小编use*_*272的帖子

当我尝试解析JSON时,我收到一个非常奇怪的错误.实际上,文件非常简单,由一个简单的对象组成如下:

{
    "registered":false,
    "firstname":"xxx",
    "name":"yyyy",
    "email":"yyyy.xxx@gmail.com",
    "picture":"xxxxx.jpg",
    "username":"xxxy"
}

为了解析这个文件,我使用了以下代码,其灵感来自Android SDK的示例:

public static boolean isRegistered(int nmb) {
    boolean toReturn = true;
    JsonReader reader = null;
    try {
        reader = new JsonReader(new InputStreamReader(new URL("xxx").openConnection().getInputStream()));
        reader.beginObject();
        while(reader.hasNext()) {
            String name = reader.nextName();
            Log.i("Next value", name);
            switch (name) {
                case "registered":
                    toReturn = reader.nextBoolean();
                    break;
                case "firstname":
                    ProfileManager.getInstance().setFirstname(reader.nextString());
                    break;
                case "name":
                    ProfileManager.getInstance().setName(reader.nextString());
                    break;
                case "email":
                    break;
                case "picture":
                    break;
                case "username":
                    break;
            }
        }
        reader.endObject();
    } catch (FileNotFoundException e) {
        e.printStackTrace();
    } catch …

我对按位运算所需的周期数有疑问,或者更确切地说,是XOR运算.在我的程序中,我有两个uint8_t变量的一维数组,固定大小为8.我想对两个数组进行异或,我想知道最有效的方法是什么.这是一个总结我发现的选项的代码:

int main() {
    uint8_t tab[4] = {1,0,0,2};
    uint8_t tab2[4] = {2,3,4,1};

    /* First option */

    uint8_t tab3[4] = {tab[0]^tab2[0], tab[1]^tab2[1], tab[2]^tab2[2], tab[3]^tab2[3]};

    /* Second option */

    uint32_t* t = tab; 
    uint32_t* t2 = tab2;

    uint32_t t3 = *t ^ *t2;
    uint8_t* tab4 = &t3;

    /* Comparison */ 

    printf("%d & %d\n", tab3[0], tab4[0]);
    printf("%d & %d\n", tab3[1], tab4[1]);
    printf("%d & %d\n", tab3[2], tab4[2]);
    printf("%d & %d\n", tab3[3], tab4[3]);

    return 0;
}

从循环/字节的角度来看,最佳选择是什么？