警告634：强类型不匹配（类型'bool'）相等或有条件

Question

This is part of my code base 
I am rying to run the below code and getting the warning like below code:

    #define assert_always()         assert(TRUE)

    base::derived(
        uint8 id;
    ){
        switch(id)
        {
        case one:
        ----;
        break;
        case two:
        ----;
        break;
        default:
        assert_always();
        break;
        }
    }

    Warning 634: Strong type mismatch (type 'bool') in equality or conditional  .......

The line number is pointing to `assert_always()` function call. Can you guide me to first to understand and then solve this warning?

I have checked by changing TRUE to true but stil having the same problem ...

Actually code was like this in c++ :
#define assert_always()

    base::derived(
        uint8 id;
    ){
        switch(id)
        {
        case one:
        ----;
        break;
        case two:
        ----;
        break;
        default:
        assert_always();
        break;
        }
    }

I got the warning Warning 634: Strong type mismatch (type 'bool') in equality or conditional so I introduce  

定义assert_always（）断言（TRUE）

但仍然发出警告，并在其他回复后修改为

定义assert_always（）断言（true）

but still warning is there ...I am not getting all the exact reason behind this ...

----; ----; line是一些功能...并且提到的代码是在C ++中

Answer 1

什么是TRUE？

尝试使用实际的布尔常量：

#define assert_always assert(false)

另请注意，assert(true) 永远不会断言，因为表达式（字面意思）为真。

Answer 2

使用true C ++ bool类型：

assert(true)

而不是TRUE

但它不会做你想要的（总是断言）。断言（假）会。