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.

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。

如果你指的是c风格的数组，那么你可以这样做:

int a[7];
std::cout << "Length of array = " << (sizeof(a)/sizeof(*a)) << std::endl;

这对指针不起作用(即它对以下任何一个都不起作用):

int *p = new int[7];
std::cout << "Length of array = " << (sizeof(p)/sizeof(*p)) << std::endl;

or:

void func(int *p)
{
    std::cout << "Length of array = " << (sizeof(p)/sizeof(*p)) << std::endl;
}

int a[7];
func(a);

在c++中，如果你想要这种行为，那么你应该使用容器类;可能std::向量。

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

我认为这是可行的:

for(int i=0;array[i];i++)
{
 //do_something
}  

vector有一个size()方法，该方法返回vector中元素的数量。

(是的，这是半开玩笑的回答)