c:if equal_C ++中的std

c:if equal

equal()作为STL函数 (equal() as a STL function)

Syntax:

句法：

bool equal(
InputIterator1 first1, 
InputIterator1 last1, 
InputIterator2 first2);

Where,

哪里，

InputIterator1 first = iterator to start of the first sequence range
InputIterator1 first =迭代器，开始第一个序列范围
InputIterator1 last1 = iterator to end of the first sequence range
InputIterator1 last1 =迭代到第一个序列范围的结尾
InputIterator2 first2 = iterator to start of the second sequence range
InputIterator2 first2 =迭代器开始第二个序列范围

Return type: bool

返回类型： bool

True - if all elements are the same in both the ranges
True-如果两个范围内的所有元素都相同
False - if all elements are not the same in both the ranges
False-如果两个范围内的所有元素都不相同

The above syntax is used to compare elements using standard == operator.

上面的语法用于使用标准==运算符比较元素。

We can also define our user-defined binary predicate (instead of '==') for checking whether equal or not. The syntax with user-defined binary predicate is like below:

我们还可以定义用户定义的二进制谓词(而不是'==')来检查是否相等。用户定义的二进制谓词的语法如下：

(Binary predicate is a function which takes two arguments and returns true or false only.)

(二元谓词是一个带有两个参数并且仅返回true或false的函数。)

bool equal(
InputIterator1 first1, 
InputIterator1 last1, 
InputIterator2 first2, 
BinaryPredicate pred);

Using the above syntax the elements of the corresponding ranges are checked via the predicate.

使用上述语法，可以通过谓词检查相应范围的元素。

So, as we found out the last iterator for the second range not being shared as it will compare only the same number of elements as of range 1. Also, it will check sequentially, which means [1,3,5] and [5,3,1] are not the same. So it has to be in the same order for both the range.

因此，我们发现第二个范围的最后一个迭代器没有共享，因为它只会比较范围1相同数量的元素。而且，它将顺序检查，这意味着[1、3、5]和[5] ,, 3,1]不一样。因此，两个范围的顺序必须相同。

1)使用默认的'=='/'！='运算符 (1) Using default '==' / '!=' operator)

#include <bits/stdc++.h>
using namespace std;
int main()
{
vector<int> arr1{ 3, 2, 1, 4, 5, 6, 7 };
vector<int> arr2{ 3, 2, 1, 4, 5 };
if (equal(arr1.begin(), arr1.begin() + 5, arr2.begin())) {
cout << "both ranges are exactly equal \n";
}
else
cout << "both ranges are not exactly equal \n";
vector<int> arr3{ 1, 2, 3, 4, 5, 6, 7 };
vector<int> arr4{ 1, 2, 3, 4, 5 };
if (equal(arr3.begin(), arr3.end(), arr4.begin())) {
cout << "both ranges are exactly equal \n";
}
else
cout << "both ranges are not exactly equal \n";
return 0;
}

Output:

输出：

both ranges are exactly equal
both ranges are not exactly equal

In the above program, we have checked two cases and have used the default comparator. In the first case,

在上面的程序中，我们检查了两种情况，并使用了默认的比较器。在第一种情况下，

The first range is arr1.begin() to arr1.begin()+5, i.e., only first five elements of arr1. The second range starts from arr2.begin() and it will check the first five elements only from the starting of range2. Since both are the same thus it's a match.

第一个范围是arr1.begin()到arr1.begin()+ 5 ，即arr1的仅前五个元素。第二个范围从arr2.begin()开始， 它将仅从 range2的开始检查前五个元素。由于两者相同，因此是匹配项。

[3,2,1,4,5]

In the second case since we went for the total range of arr3, thus it was a mismatch.

在第二种情况下，由于我们使用了arr3的总范围，因此这是不匹配的。

2)使用用户定义的比较器功能 (2) Using user-defined comparator function)

Here we have taken a use case where we have two vectors for student details with five students each. By using our user-defined predict we are going to check whether both of the lists are equal or not. Both list are said to be equal if each of the student's details matches. To have a match for student details, all the details (roll, name, score) need to be the same.

