這兩天翻了下 Kent? Beck 的 Test-Driven Development By Example 感覺(jué)還不錯(cuò),第一部分的例子講的挺好,本來(lái)想歸納一下,貌似太長(zhǎng)了,等有時(shí)間了再歸納下,最后看到了附錄里面的一個(gè)Fibonacci數(shù)列的例子感覺(jué)也有那么點(diǎn)TDD的味道在里面,總結(jié)分享一下:
讓我們從最簡(jiǎn)單的fibonacci(0)=0開(kāi)始吧,先寫測(cè)試如下:
import
?junit.framework.TestCase;????????
???????
public
?
class
?TestFibonacci?
extends
?TestCase?
{????????
????
public
?
void
?testFibonacci()
{????????
????????assertEquals(
0
,fibonacci(
0
));????????
????}
????????????
}
?????
添加fibonacci方法的簡(jiǎn)單實(shí)現(xiàn)即可:
private
?
int
?fibonacci(
int
?i)?
{??????????????
????????
return
?
0
;????????
????}
??
很顯然順利通過(guò)了測(cè)試;
再加一個(gè)測(cè)試實(shí)現(xiàn)fibonacci(1)=1;如下:
import
?junit.framework.TestCase;????????
???????
public
?
class
?TestFibonacci?
extends
?TestCase?
{????????
????
public
?
void
?testFibonacci()
{????????
????????assertEquals(
0
,fibonacci(
0
));????????
????????assertEquals(
1
,fibonacci(
1
));????????
????}
????????????
}
???????
為了順利通過(guò)測(cè)試,只需簡(jiǎn)單的改動(dòng)下fibonacci方法實(shí)現(xiàn)即可;
private
?
int
?fibonacci(
int
?i)?
{??????????????
????????
if
(i
==
0
)?
return
?
0
;????????
????????
else
?
return
?
1
;????????
????}
???
在添加第三個(gè)測(cè)試條件fibonacci(2)=1時(shí)候發(fā)現(xiàn),隨著測(cè)試條件的增多無(wú)疑帶來(lái)了大量的斷言重復(fù)代碼,通過(guò)分析我們可以將testFibonacci方法改寫如下:
public
?
void
?testFibonacci()
{????
????????
int
?cases[][]
=
{
{
0
,
0
}
,
{
1
,
1
}
}
;????
????????
for
(
int
?i
=
0
;i?
????????????assertEquals(cases[i][
1
],fibonacci(cases[i][
0
]));???????
????}
???
這樣我們就可以很容易的添加測(cè)試條件:fibonacci(2)=1;
import
?junit.framework.TestCase;????
???
public
?
class
?TestFibonacci?
extends
?TestCase?
{????
????
public
?
void
?testFibonacci()
{????
????????
int
?cases[][]
=
{
{
0
,
0
}
,
{
1
,
1
}
,
{
2
,
1
}
}
;????
????????
for
(
int
?i
=
0
;i?
????????????assertEquals(cases[i][
1
],fibonacci(cases[i][
0
]));???????
????}
????
???
????
private
?
int
?fibonacci(
int
?i)?
{??????????
????????
if
(i
==
0
)?
return
?
0
;????
????????
else
?
return
?
1
;????
????}
????
}
???
嘿嘿,很順利,狀態(tài)條依然出現(xiàn)的是綠色,那我們就不管代碼到底合不合理了,出現(xiàn)綠色就是王道,繼續(xù)添加下一個(gè)測(cè)試條件:fibonacci(3)=2;
import
?junit.framework.TestCase;????
???
public
?
class
?TestFibonacci?
extends
?TestCase?
{????
????
public
?
void
?testFibonacci()
{????
????????
int
?cases[][]
=
{
{
0
,
0
}
,
{
1
,
1
}
,
{
2
,
1
}
,
{
3
,
2
}
}
;????
????????
for
(
int
?i
=
0
;i?
????????????assertEquals(cases[i][
1
],fibonacci(cases[i][
0
]));???????
????}
????
???
????
private
?
int
?fibonacci(
int
?i)?
{??????????
????????
if
(i
==
0
)?
return
?
0
;????
????????
else
?
return
?
1
;????
????}
????
}
???
終于又出現(xiàn)紅色了,it fails .說(shuō)明我們又將快有新進(jìn)展了.為了滿足fibonacci(3)=2,也很簡(jiǎn)單,簡(jiǎn)單的修改下fibonacci方法即可,如下:
private?int?fibonacci(int?i)?{??????????
????????if(i==0)?return?0;????
????????if(i<=2)?return?1;????
????????else?return?2;????
????}?? 又出現(xiàn)綠色了吧! 這里返回的2可以表示為1+1;
我們可以寫成:
private?int?fibonacci(int?i)?{??????????
????????if(i==0)?return?0;????
????????if(i<=2)?return?1;????
????????else?return?fibonacci(i-1)+1;????
????}???
第二個(gè)1可以寫成fibonacci(i-2):
import?junit.framework.TestCase;????
???
public?class?TestFibonacci?extends?TestCase?{????
????public?void?testFibonacci(){????
????????int?cases[][]={{0,0},{1,1},{2,1},{3,2}};????
????????for(int?i=0;i?
????????????assertEquals(cases[i][1],fibonacci(cases[i][0]));???????
????}????
???
????private?int?fibonacci(int?i)?{??????????
????????if(i==0)?return?0;????
????????if(i==1)?return?1;????
????????else?return?fibonacci(i-1)+fibonacci(i-2);????
????}????
}???
再添加測(cè)試條件:fibonacci(4)=3;fibonacci(5)=5:
import?junit.framework.TestCase;????
???
public?class?TestFibonacci?extends?TestCase?{????
????public?void?testFibonacci(){????
????????int?cases[][]={{0,0},{1,1},{2,1},{3,2},{4,3},{5,5}};????
????????for(int?i=0;i?
????????????assertEquals(cases[i][1],fibonacci(cases[i][0]));???????
????}????
???
????private?int?fibonacci(int?i)?{??????????
????????if(i==0)?return?0;????
????????if(i==1)?return?1;????
????????else?return?fibonacci(i-1)+fibonacci(i-2);????
????}????
}? 依然是綠色,呵呵,我們得到了正確的Fibonacci數(shù)列,驅(qū)動(dòng)源于測(cè)試.
附:
Fibonacci數(shù)列簡(jiǎn)介:Fibonacci數(shù)列從第0項(xiàng)開(kāi)始依次為0,1,1,2,3,5,8……的數(shù)列，它存在如下特點(diǎn):第0,1個(gè)數(shù)為0,1。從第2個(gè)數(shù)開(kāi)始，該數(shù)是前面兩個(gè)數(shù)之和。即該數(shù)列第n項(xiàng)的值滿足如下公式：
