C++ “表达式必须具有类类型”错误是什么意思？
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What does "expression must have class type" error mean?
提问by Robert Spratlin
#include <cstdlib>
using namespace std;
int main()
{
int arrayTest[512];
int size = arrayTest.size();
for(int a = 0;a<size;a++)
{
//stuff will go here
}
}
What am I doing wrong here becuase the plan is to just fill the array with some numbers
我在这里做错了什么，因为计划只是用一些数字填充数组
回答by Nawaz
Do this:
做这个：
int arrayTest[512];
int size = sizeof(arrayTest)/sizeof(*arrayTest);
C-style arrays don't have member function. They don't have any notion of class.
C 风格的数组没有成员函数。他们没有阶级观念。
Anyway, betteruse std::array:
无论如何，更好地使用std::array：
#include <array>
std::array<int,512> arrayTest;
int size = arrayTest.size(); //this line is exactly same as you wrote!
which looks like what you wanted. Now you can use index ito access elements of arrayTestas arrayTest[i]where ican vary from 0to size-1(inclusive).
这看起来像你想要的。现在您可以使用 indexi来访问arrayTestas arrayTest[i]where 的元素，i可以从0到size-1（包括）变化。
回答by Shafik Yaghmour
arrayTestis not a class or struct but an array and it does not have member functions, in this case this will get your the size of the array:
arrayTest不是一个类或结构，而是一个数组，它没有成员函数，在这种情况下，这将获得数组的大小：
size_t size = sizeof(arrayTest)/sizeof(int);
although if your compiler supports C++11than using std::arraywould be better:
尽管如果您的编译器支持C++11，那么使用std::array会更好：
#include <array>
std::array<int,512> arrayTest ;
size_t size = arrayTest.size() ;
as the document linked above shows you can also use range for loop to iterate over the elements of a std::array:
正如上面链接的文档所示，您还可以使用 range for 循环来迭代std::array的元素：
for( auto &elem : arrayTest )
{
//Some operation here
}
回答by Dark Falcon
Arrays don't have members. You have to use something like:
数组没有成员。你必须使用类似的东西：
int size = sizeof(arrayTest) / sizeof(arrayTest[0]);
Better yet, if you must work with normal arrays instead of std::array, use a helper function. This also has the advantage of not breaking when you attempt it on a pointer instead of an array:
更好的是，如果您必须使用普通数组而不是std::array，请使用辅助函数。这还具有在您尝试使用指针而不是数组时不会中断的优点：
template<int N, typename T> int array_size(T (&)[N]) {return N;}
int size = array_size(arrayTest);
回答by jimifiki
If you are stuck with arrays you can define your getArraySize function:
如果你被数组卡住了，你可以定义你的 getArraySize 函数：
template <typename T,unsigned S>
inline unsigned getArraySize(const T (&v)[S]) { return S; }
seen here: http://www.cplusplus.com/forum/general/33669/#msg181103
在这里看到：http: //www.cplusplus.com/forum/general/33669/#msg181103
std::array remains the better solution.
std::array 仍然是更好的解决方案。