在这里，我们以一个用例为例，我们有两个向量用于学生详细信息，每个向量有五个学生。通过使用用户定义的预测，我们将检查两个列表是否相等。如果每个学生的详细信息都匹配，则两个列表都相等。要使学生的详细信息匹配，所有详细信息(名次，姓名，分数)必须相同。

#include <bits/stdc++.h>
using namespace std;
class student {
int score;
int roll;
string name;
public:
student()
{
score = 0;
roll = 0;
name = "";
}
student(int sc, int ro, string nm)
{
score = sc;
roll = ro;
name = nm;
}
int get_score()
{
return score;
}
int get_roll()
{
return roll;
}
string get_name()
{
return name;
}
};
bool pred(student a, student b)
{
//if all details are same return true else false
if (a.get_name() == b.get_name() && a.get_score() == b.get_score() && a.get_roll() == b.get_roll())
return true;
return false;
}
int main()
{
//1st list
vector<student> arr1(5);
//1st student
arr1[0] = student(80, 5, "XYZ"); //roll 5, marks 80
//2nd student
arr1[1] = student(70, 10, "INC"); //roll 10, marks 70
//3rd student
arr1[2] = student(85, 7, "HYU"); //roll 7, marks 85
//4th student
arr1[3] = student(83, 1, "EFG"); //roll 1,  marks 83
//5th student
arr1[4] = student(81, 11, "ABC"); //roll 11,  marks 81
//2nd list
vector<student> arr2(5);
//1st student
arr2[0] = student(80, 5, "XYZ"); //roll 5, marks 80
//2nd student
arr2[1] = student(70, 10, "INC"); //roll 10, marks 70
//3rd student
arr2[2] = student(85, 7, "HYU"); //roll 7, marks 85
//4th student
arr2[3] = student(83, 1, "EFG"); //roll 1,  marks 83
//5th student
arr2[4] = student(81, 11, "ABC"); //roll 11,  marks 81
//find check both lists are equal or not 
//based on user-defined predicate
if (equal(arr1.begin(), arr1.end(), arr2.begin(), pred))
cout << "Both lists arr1,arr2 are equal\n";
else
cout << "Both lists arr1,arr2 are not equal\n";
//3rd list
vector<student> arr3(5);
//1st student
arr3[0] = student(89, 5, "PVR"); //roll 5, marks 89
//2nd student
arr3[1] = student(70, 10, "INC"); //roll 10, marks 70
//3rd student
arr3[2] = student(85, 7, "HYU"); //roll 7, marks 85
//4th student
arr3[3] = student(83, 1, "EFG"); //roll 1,  marks 83
//5th student
arr3[4] = student(81, 11, "ABC"); //roll 11,  marks 81
//find check both lists are equal or not based 
//on user-defined predicate
if (equal(arr1.begin(), arr1.end(), arr3.begin(), pred))
cout << "Both lists arr1,arr3 are equal\n";
else
cout << "Both lists arr1,arr3 are not equal\n";
return 0;
}

Output:

输出：

Both lists arr1,arr2 are equal
Both lists arr1,arr3 are not equal

Here we have first created two lists with the same elements. Since both lists are of equal size that's why we found equal to return true. For the second case, we have altered an element in arr3 to make unequal and the same reflected in the output. In our user-defined predicate, we have returned true all if all details matched for two students.

在这里，我们首先创建了两个具有相同元素的列表。由于两个列表的大小相等，因此我们发现等于返回true。对于第二种情况，我们更改了arr3中的元素以使其不相等，并且在输出中反映出相同的内容。在我们用户定义的谓词中，如果所有细节都匹配两个学生，则我们返回true。

So in this article, you saw how efficiently we can use equal to check two ranges are equal or not. An application of this can be checking whether an array is subarray of the other one or not.

因此，在本文中，您看到了我们可以使用equal来检查两个范围是否相等的效率。这种方法的应用可以是检查一个数组是否是另一个数组的子数组。