Array[n] vs Array[10] - 用变量 vs 实数初始化数组
我的代码存在以下问题:
I am having the following issue with my code:
int n = 10;
double tenorData[n] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
返回以下错误:
error: variable-sized object 'tenorData' may not be initialized
而使用 double tenorData[10] 有效.
有人知道为什么吗?
推荐答案
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 关于遵循 C++ 标准)，您可以这样做:
In C++, variable length arrays are not legal. G++ allows this as an "extension" (because C allows it), so in G++ (without being -pedantic about following the C++ standard), you can do:
int n = 10;
double a[n]; // Legal in g++ (with extensions), illegal in proper C++
如果您想要一个可变长度数组"(在 C++ 中更好地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)，您要么必须自己动态分配内存:
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:
int n = 10;
double* a = new double[n]; // Don't forget to delete [] a; when you're done!
或者，更好的是，使用标准容器:
Or, better yet, use a standard container:
int n = 10;
std::vector<double> a(n); // Don't forget to #include <vector>
如果你仍然想要一个合适的数组，你可以在创建它时使用常量，而不是变量:
If you still want a proper array, you can use a constant, not a variable, when creating it:
const int n = 10;
double a[n]; // now valid, since n isn't a variable (it's a compile time constant)
同样，如果你想从 C++11 中的函数中获取大小，你可以使用 constexpr:
Similarly, if you want to get the size from a function in C++11, you can use a constexpr:
constexpr int n()
{
return 10;
}
double a[n()]; // n() is a compile time constant expression
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