Java打印整數的二進制表示(代碼與解析)

int a=-99;
for(int i=0;i<32;i++){
        int t=(a & 0x80000000>>>i)>>>(31-i);
        System.out.print(t);
}

拆分一下這段代碼，要理解這個移位輸出的問題還需要理解的以下內容：
* 0x80000000是數的十六進制表示，轉成二進制表示為10000000000000000000000000000000
* 運算的優先級，移位運算高於邏輯運算，>>>高於&
* 位邏輯與運算 1&1 = 1 ，0&1 = 0
* >>>無符號右移，移出部分舍棄，左邊位補0；

for循環的語句塊的執行順序：
1 0x80000000 無符號右移 i位；
2 a 和 1的結果做按位與；
3 2的結果無符號右移31-i位
4 輸出3的結果

-99的機器數（補碼）表示
11111111111111111111111110011101
循環過程演示：
i = 0
1 10000000000000000000000000000000 >>> 0 = 10000000000000000000000000000000
2 11111111111111111111111110011101 & 10000000000000000000000000000000 = 10000000000000000000000000000000
3 10000000000000000000000000000000 >>> (31 - 0) = 00000000000000000000000000000001
4 輸出00000000000000000000000000000001，屏顯“1”

i = 1
1 10000000000000000000000000000000 >>> 1 = 01000000000000000000000000000000
2 11111111111111111111111110011101 & 01000000000000000000000000000000 = 01000000000000000000000000000000
3 10000000000000000000000000000000 >>> (31 - 1) = 00000000000000000000000000000001
4 輸出00000000000000000000000000000001，屏顯“1”

……

i = 30
1 10000000000000000000000000000000 >>> 30 = 00000000000000000000000000000010
2 11111111111111111111111110011101 & 00000000000000000000000000000010 = 00000000000000000000000000000000
3 00000000000000000000000000000000 >>> (31 - 30) = 00000000000000000000000000000000
4 輸出00000000000000000000000000000000，屏顯“0”

i = 31
1 10000000000000000000000000000000 >>> 31 = 00000000000000000000000000000001
2 11111111111111111111111110011101 & 00000000000000000000000000000001 = 00000000000000000000000000000001
3 00000000000000000000000000000001 >>> (31 - 31) = 00000000000000000000000000000001
4 輸出00000000000000000000000000000001，屏顯“1”

根據循環過程可以看出
(0x80000000 >>> i)的作用是mask，每次循環都與a做邏輯與操作，取a的第i位；
對上條結果>>> 31 - i作用是把取出的a的第i位移到最末位。