如何在 C + + 中逐个元素比较等式元素的两个向量？

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最佳答案

Check std::mismatch method of C++.

comparing vectors has been discussed on DaniWeb forum and also answered.

C++: Comparing two vectors

Check the below SO post. will helpful for you. they have achieved the same with different-2 method.

Compare two vectors C++

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If they really absolutely have to remain unsorted (which they really don't.. and if you're dealing with hundreds of thousands of elements then I have to ask why you would be comparing vectors like this), you can hack together a compare method which works with unsorted arrays.

The only way I though of to do that was to create a temporary vector3 and pretend to do a set_intersection by adding all elements of vector1 to it, then doing a search for each individual element of vector2 in vector3 and removing it if found. I know that sounds terrible, but that's why I'm not writing any C++ standard libraries anytime soon.

Really, though, just sort them first.

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Your code (vector1 == vector2) is correct C++ syntax. There is an == operator for vectors.

If you want to compare short vector with a portion of a longer vector, you can use theequal() operator for vectors. (documentation here)

Here's an example:

using namespace std;


if( equal(vector1.begin(), vector1.end(), vector2.begin()) )
DoSomething();

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According to the discussion here you can directly compare two vectors using

==

if (vector1 == vector2){
//true
}
else{
//false
}

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C++11 standard on == for std::vector

Others have mentioned that operator== does compare vector contents and works, but here is a quote from the C++11 N3337 standard draft which I believe implies that.

We first look at Chapter 23.2.1 "General container requirements", which documents things that must be valid for all containers, including therefore std::vector.

That section Table 96 "Container requirements" which contains an entry:

Expression   Operational semantics
===========  ======================
a == b       distance(a.begin(), a.end()) == distance(b.begin(), b.end()) &&
equal(a.begin(), a.end(), b.begin())

The distance part of the semantics means that the size of both containers are the same, but stated in a generalized iterator friendly way for non random access addressable containers. distance() is defined at 24.4.4 "Iterator operations".

Then the key question is what does equal() mean. At the end of the table we see:

Notes: the algorithm equal() is defined in Clause 25.

and in section 25.2.11 "Equal" we find its definition:

template<class InputIterator1, class InputIterator2>
bool equal(InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2);


template<class InputIterator1, class InputIterator2,
class BinaryPredicate>
bool equal(InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, BinaryPredicate pred);
1 Returns: true if for every iterator i in the range [first1,last1) the following corresponding conditions hold: *i == *(first2 + (i - first1)), pred(*i, *(first2 + (i - first1))) != false. Otherwise, returns false.

In our case, we care about the overloaded version without BinaryPredicate version, which corresponds to the first pseudo code definition *i == *(first2 + (i - first1)), which we see is just an iterator-friendly definition of "all iterated items are the same".