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为什么在“单项......”中“t”;t+=j;}while(j<5);”返回15? [闭门]
原标题:Why does "t" in "do{j++; t+=j;}while(j<5);" return 15? [closed]
  • 时间:2011-10-31 01:22:21
  •  标签:
  • javascript
最佳回答

Let:

var j = 0;
var t = 0;

do {
  j++;
  t += j;
} while (j < 5);

alert(t);

如果你人工扩大(而且我会这样做),这是无漏洞的等同法:

var j = 0;
var t = 0;

j++; // j = 1
t += j;

j++; // j = 2
t += j;

j++; // j = 3
t += j;

j++; // j = 4
t += j;

j++; // j = 5
t += j;

alert(t);

证明这一点:

var t = 0;

t += 1;
t += 2;
t += 3;
t += 4;
t += 5;

alert(t);

其结果如下:

var t = 1 + 2 + 3 + 4 + 5;

alert(t);

我认为你可以做简单的算术:

var t = 15;

alert(t);
问题回答

由于在评估试验之前 j缩了,因此 lo将被用于下列j值:

j = 1, 2, 3, 4, 5

最终价值为15(1+2+3+4+5)。

可能有三个陷阱:

  1. do block is always executed once - independent of the condition
  2. it uses t+=j so it adds j up to the previous t value
  3. when j becomes 5, the do-loop is still executed

页: 1

loop #0: j=1 t=1

loop #1: j=2 t=3

loop #2: j=3 t=6

loop #3: j=4 t=10

loop #4: j=5 t=10

这只是一纸空洞的 st脚石,你应该能够跟随我在这里所做的事情。

j = 1; t = 1
j = 2; t = 3
j = 3; t = 6
j = 4; t = 10
j = 5; t = 15

每当休息时间,都会增加。 在加固后,拖拉机就添加了。 第一个通行证为1。 之后, j变成了2,t是3,等等。

j=1, t=1
j=2, t=3
j=3, t=6
j=4, t=10
j=5, t=15 (j is not less than 5, loop ends).




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