C++ Array[n] vs Array[10] - 用变量 vs 实数初始化数组
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Array[n] vs Array[10] - Initializing array with variable vs real number
提问by msmf14
I am having the following issue with my code:
我的代码存在以下问题：
int n = 10;
double tenorData[n] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
Returns the following error:
返回以下错误：
error: variable-sized object 'tenorData' may not be initialized
Whereas using double tenorData[10]works.
而使用double tenorData[10]作品。
Anyone know why?
有谁知道为什么？
回答by Cornstalks
In C++, variable length arrays are not legal. G++ allows this as an "extension" (because C allows it), so in G++ (without being -pedanticabout following the C++ standard), you can do:
在 C++ 中，变长数组是不合法的。G++ 允许将其作为“扩展”（因为 C 允许），因此在 G++ 中（无需-pedantic遵循 C++ 标准），您可以执行以下操作：
int n = 10;
double a[n]; // Legal in g++ (with extensions), illegal in proper C++
If you want a "variable length array" (better called a "dynamically sized array" in C++, since proper variable length arrays aren't allowed), you either have to dynamically allocate memory yourself:
如果您想要一个“可变长度数组”（在 C++ 中最好称为“动态大小的数组”，因为不允许使用适当的可变长度数组），您必须自己动态分配内存：
int n = 10;
double* a = new double[n]; // Don't forget to delete [] a; when you're done!
Or, better yet, use a standard container:
或者，更好的是，使用标准容器：
int n = 10;
std::vector<double> a(n); // Don't forget to #include <vector>
If you still want a proper array, you can use a constant, not a variable, when creating it:
如果你仍然想要一个合适的数组，你可以在创建它时使用一个constant，而不是一个variable：
const int n = 10;
double a[n]; // now valid, since n isn't a variable (it's a compile time constant)
Similarly, if you want to get the size from a function in C++11, you can use a constexpr:
同样，如果要从 C++11 中的函数获取大小，可以使用constexpr：
constexpr int n()
{
return 10;
}
double a[n()]; // n() is a compile time constant expression
