还有TR1/ c++ 11/ c++ 17方式(参见Coliru Live):

const std::string s[3] = { "1"s, "2"s, "3"s };
constexpr auto n       = std::extent<   decltype(s) >::value; // From <type_traits>
constexpr auto n2      = std::extent_v< decltype(s) >;        // C++17 shorthand

const auto     a    = std::array{ "1"s, "2"s, "3"s };   // C++17 class template arg deduction -- http://en.cppreference.com/w/cpp/language/class_template_argument_deduction
constexpr auto size = std::tuple_size_v< decltype(a) >;

std::cout << n << " " << n2 << " " << size << "\n"; // Prints 3 3 3

只是一个想法，但只是决定创建一个计数器变量，并将数组大小存储在位置[0]。我删除了函数中的大部分代码，但你会看到退出循环后，prime[0]被分配为'a'的最终值。我尝试使用矢量，但VS Express 2013不太喜欢。还要注意，'a'从1开始，以避免覆盖[0]，并在开始时初始化以避免错误。我不是专家，只是想分享一下。

int prime[] = {0};
int primes(int x, int y){
    using namespace std; int a = 1;
    for (int i = x; i <= y; i++){prime[a] = i; a++; }
    prime[0] = a; return 0;
}

假设在页面顶部声明了一个全局数组

int global[] = { 1, 2, 3, 4 };

要找出数组中有多少个元素(在c++中)，输入以下代码:

sizeof(global) / 4;

sizeof(NAME_OF_ARRAY) / 4将返回给定数组名的元素数量。

给你:

#include <iostream>
using namespace std;

int main() {
 int arr[] = {10,20,30,40,50,60};
 int arrSize = sizeof(arr)/sizeof(arr[0]);
 cout << "The size of the array is: " << arrSize;
return 0;
}

下面是谷歌Protobuf中ArraySize的一个实现。

#define GOOGLE_ARRAYSIZE(a) \
  ((sizeof(a) / sizeof(*(a))) / static_cast<size_t>(!(sizeof(a) % sizeof(*(a)))))

// test codes...
char* ptr[] = { "you", "are", "here" };
int testarr[] = {1, 2, 3, 4};
cout << GOOGLE_ARRAYSIZE(testarr) << endl;
cout << GOOGLE_ARRAYSIZE(ptr) << endl;

ARRAYSIZE(arr) works by inspecting sizeof(arr) (the # of bytes in the array) and sizeof(*(arr)) (the # of bytes in one array element). If the former is divisible by the latter, perhaps arr is indeed an array, in which case the division result is the # of elements in the array. Otherwise, arr cannot possibly be an array, and we generate a compiler error to prevent the code from compiling. Since the size of bool is implementation-defined, we need to cast !(sizeof(a) & sizeof(*(a))) to size_t in order to ensure the final result has type size_t. This macro is not perfect as it wrongfully accepts certain pointers, namely where the pointer size is divisible by the pointee size. Since all our code has to go through a 32-bit compiler, where a pointer is 4 bytes, this means all pointers to a type whose size is 3 or greater than 4 will be (righteously) rejected.

正如其他人所说，您可以使用sizeof(arr)/sizeof(*arr)，但这将为非数组的指针类型提供错误的答案。

template<class T, size_t N>
constexpr size_t size(T (&)[N]) { return N; }

这具有对非数组类型编译失败的良好属性(Visual Studio有_countof可以做到这一点)。constexpr使它成为编译时表达式，因此它没有任何宏的缺点(至少我不知道)。

你也可以考虑使用c++ 11中的std::array，它公开了它的长度，而不增加C原生数组的开销。

c++ 17在<iterator>头文件中有std::size()，它具有相同的功能，也适用于STL容器(感谢@Jon C)。