問題描述

我的代碼存在以下問題:

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};

返回以下錯誤:

error: variable-sized object 'tenorData' may not be initialized

而使用 double tenorData[10] 有效.

有人知道為什么嗎?

推薦答案

在 C++ 中，變長數(shù)組是不合法的.G++ 允許將其作為擴展"(因為 C 允許)，因此在 G++ 中(無需 -pedantic 關于遵循 C++ 標準)，您可以這樣做:

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++

如果您想要一個可變長度數(shù)組"(在 C++ 中更好地稱為動態(tài)大小的數(shù)組"，因為不允許使用適當?shù)目勺冮L度數(shù)組)，您要么必須自己動態(tài)分配內(nèi)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，使用標準容器:

Or, better yet, use a standard container:

int n = 10;
std::vector<double> a(n); // Don't forget to #include <vector>

如果你仍然想要一個合適的數(shù)組，你可以在創(chuàng)建它時使用常量，而不是變量:

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)

同樣，如果你想從 C++11 中的函數(shù)中獲取大小，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

這篇關于Array[n] vs Array[10] - 用變量 vs 實數(shù)初始化數(shù)組的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！