C++ 数组 - 表达式必须有一个常量值
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9219712/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
c++ array - expression must have a constant value
提问by Nicholas Kong
I get an error when I try to create an array from the variables I declared.
当我尝试从我声明的变量创建数组时出现错误。
int row = 8;
int col= 8;
int [row][col];
Why do I get this error:
为什么我会收到此错误：
expression must have a constant value.
表达式必须有一个常量值。
回答by Foggzie
When creating an array like that, its size must be constant. If you want a dynamically sized array, you need to allocate memory for it on the heap and you'll also need to free it with deletewhen you're done:
创建这样的数组时，它的大小必须是恒定的。如果你想要一个动态大小的数组，你需要在堆上为它分配内存，你还需要delete在完成后释放它：
//allocate the array
int** arr = new int*[row];
for(int i = 0; i < row; i++)
arr[i] = new int[col];
// use the array
//deallocate the array
for(int i = 0; i < row; i++)
delete[] arr[i];
delete[] arr;
If you want a fixed size, then they must be declared const:
如果您想要固定大小，则必须将它们声明为 const：
const int row = 8;
const int col = 8;
int arr[row][col];
Also,
还，
int [row][col];
doesn't even provide a variable name.
甚至不提供变量名。
回答by Nicholas Kong
The standard requires the array length to be a value that is computable at compile time so that the compiler is able to allocate enough space on the stack. In your case, you are trying to set the array length to a value that is unknown at compile time. Yes, i know that it seems obvious that it should be known to the compiler, but this is not the case here. The compiler cannot make any assumptions about the contents of non-constant variables. So go with:
该标准要求数组长度是一个在编译时可计算的值，以便编译器能够在堆栈上分配足够的空间。在您的情况下，您试图将数组长度设置为编译时未知的值。是的，我知道编译器应该知道它似乎很明显，但这里的情况并非如此。编译器不能对非常量变量的内容做出任何假设。所以去吧：
const int row = 8;
const int col= 8;
int a[row][col];
UPD: some compilers will actually allow you to pull this off. IIRC, g++ has this feature. However, never use it because your code will become un-portable across compilers.
UPD：一些编译器实际上会允许你把它拉下来。IIRC，g++ 有这个功能。但是，永远不要使用它，因为您的代码将变得不可跨编译器移植。
回答by Mark Ransom
C++ doesn't allow non-constant values for the size of an array. That's just the way it was designed.
C++ 不允许数组大小的非常量值。这就是它的设计方式。
C99 allows the size of an array to be a variable, but I'm not sure it is allowed for two dimensions. Some C++ compilers (gcc) will allow this as an extension, but you may need to turn on a compiler option to allow it.
C99 允许数组的大小是一个变量，但我不确定它是否允许用于二维。某些 C++ 编译器 (gcc) 将允许将其作为扩展，但您可能需要打开编译器选项以允许它。
And I almost missed it - you need to declare a variable name, not just the array dimensions.
我几乎错过了 - 你需要声明一个变量名，而不仅仅是数组维度。
回答by sjtupuzhao
You can use #define as an alternative solution, which do not introduce vector and malloc, and you are still using the same syntax when defining an array.
您可以使用 #define 作为替代解决方案，它不会引入 vector 和 malloc，并且您在定义数组时仍然使用相同的语法。
#define row 8
#define col 8
int main()
{
int array_name[row][col];
}
回答by arrdem
When you declare a variable as here
当你在这里声明一个变量时
int a[10][10];
you are telling the C++ compiler that you want 100 consecutive integers allocated in the program's memory at runtime. The compiler will then provide for your program to have that much memory available and all is well with the world.
您告诉 C++ 编译器您希望在运行时在程序内存中分配 100 个连续整数。然后编译器将为您的程序提供那么多可用内存，并且一切都很好。
If however you tell the compiler
但是，如果您告诉编译器
int x = 9001;
int y = 5;
int a[x][y];
the compiler has no way of knowing how much memory you are actually going to need at run time without doing a lot of very complex analysis to track down every last place where the values of x and y changed [if any]. Rather than support such variable size arrays, C++ and C strongly suggest if not outright demand that you use malloc() to manually allocate the space you want.
编译器无法知道您在运行时实际需要多少内存，而无需进行大量非常复杂的分析来追踪 x 和 y 的值发生变化的每个最后位置 [如果有的话]。C++ 和 C 强烈建议您使用 malloc() 手动分配所需的空间，而不是支持这种可变大小的数组。
TL;DR
TL; 博士
int x = 5;
int y = 5;
int **a = malloc(x*sizeof(int*));
for(int i = 0; i < y; i++) {
a[i] = malloc(sizeof(int*)*y);
}
a is now a 2D array of size 5x5 and will behave the same as int a[5][5]. Because you have manually allocated memory, C++ and C demand that you delete it by hand too...
a 现在是大小为 5x5 的二维数组，其行为与 int a[5][5] 相同。因为您手动分配了内存，C++ 和 C 也要求您手动删除它...
for(int i = 0; i < x; i++) {
free(a[i]); // delete the 2nd dimension array
}
free(a); // delete a itself
回答by BigChief
one could also use a fixed lengths vector and access it with indexing
还可以使用固定长度的向量并通过索引访问它
int Lcs(string a, string b)
{
int x = a.size() + 1;
int y = b.size() + 1;
vector<vector<int>> L(x, vector<int>(y));
for (int i = 1; i < x; i++)
{
for (int j = 1; j < y; j++)
{
L[i][j] = a[i - 1] == b[j - 1] ?
L[i - 1][j - 1] + 1 :
max(L[i - 1][j], L[i][j - 1]);
}
}
return L[a.size()][b.size()];
}
回答by BigChief
No it doesn't need to be constant, the reason why his code above is wrong is because he needs to include a variable name before the declaration.
不，它不需要是常量，他上面的代码错误的原因是因为他需要在声明之前包含一个变量名。
int row = 8;
int col= 8;
int x[row][col];
In Xcode that will compile and run without any issues, in M$ C++ compiler in .NET it won't compile, it will complain that you cannot use a non const literal to initialize array, the size needs to be known at compile time
在编译和运行没有任何问题的 Xcode 中，在 .NET 中的 M$ C++ 编译器中它不会编译，它会抱怨你不能使用非 const 文字来初始化数组，需要在编译时知道大小
