在C ++中返回已知大小的数组？

If I can pass in an array of a known size:

如果我可以传入已知大小的数组:

void fn(int(*intArray)[4])
{
    (*intArray)[0] = 7;
}

why can't I return one:

为什么我不能归还一个:

int intArray[4] = {0};
int(*)[4] fn()
{
    return &intArray;
}

here, the ")" in "(*)" generates "syntax error : )".

这里,“(*)”中的“)”生成“语法错误:)”。

3 个解决方案

#1

The [4] goes after the function name, just like it goes after the variable name in a variable definition:

[4]在函数名之后,就像它在变量定义中的变量名之后:

int (*fn())[4]
{
    return &intArray;
}

Since this is very obscure syntax, prone to be confusing to everybody who reads it, I would recommend to return the array as a simple int*, if you don't have any special reason why it has to be a pointer-to-array.

由于这是非常模糊的语法,容易让每个读它的人感到困惑,我建议将数组作为一个简单的int *返回,如果你没有任何特殊的原因,为什么它必须是一个指向数组的指针。

You could also simplify the function definition with a typedef:

您还可以使用typedef简化函数定义:

typedef int intarray_t[4];

intarray_t* fn() { ... }

#2

Its not allowed to return arrays from functions, but there are workarounds.

它不允许从函数返回数组,但有解决方法。

For example, the following code:

例如,以下代码:

int fn()[4] {
   ...

Doesn't get accepted by the various online compilers; I tried it on the Comeau online compiler, which is considered to be one of the most standard-following of compilers, and even in relaxed mode it says:

不被各种在线编译器接受;我在Comeau在线编译器上尝试过它,它被认为是最标准的编译器之一,甚至在放松模式下它说:

error: function returning array is not allowed

错误:不允许函数返回数组

Another poster suggests returning int**; this will work, but be very careful that you return heap-allocated memory and not stack-allocated.

另一张海报建议返回int **;这将工作,但要小心你返回堆分配的内存而不是堆栈分配。

#3

You can return int**

你可以返回int **

#1