector lt int gt   iterator it   myvector begin     it    myvector end      it        it   j

User · Answer

My first choice  even in C  11  would be boost  counting iterator    std  vector lt int gt  ivec  boost  counting iterator lt int gt   0                           boost  counting iterator lt int gt   n        or if the vector was already constructed   std  copy  boost  counting iterator lt int gt   0               boost  counting iterator lt int gt   ivec size                 ivec begin        If you can t use Boost  either std  generate  as suggested in other answers   or implement counting iterator yourself  if you need it in various places    With Boost  you can use a transform iterator of a counting iterator to create all sorts of interesting sequences   Without Boost  you can do a lot of this by hand  either in the form of a generator object type for std  generate  or as something you can plug into a hand written counting iterator

User · Answer

brainsandwich and underscore d gave very good ideas  Since to fill is to change content  for each    the simplest among the STL algorithms  should also fill the bill   std  vector lt int gt  v 10   std  for each v begin    v end     i 0   int amp  x  mutable  x   i             The generalized capture  i o  imparts the lambda expression with an invariant and initializes it to a known state  in this case 0   the keyword mutable allows this state to be updated each time lambda is called   It takes only a slight modification to get a sequence of squares   std  vector lt int gt  v 10   std  for each v begin    v end     i 0   int amp  x  mutable  x   i i  i         To generate R-like seq is no more difficult  but note that in R  the numeric mode is actually double  so there really isn t a need to parametrize the type  Just use double

User · Answer

We can use generate function which exists in algorithm header file   Code Snippet     include lt bits stdc   h gt  using namespace std    int main         ios  sync with stdio false        vector lt int gt v 10        int n 0       generate v begin    v end      amp n    return n            for auto item   v              cout lt  lt item lt  lt                cout lt  lt endl       return 0

User · Answer

I ve seen the answers with std  generate but you can also  improve  that by using static variables inside the lambda  instead of declaring a counter outside of the function or creating a generator class    std  vector lt int gt  vec  std  generate vec begin    vec end             static int i   0      return i          I find it a little more concise

[c++] use std::fill to populate vector with increasing numbers

Examples related to c++

Examples related to stl