初始化程序列表作为数组的函数参数传递 [英] Initialiser list passed as function parameter for array

问题描述

我如何进行这项工作：

void foo(uint8_t a[]) { ... }

foo({0x01, 0x02, 0x03});

这给了我一个错误：

error: cannot convert '<brace-enclosed initializer list>' to 'uint8_t* {aka unsigned char*}' for argument '1'
                                                     ^

推荐答案

到目前为止，答案尚未解决以下主要问题：签名

The answers so far haven't addressed the main problem with the question: In the signature

void foo(uint8_t a[])

a 不是数组，而是指向 uint8_t 的指针。尽管事实上 a 的声明使它看起来像一个数组。错误消息甚至指出了这一点：

a is not an array, but a pointer to a uint8_t. This is despite the fact that the declaration of a makes it look like an array. This is even pointed out by the error message:

cannot convert '<brace-enclosed initializer list>' to 'uint8_t* {aka unsigned char*}'

因此，同样不允许您执行以下操作：

So, in the same way you are not allowed to do this:

uint8_t *a = {0x01, 0x02, 0x03}; // Eek! Error

您不能调用 foo（{0x01，0x02，0x03} ）; 具有上面的签名。

You can't call foo({0x01, 0x02, 0x03}); With the signature above.

我建议您花一些时间阅读C风格的数组以及它们的用法。

I suggest you take some time to read up on C-style arrays and how they are not first-class citizens in C++.

从您对自己的问题发布的答案来看，您似乎正在寻找一种适用于固定大小数组的函数。但是，不要通过价值来传递它！我建议使用以下声明：

From the answer you posted to your own question, it seems that you are looking for a function that works for fixed-size arrays. But don't pass it by value! I recommend using the following declaration:

void foo(std::array<uint8_t, 3> const &a);

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