英文:
overlapping ifs refactoring of java
问题
String returnStr = "";
if (!a.isEmpty() && !b.isEmpty() && !c.isEmpty()) {
returnStr = "1";
} else {
if (!a.isEmpty() && !b.isEmpty()) {
returnStr = "2";
} else if (!a.isEmpty() && !c.isEmpty()) {
returnStr = "3";
} else if (!b.isEmpty() && !c.isEmpty()) {
returnStr = "4";
} else {
if (!a.isEmpty()) {
returnStr = "5";
} else if (!b.isEmpty()) {
returnStr = "6";
} else if (!c.isEmpty()) {
returnStr = "7";
} else {
returnStr = "8";
}
}
}
return returnStr;
英文:
I'm using java8.
I made the code below from the sample.
I want to refactoring this, but I don't know how.
Let me know if you have any good ideas.
Add Description :
There was some error in coding, so I corrected it.
String returnStr = "";
if(a != "" && b != "" && c != ""){
returnStr = "1";
}else{
if(a != "" && b != ""){
returnStr = "2";
}else if(a != "" && c != ""){
returnStr = "3";
}else if(b != "" && c != ""){
returnStr = "4";
}else{
if(a != ""){
returnStr = "5";
}else if(b != ""){
returnStr = "6";
}else if(c != ""){
returnStr = "7";
}else{
returnStr = "8";
}
}
}
return returnStr;
答案1
得分: 1
从“空”检查中实际暗示的3个位构造一个int：
int i = (a != "" ? 4 : 0) | (b != "" ? 2 : 0) | (c != "" ? 1 : 0);
（是的，你应该使用equals或isEmpty()，而不是!=。这不是真正的重点）。
这将得到一个范围在0到7之间的数字。
构造一个包含8个元素的列表或数组，其中数字是您想要返回的值。
然后，使用上面的i来从该列表/数组中选择一个元素：
return list.get(i); // 或者 array[i]
英文:
Construct an int from the 3 bits effectively implied by the "emptiness" checks:
int i = (a != "" ? 4 : 0) | (b != "" ? 2 : 0) | (c != "" ? 1 : 0);
(Yes, you should be using equals, or isEmpty(), instead of !=. That's not really the point).
This gives a number in the range 0 to 7, inclusive.
Construct an 8-element list or array, where the number is the value you want to return.
Then, use i from above to select an element from that list/array:
return list.get(i); // or array[i]
答案2
得分: 0
你不需要那些 else，你可以将代码写得更简洁明了，像这样：
if (a != "" && b != "" && c != "") return 1;
if (a != "" && b != "") return 2;
if (a != "" && c != "") return 3;
if (b != "" && c != "") return 4;
if (a != "") return 5;
if (b != "") return 6;
if (c != "") return 7;
return 8;
英文:
You don't need all those else and you could make it more concise and clear like this:
if(a != "" && b != "" && c != "") return 1;
if(a != "" && b != "") return 2;
if(a != "" && c != "") return 3;
if(b != "" && c != "") return 4;
if(a != "") return 5;
if(b != "") return 6;
if(c != "") return 7;
return 8;
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