C++ 如何检查数组中是否存在给定的 int？
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How can I check if given int exists in array?
提问by rsk82
For example, I have this array:
例如，我有这个数组：
int myArray[] = { 3, 6, 8, 33 };
How to check if given variable x is in it?
如何检查给定的变量 x 是否在其中？
Do I have to write my own function and loop the array, or is there in modern c++ equivalent to in_arrayin PHP?
我是否必须编写自己的函数并循环数组，或者在现代 C++ 中是否有与in_arrayPHP等效的函数？
回答by juanchopanza
You can use std::findfor this:
您可以std::find为此使用：
#include <algorithm> // for std::find
#include <iterator> // for std::begin, std::end
int main ()
{
int a[] = {3, 6, 8, 33};
int x = 8;
bool exists = std::find(std::begin(a), std::end(a), x) != std::end(a);
}
std::findreturns an iterator to the first occurrence of x, or an iterator to one-past the end of the range if xis not found.
std::findx如果x未找到，则返回指向第一次出现的的迭代器，或者返回超出范围末尾的迭代器。
回答by Zac Howland
I think you are looking for std::any_of, which will return a true/false answer to detect if an element is in a container (array, vector, deque, etc.)
我认为您正在寻找std::any_of，它将返回真/假答案以检测元素是否在容器中（数组、向量、双端队列等）
int val = SOME_VALUE; // this is the value you are searching for
bool exists = std::any_of(std::begin(myArray), std::end(myArray), [&](int i)
{
return i == val;
});
If you want to know where the element is, std::findwill return an iterator to the first element matching whatever criteria you provide (or a predicate you give it).
如果您想知道元素在哪里，std::find将返回一个迭代器到匹配您提供的任何条件（或您提供的谓词）的第一个元素。
int val = SOME_VALUE;
int* pVal = std::find(std::begin(myArray), std::end(myArray), val);
if (pVal == std::end(myArray))
{
// not found
}
else
{
// found
}
回答by John Dibling
You almost never have to write your own loops in C++. Here, you can use std::find.
您几乎不需要用 C++ 编写自己的循环。在这里，您可以使用std::find。
const int toFind = 42;
int* found = std::find (myArray, std::end (myArray), toFind);
if (found != std::end (myArray))
{
std::cout << "Found.\n"
}
else
{
std::cout << "Not found.\n";
}
std::endrequires C++11. Without it, you can find the number of elements in the array with:
std::end需要 C++11。没有它，您可以使用以下命令找到数组中的元素数：
const size_t numElements = sizeof (myArray) / sizeof (myArray[0]);
...and the end with:
...最后是：
int* end = myArray + numElements;
回答by Duncan Smith
Try this
尝试这个
#include <iostream>
#include <algorithm>
int main () {
int myArray[] = { 3 ,6 ,8, 33 };
int x = 8;
if (std::any_of(std::begin(myArray), std::end(myArray), [=](int n){return n == x;})) {
std::cout << "found match/" << std::endl;
}
return 0;
}
}
回答by Neil Kirk
int index = std::distance(std::begin(myArray), std::find(begin(myArray), end(std::myArray), VALUE));
Returns an invalid index (length of the array) if not found.
如果未找到，则返回无效索引（数组的长度）。
回答by zirror
You do need to loop through it. C++ does not implement any simpler way to do this when you are dealing with primitive type arrays.
你确实需要遍历它。当您处理原始类型数组时，C++ 没有实现任何更简单的方法来执行此操作。
also see this answer: C++ check if element exists in array
另请参阅此答案：C++ 检查数组中是否存在元素
