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Home/ Questions/Q 951333
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T23:43:08+00:00 2026-05-15T23:43:08+00:00

Perhaps this question is trivial, but thinking a second time about it I wonder

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Perhaps this question is trivial, but thinking a second time about it I wonder how to do the following really the right way:

std::vector<K> v = ...;
for(T i=0; i<v.size(); ++i) {
  const K& t = v[i];
  // use t *and i*
}

What type should T be? int, unsigned int, int32_t, size_t (which would be the type of v.size()) or any other suggestions? Please try to consider portability, error-proneness and performance and be objective in your answer.

Edit: I did not choose iterators, because it would also like to use the index number i explicitly.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T23:43:09+00:00Added an answer on May 15, 2026 at 11:43 pm

    The type of i should be the same as the return value of size(), which is std::vector<K>::size_type. However, in practice, size_t will do just fine. If you use a signed integer type, then your compiler will probably warn about a signed/unsigned mismatch in the less-than comparison.

    Normally you would use an iterator for this anyway:

    std::vector<K> v = ...;
    for (std::vector<K>::iterator i = v.begin(); i != v.end(); ++i) {
      const K& t = *i;
      // use t
    }
    

    Or, in C++0x:

    std::vector<K> v = ...;
    for (auto i = v.begin(); i != v.end(); ++i) {
      const K& t = *i;
      // use t
    }
    

    In response to your comment about using the vector index with iterators, consider the std::distance() function which is constant time operation for vector iterators:

    std::vector<K> v = ...;
    for (auto i = v.begin(); i != v.end(); ++i) {
      const K& t = *i;
      size_t index = std::distance(v.begin(), i);
      // use t and index
    }
    
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