c++ - 整数的按位运算在项目的 DEBUG 和 RELEASE 版本中表现不同

Question

我在 Windows 8 64 位上使用 Visual Studio 2012。我有这两种方法:

void Node::setState(States state, bool value) {
    if (value) {
        states_ |= 1 << state;
    }
    else {
        states_ &= ~(1 << state);
    }
}

bool Node::getState(States state) {
    return (states_ & (1 << state)) != 0;
}

状态枚举:

enum States {
    UPDATABLE = 0x01, // Node data update.
    RENDERABLE = 0x02, // Node rendering on screen.
    TRANSFORMABLE = 0x04, // Position, rotation and scaling update.
    POSITION = 0x08, // Position update.
    ROTATION = 0x10, // Rotation update.
    SCALING = 0x20 // Scaling update.
};

声明整数初始化:

states_(TRANSFORMABLE | RENDERABLE |
            UPDATABLE | POSITION | ROTATION | SCALING)

现在问题出在这部分代码:

LOGI("SETTING SCALING FOR NODE: %s", node->getName().c_str());
node->setState(Node::SCALING, true);
if (node->getState(Node::SCALING)) {
    LOGI("NODE WILL BE UPDATED.");
}
else {
    LOGI("NODE WILL NOT BE UPDATED.");
}

在 Debug模式下我得到“节点将被更新”，在发布时我得到“节点将不会被更新”。这是什么原因造成的？

Answer 1

SCALING 是一个 32 位有符号整数，值为 0x20。您将 1 左移 0x20 (32) 位位置。这意味着您将它移动得太远，导致未定义的行为。

标准相关引述:

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1× 2E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1× 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.