c++ - 将 4 个值编码为 32 位整数

Question

我正在尝试创建一个将 4 个值编码为一个 32 位整数的类。到目前为止，它保存并输出了所有值而没有丢失任何东西，但是由于某种原因，顺序错误。

struct encoder {
    uint32_t val;

    encoder(uint32_t _val = 0) : val(_val) {}

    uint32_t first(uint32_t v = 0) {
        if (!v) return (val << (8*3)) >> (8*3);

        val |= v;
    }

    uint32_t second(uint32_t v = 0) {
        if (!v) return (val << (8*2)) >> (8*3);

        encoder _backupval(val);

        val = (val >> (8*1));
        val |= v;
        val = (val << (8*1));
        val |= _backupval.first();
    }

    uint32_t third(uint32_t v = 0) {
        if (!v) return (val << (8*1)) >> (8*3);

        encoder _backupval(val);

        val = (val >> (8*2));
        val |= v;
        //now restore
        val = val << 8;
        val |= _backupval.second();
        val = val << 8;
        val |= _backupval.first();
    }

    uint32_t fourth(uint32_t v = 0) {
        if (!v) return (val << (8*0)) >> (8*3);

        encoder _backupval(val);

        val = (val >> (8*3));
        val |= v;
        //now restore
        val = val << 8;
        val |= _backupval.second();
        val = val << 8;
        val |= _backupval.first();
        val = val << 8;
        val |= _backupval.third();
    }
};

int main() {
    encoder t;
    t.first(6);
    t.second(42);
    t.third(212);
    t.fourth(23);

    cout << "first number: " << t.first()
         << "\nsecond number: " << t.second()
         << "\nthird number: " << t.third()
         << "\nfourth number: " << t.fourth() << endl;

}

请注意，我这样做是作为练习，并且该程序不会在实际程序中使用。所以不要提出替代方法，也不要指出缺陷（比如如果数字的大小超过 256，它会破坏另一个整数等）

Answer 1

您混合了函数的顺序fourth：

uint32_t fourth(uint32_t v = 0) {
    if (!v) return (val << (8*0)) >> (8*3);

    encoder _backupval(val);

    val = (val >> (8*3));
    val |= v;
    //now restore
    val = val << 8;
    val |= _backupval.second();
    val = val << 8;
    val |= _backupval.first();
    val = val << 8;
    val |= _backupval.third();  // this should be called first!
}

这将导致加扰的返回值。只需third先打电话，它就会工作：

uint32_t fourth(uint32_t v = 0) {
    if (!v) return (val << (8*0)) >> (8*3);

    encoder _backupval(val);

    val = (val >> (8*3));
    val |= v;
    //now restore
    val = val << 8;
    val |= _backupval.third();
    val = val << 8;
    val |= _backupval.second();
    val = val << 8;
    val |= _backupval.first();
}