如何让 gcc 警告将错误的枚举传递给函数 [英] How to make gcc warn about passing wrong enum to a function

问题描述

gcc 似乎没有用以下代码产生警告.我怎样才能让它产生警告?

gcc doesn't seem to produce a warning with the following code. How can I get it to produce a warning?

typedef enum
{
    REG8_A,
    REG8_B,
    REG8_C
}REG8;

typedef enum
{
    REG16_A,
    REG16_B,
    REG16_C
}REG16;

void function(REG8 reg8)
{

}

int main(void)
{
    function(REG16_A);    // Should warn about wrong enum
}

推荐答案

这种行为的原因是您使用的是 C 编译器而不是 C++.并且 在 C 中 枚举类型并不是真正的类型，C 中的枚举仅包含 int 常量和它们可以与任何整数和任何算术自由混合.

The reason of such behaviour is that you are using C compiler rather than C++. And in C enum types are not really types, enums in C just hold int constants and they can be freely mixed with whatever integers and whatever arithmetic.

在 C++ 中，您有真正的枚举，正如您所想的那样，并且按照语言标准的规定进行所需的类型检查.

In C++ instead you have real enums as you would like to think of them, and the needed typechecking occurs as conveyed by the language standard.

您的问题可以通过两种方式解决:

Your problem can be resolved in two ways:

• 使用 C++ 编译器.

• Use C++ compiler.

通过这种方式，您将拥有真正的枚举，如您所愿.

This way you will have real enums, as you want them.

将您的代码更改为纯 C 风格，即不使用枚举，因为在 C 中它们只是常量集，编译器仅帮助您对常量值进行排序.而在 C 中，you 将负责保持类型"传递的常量一致.再说一次:对于 C，枚举成员只是 int 常量，你不能让它们输入.

Change your code to be in pure-C style, i.e. not using enums, as in C they are merely constant sets where the compiler only helps you to order the constant values. And in C you will be the one responsible to keep the "types" of passed constants consistent. Once again: for C, enum members are just int constants, you can't make them typed.

#define REG8_A 0
#define REG8_B 1
#define REG8_C 2

#define REG16_A 0
#define REG16_B 1
#define REG16_C 2

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