在Debug模式下中断, 在Release模式下跳出当前函数的断言

Posted 2020-10-02 claireyuancy

在Debug模式下中断, 在Release模式下跳出当前函数的断言

#ifdef DEBUG
#define __breakPoint_on_debug asm("int3")
#else
#define __breakPoint_on_debug
#endif

// 验证
#define UXY_ASSERT_RETURN_ON_RELEASE( __condition, __desc, ... ) \
        metamacro_if_eq(0, metamacro_argcount(__VA_ARGS__))         (__XY_ASSERT_1(__condition, __desc, __VA_ARGS__))            (__XY_ASSERT_2(__condition, __desc, __VA_ARGS__))

#define __XY_ASSERT_1( __condition, __desc ) \
        if ( !(__condition) ) __breakPoint_on_debug;           else return;

#define __XY_ASSERT_2( __condition, __desc, __returnedValue ) \
        if ( !(__condition) ) __breakPoint_on_debug;           else return __returnedValue;