Common trap questions and answers in Java

1. Find odd numbers:

public static boolean isOdd(int i){
 return i % 2 == 1;
 }

Can the above method really find all odd numbers?

A: The problem of negative numbers is not taken into account. If i is negative, it is incorrect. Should return i%2 == 0

2. Floating point subtraction

System.out.println(2.0-1.9);

A: The simple floating point types float and double in Java cannot be operated. Not only Java, but also many other programming languages ​​have this problem. In most cases, the calculated results are accurate, but if you try a few times (you can make a loop), you can make errors like the above. Of course, there may be problems with addition, subtraction, multiplication and division.

For example:

System.out.println(0.05+0.01);
System.out.println(1.0-0.42);
System.out.println(4.015*100);
System.out.println(123.3/100);

This is because some decimals with limited digits may become infinitely looping decimals in binary. In floating point numbers, cannot be represented and the accuracy is impaired.

Solution:

1. If you want to determine whether a-b is equal to c, or whether a+b is equal to c, you can use

if(0.05+0.01-0.06 < 0.0000001)
{
}

2. In "Effective Java" This book mentions a principle that float and double can only be used for scientific calculations or engineering calculations. In commercial calculations, we have to use java.math.BigDecimal to solve

System.out.println((new BigDecimal("2.0")).subtract(
 new BigDecimal("1.9")).doubleValue());

3. Infinite loop

public static final int END = Integer.MAX_VALUE;
 public static final int START = END - 2;
 
 public static void main(String[] args) {
 int count = 0;
 for (int i = START; i <= END; i++)
 count++;
 System.out.println(count);
 }

A: The reason for the infinite loop here is that when i is Integer.MAX_VALUE, the for loop first ++, then determines whether i is <=END, when i is Integer.MAX_VALUE and then ++ , i becomes a negative number. So the cycle continues.
The reason why it becomes a negative number is that int overflows. Here, changing <=END to

4. What exactly is returned?

public static boolean decision() {
 try {
 return true;
 } finally {
 return false;
 }
}

A: Return false. At this time, return true is an unreachable statement and will be optimized and removed during the compilation phase.

3. Let’s share a trap question that you may encounter during the interview

Look at the code:

int a=5;
  System.out.println("value is"+((a<5)? 10.9:9 ));

The output result is:

A .Compilation error B10.9 C.9 D None of the above answers are correct.

The execution result is:

value is9.0

Because ((a<5) ? 10.9) has a 10.9java automatic transformation based on operator precision. Therefore, the following 9 will also become 9.0.

So choose D.

a

StringBuffer str1=new StringBuffer("123");
  StringBuffer str2=new StringBuffer("123");
  if(str1.equals(str2)){
   System.out.println("str1.equalstr2");
  }else{
   System.out.println("str1.notequalstr2");
  }

The result is: str1.notequalsstr2 This shows that StringBuffer does not override the equals method.

Float fa=new Float(0.9f);
  Float fb=new Float(0.9f); //Float fb=new Float("0.9f");
  Double db=new Double(0.9f);
  if(fa==fb){ //false
   System.out.println("fa==fb");
  }else{
   System.out.println("fa!=fb");
  }
  if(fa.equals(fb)){ //true
   System.out.println("fa.equalfb");
  }else{
   System.out.println("fa!equalfb");
  }
  if(db.equals(fb)){ //false
   System.out.println("db.equalfb");
  }else{
   System.out.println("db!equalfb");
  }

The result is:

fa!=fb
fa.equalfb
db!equalfb Float 型与Double 型肯定不相等

If you have anything else, welcome to add.

Reference:

1. http://blog.csdn.net/ol_beta/article/details/5598867

2. http://zhidao. baidu.com/link?url=0UyDU42L7DXZitdydJMG3IIUDIf3xidFCRAObZAq6SHFCEaNnp2Oyuq1KVwBvmlR0UZGHSjD4f6A1yD0d65JL_