定义不评估 POD？
我正在在线查看 C++ FAQ Lite。我再次浏览内联，因为我还没有找到它们的用途，并且想知道如何停止循环依赖，如 这个答案。我首先尝试做“为什么内联比定义更好”。具有以下代码的示例：
#define unsafe(i) \
( (i) >= 0 ? (i) : -(i) )
inline
int safe(int i)
{
return i >= 0 ? i : -(i);
}
int f();
int main(void)
{
int x(5);
int ans;
ans = unsafe(x++);
cout << ans << endl;
ans = unsafe(++x);
cout << ans << endl;
ans = safe(x++);
cout << ans << endl;
ans = safe(++x);
cout << ans << endl;
std::cin.get();
return 0;
}
编辑：
太棒了。解决了拼写错误。并不是说我因为没有发现这样的错误或任何东西而感到痛苦。
现在输出为 6, 9, 9, 11。
然而，即使预自增，第一个值不应该结果为 7 吗？
如果宏被调用两次，那么它不是像这样：
unsafe(x) // 预递增在调用时不会修改值。
unsafe(++x) // 出于所有意图和目的，增量发生在第二次调用之前，因此 ++x.这是针对第一个 ans = unsafe(x++)（如果它被调用两次的话）。
当我们到达第二个 ans = unsafe(++x) 时，x 不应该增加两次吗？一次是通过双重调用，一次是在第一次双重调用完成时？
I was going over the C++ FAQ Lite online. I was browsing inlines again since I haven't found a use for them and wanted to know how the stopped the circular dependency as showed in this answer. I first tried to do the, "Why inlines are better than defines." example with the following code:
#define unsafe(i) \
( (i) >= 0 ? (i) : -(i) )
inline
int safe(int i)
{
return i >= 0 ? i : -(i);
}
int f();
int main(void)
{
int x(5);
int ans;
ans = unsafe(x++);
cout << ans << endl;
ans = unsafe(++x);
cout << ans << endl;
ans = safe(x++);
cout << ans << endl;
ans = safe(++x);
cout << ans << endl;
std::cin.get();
return 0;
}
EDIT:
Great. Got the typo out of the way. Not that I'm bitter that I don't find such errors or anything.
The output is now 6, 9, 9, 11.
However, even with pre-incrementation, shouldn't the first value result in 7?
If the macro is being called twice, then doesn't it go like this:
unsafe(x) // pre-incrementation doesn't modify the value when called.
unsafe(++x) // for all intents and purposes, the incrementation happens before the second call, so the ++x. This is for the first ans = unsafe(x++) if it's being called twice.
By the time we reach the second ans = unsafe(++x), shouldn't the x have been incremented twice? Once by the double call and once when the first double call was finished?
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而不是：
我认为您想要：
响应您的编辑：
在第一次调用
unsafe(x++)后，x是 7，即使ans< /code> 是 6。这是因为您有这样的语句：ans被分配给最左边的之后的中间被评估。结果是，x++x++ans == 6但x == 7。与unsafe(++x)的区别在于，ans被赋值给++x，意味着结果是ans == x == 9 。Instead of:
I think you want:
In response to your edit:
After the first call to
unsafe(x++),xis 7, even though theansis 6. This is because you have the statement:ansis assigned to the middlex++after the left-mostx++is evaluated. As a result,ans == 6butx == 7. The difference withunsafe(++x)is thatansis assigned to++x, meaning the result isans == x == 9.