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How to get the index of the smallest element in a vector?

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2

I want to know which element of vector is the minimum, but min_element returns an iterator to the element. So I tried this:

vector<int> vec = 4,5,0,1,2,3 ;
int min_element_index = min_element(vec.begin(), vec.end()) - vec.begin();

However, I'm unsure this will always work. I don't know how vectors are stored in memory, and I'm worried this will break when I use push_back.

c++ vector

share|improve this question

edited Mar 22 at 17:51

NathanOliver

100k16139221

asked Mar 22 at 17:23

M12421KM12421K

167

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I'm not sure it will always work .Are vector elements stored continuously ?
  
  – M12421K
  Mar 22 at 17:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Vector elements are stored continuously in memory. But you don't need to care unless you use vector::data which most of the time you shouldn't do anyway.
  
  – Jabberwocky
  Mar 22 at 17:30
  
  

  
  
  
  

add a comment | 

2

I want to know which element of vector is the minimum, but min_element returns an iterator to the element. So I tried this:

vector<int> vec = 4,5,0,1,2,3 ;
int min_element_index = min_element(vec.begin(), vec.end()) - vec.begin();

However, I'm unsure this will always work. I don't know how vectors are stored in memory, and I'm worried this will break when I use push_back.

c++ vector

share|improve this question

edited Mar 22 at 17:51

NathanOliver

100k16139221

asked Mar 22 at 17:23

M12421KM12421K

167

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I'm not sure it will always work .Are vector elements stored continuously ?
  
  – M12421K
  Mar 22 at 17:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Vector elements are stored continuously in memory. But you don't need to care unless you use vector::data which most of the time you shouldn't do anyway.
  
  – Jabberwocky
  Mar 22 at 17:30
  
  

  
  
  
  

add a comment | 

I want to know which element of vector is the minimum, but min_element returns an iterator to the element. So I tried this:

vector<int> vec = 4,5,0,1,2,3 ;
int min_element_index = min_element(vec.begin(), vec.end()) - vec.begin();

However, I'm unsure this will always work. I don't know how vectors are stored in memory, and I'm worried this will break when I use push_back.

c++ vector

share|improve this question

edited Mar 22 at 17:51

NathanOliver

100k16139221

asked Mar 22 at 17:23

M12421KM12421K

167

vector<int> vec = 4,5,0,1,2,3 ;
int min_element_index = min_element(vec.begin(), vec.end()) - vec.begin();

share|improve this question

edited Mar 22 at 17:51

NathanOliver

100k16139221

asked Mar 22 at 17:23

M12421KM12421K

167

share|improve this question

edited Mar 22 at 17:51

NathanOliver

100k16139221

asked Mar 22 at 17:23

M12421KM12421K

167

edited Mar 22 at 17:51

NathanOliver

100k16139221

edited Mar 22 at 17:51

NathanOliver

100k16139221

asked Mar 22 at 17:23

M12421KM12421K

167

asked Mar 22 at 17:23

M12421KM12421K

167

2 Answers
2

active

oldest

votes

10

You can also do this:

std::vector<int>::iterator it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or even simpler:

auto it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or:

std::cout << "index of smallest element: " <<
std::distance(std::begin(v), std::min_element(std::begin(v), std::end(v)))

share|improve this answer

edited Mar 22 at 17:47

answered Mar 22 at 17:36

JabberwockyJabberwocky

28.5k103875

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  I like this answer the most since it is more generic and will work with non random access containers as well.
  
  – NathanOliver
  Mar 22 at 17:38
  

  
  
  
  

add a comment | 

3

Vector members are guaranteed to be stored in contiguous memory, which doesn't really matter here.

The vector iterators are random access iterators, so subtracting vector iterators is well defined (This will always work with a standards compliant std::vector)

To convert a vector iterator into a pointer, you can do &*(vector-iterator), but there is no reason to here.

share|improve this answer

answered Mar 22 at 17:35

ArtyerArtyer

5,898829

add a comment | 

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10

You can also do this:

std::vector<int>::iterator it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or even simpler:

auto it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or:

std::cout << "index of smallest element: " <<
std::distance(std::begin(v), std::min_element(std::begin(v), std::end(v)))

share|improve this answer

edited Mar 22 at 17:47

answered Mar 22 at 17:36

JabberwockyJabberwocky

28.5k103875

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  I like this answer the most since it is more generic and will work with non random access containers as well.
  
  – NathanOliver
  Mar 22 at 17:38
  

  
  
  
  

add a comment | 

10

You can also do this:

std::vector<int>::iterator it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or even simpler:

auto it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or:

std::cout << "index of smallest element: " <<
std::distance(std::begin(v), std::min_element(std::begin(v), std::end(v)))

share|improve this answer

edited Mar 22 at 17:47

answered Mar 22 at 17:36

JabberwockyJabberwocky

28.5k103875

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  I like this answer the most since it is more generic and will work with non random access containers as well.
  
  – NathanOliver
  Mar 22 at 17:38
  

  
  
  
  

add a comment | 

You can also do this:

std::vector<int>::iterator it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or even simpler:

auto it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

or:

std::cout << "index of smallest element: " <<
std::distance(std::begin(v), std::min_element(std::begin(v), std::end(v)))

share|improve this answer

edited Mar 22 at 17:47

answered Mar 22 at 17:36

JabberwockyJabberwocky

28.5k103875

std::vector<int>::iterator it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

auto it = std::min_element(std::begin(vec), std::end(vec));
std::cout << "index of smallest element: " << std::distance(std::begin(vec), it);

std::cout << "index of smallest element: " <<
std::distance(std::begin(v), std::min_element(std::begin(v), std::end(v)))

share|improve this answer

edited Mar 22 at 17:47

answered Mar 22 at 17:36

JabberwockyJabberwocky

28.5k103875

answered Mar 22 at 17:36

JabberwockyJabberwocky

28.5k103875

answered Mar 22 at 17:36

JabberwockyJabberwocky

28.5k103875

3

Vector members are guaranteed to be stored in contiguous memory, which doesn't really matter here.

The vector iterators are random access iterators, so subtracting vector iterators is well defined (This will always work with a standards compliant std::vector)

To convert a vector iterator into a pointer, you can do &*(vector-iterator), but there is no reason to here.

share|improve this answer

answered Mar 22 at 17:35

ArtyerArtyer

5,898829

add a comment | 

3

Vector members are guaranteed to be stored in contiguous memory, which doesn't really matter here.

The vector iterators are random access iterators, so subtracting vector iterators is well defined (This will always work with a standards compliant std::vector)

To convert a vector iterator into a pointer, you can do &*(vector-iterator), but there is no reason to here.

share|improve this answer

answered Mar 22 at 17:35

ArtyerArtyer

5,898829

add a comment | 

Vector members are guaranteed to be stored in contiguous memory, which doesn't really matter here.

The vector iterators are random access iterators, so subtracting vector iterators is well defined (This will always work with a standards compliant std::vector)

To convert a vector iterator into a pointer, you can do &*(vector-iterator), but there is no reason to here.

share|improve this answer

answered Mar 22 at 17:35

ArtyerArtyer

5,898829

share|improve this answer

answered Mar 22 at 17:35

ArtyerArtyer

5,898829

answered Mar 22 at 17:35

ArtyerArtyer

5,898829

answered Mar 22 at 17:35

ArtyerArtyer

5,898829