為什么 Java 溢出返回 0 作為值

Question

我有以下代碼片段

public class DN1{
  public static void main(String argv[]){
    int a = 869;
    int b = 85;
    for(int i = 0; i < a; i++){
      b += b;
    }
    System.out.println(b);
  }
}

這段代碼在i = 24處明顯溢出

我想知道為什么在幾個循環之后b設置為 0。這種溢出行為是什么以及它在 Java 中究竟是如何工作的？

我預計 output 會在-2^31 to +2^31之間變化，但事實並非如此，這是為什么呢？

Answer 1

只需在 for 循環中添加一個 print 語句，如下所示：

public static void main(String argv[]){
    int a = 869;
    int b = 85;
    for(int i = 0; i < a; i++){
      b += b;
      System.out.println(b);          <----- Add this to see 
    }
    System.out.println(b);
  }

Output：

170
340
680
1360
2720
5440
10880
21760
43520
87040
174080
348160
696320
1392640
2785280
5570560
11141120
22282240
44564480
89128960
178257920
356515840
713031680
1426063360
-1442840576
1409286144
-1476395008
1342177280
-1610612736
1073741824
-2147483648
0
0
0
0
0
0
.....

依此類推，直到循環結束。

請注意，您設置的限制已被提前超過。 現在你從這里開始看到 0 的原因是因為這里和這里提到的鏈接

請詳細記下答案，即共享的第二個鏈接。 這將幫助您理解為什么它的行為方式如此，我認為它不僅限於int ，還適用於long 、 double和所有其他numeric類型。

希望有幫助！

Answer 2

更准確地說，output 在您指定的范圍內。 0 有資格在Integer.MIN_VALUE和Integer.MAX_VALUE之間。

稍微修改一下你的代碼，你可以看看發生了什么並打印出實際的二進制文件

int a = 869;
int b = 85;
for(int i = 0; i < a; i++){
    System.out.printf(
            "Iteration %d: b = %d; b+b = %d; bin(b) = %s; bin(b+b) = %s%n",
            i, b, (b+b), Integer.toBinaryString(b), Integer.toBinaryString(b+b));
            b += b;
}
System.out.println(b);

這給出了以下 output

Iteration 0: b = 85; b+b = 170; bin(b) = 1010101; bin(b+b) = 10101010
Iteration 1: b = 170; b+b = 340; bin(b) = 10101010; bin(b+b) = 101010100
Iteration 2: b = 340; b+b = 680; bin(b) = 101010100; bin(b+b) = 1010101000
Iteration 3: b = 680; b+b = 1360; bin(b) = 1010101000; bin(b+b) = 10101010000
Iteration 4: b = 1360; b+b = 2720; bin(b) = 10101010000; bin(b+b) = 101010100000
Iteration 5: b = 2720; b+b = 5440; bin(b) = 101010100000; bin(b+b) = 1010101000000
Iteration 6: b = 5440; b+b = 10880; bin(b) = 1010101000000; bin(b+b) = 10101010000000
Iteration 7: b = 10880; b+b = 21760; bin(b) = 10101010000000; bin(b+b) = 101010100000000
Iteration 8: b = 21760; b+b = 43520; bin(b) = 101010100000000; bin(b+b) = 1010101000000000
Iteration 9: b = 43520; b+b = 87040; bin(b) = 1010101000000000; bin(b+b) = 10101010000000000
Iteration 10: b = 87040; b+b = 174080; bin(b) = 10101010000000000; bin(b+b) = 101010100000000000
Iteration 11: b = 174080; b+b = 348160; bin(b) = 101010100000000000; bin(b+b) = 1010101000000000000
Iteration 12: b = 348160; b+b = 696320; bin(b) = 1010101000000000000; bin(b+b) = 10101010000000000000
Iteration 13: b = 696320; b+b = 1392640; bin(b) = 10101010000000000000; bin(b+b) = 101010100000000000000
Iteration 14: b = 1392640; b+b = 2785280; bin(b) = 101010100000000000000; bin(b+b) = 1010101000000000000000
Iteration 15: b = 2785280; b+b = 5570560; bin(b) = 1010101000000000000000; bin(b+b) = 10101010000000000000000
Iteration 16: b = 5570560; b+b = 11141120; bin(b) = 10101010000000000000000; bin(b+b) = 101010100000000000000000
Iteration 17: b = 11141120; b+b = 22282240; bin(b) = 101010100000000000000000; bin(b+b) = 1010101000000000000000000
Iteration 18: b = 22282240; b+b = 44564480; bin(b) = 1010101000000000000000000; bin(b+b) = 10101010000000000000000000
Iteration 19: b = 44564480; b+b = 89128960; bin(b) = 10101010000000000000000000; bin(b+b) = 101010100000000000000000000
Iteration 20: b = 89128960; b+b = 178257920; bin(b) = 101010100000000000000000000; bin(b+b) = 1010101000000000000000000000
Iteration 21: b = 178257920; b+b = 356515840; bin(b) = 1010101000000000000000000000; bin(b+b) = 10101010000000000000000000000
Iteration 22: b = 356515840; b+b = 713031680; bin(b) = 10101010000000000000000000000; bin(b+b) = 101010100000000000000000000000
Iteration 23: b = 713031680; b+b = 1426063360; bin(b) = 101010100000000000000000000000; bin(b+b) = 1010101000000000000000000000000
Iteration 24: b = 1426063360; b+b = -1442840576; bin(b) = 1010101000000000000000000000000; bin(b+b) = 10101010000000000000000000000000
Iteration 25: b = -1442840576; b+b = 1409286144; bin(b) = 10101010000000000000000000000000; bin(b+b) = 1010100000000000000000000000000
Iteration 26: b = 1409286144; b+b = -1476395008; bin(b) = 1010100000000000000000000000000; bin(b+b) = 10101000000000000000000000000000
Iteration 27: b = -1476395008; b+b = 1342177280; bin(b) = 10101000000000000000000000000000; bin(b+b) = 1010000000000000000000000000000
Iteration 28: b = 1342177280; b+b = -1610612736; bin(b) = 1010000000000000000000000000000; bin(b+b) = 10100000000000000000000000000000
Iteration 29: b = -1610612736; b+b = 1073741824; bin(b) = 10100000000000000000000000000000; bin(b+b) = 1000000000000000000000000000000
Iteration 30: b = 1073741824; b+b = -2147483648; bin(b) = 1000000000000000000000000000000; bin(b+b) = 10000000000000000000000000000000
Iteration 31: b = -2147483648; b+b = 0; bin(b) = 10000000000000000000000000000000; bin(b+b) = 0

您實際上是在向左移動位，邏輯上相當於在每次迭代時乘以 2。 在第 31 次迭代中，您最終得到二進制10000000000000000000000000000000並將其向左移動會將1從序列中推出，因此您將得到00000000000000000000000000000000 。 這發生在第 31 次迭代，無論您的 b 起始值如何（您只有 32 位可使用，位置 0-31）。

附錄：其他寫法b += b ：

b <<= 1
b *= 2