sonar常见bug类型及解决方法（持续更新）
NullPointerException might be thrown as ‘XXX’ is nullable here
说明：未做非空校验，可能产生空指针
解决方案：加上非空校验
Use a “double” or “BigDecimal” instead.
说明：两个float计算会产生不精确的结果
解决方案：将float换为double或者BigDecimal计算
举例：
float a = 16777216.0f;
float b = 1.0f;
float c = a + b; // Noncompliant; yields 1.6777216E7 not 1.6777217E7
double d = a + b; // Noncompliant; addition is still between 2 floats
换为
float a = 16777216.0f;
float b = 1.0f;
BigDecimal c = BigDecimal.valueOf(a).add(BigDecimal.valueOf(b));
double d = (double)a + (double)b;
Cast one of the operands of this multiplication operation to a “long”
说明：int数运算最终再把结果转为long将有可能产生溢出
解决方案：转换为long型预算
举例：
long bigNum = Integer.MAX_VALUE + 2; // Noncompliant. Yields -2147483647
换为
long bigNum = Integer.MAX_VALUE + 2L;
Close this “XXX”.
说明：流没有显示的关闭。
解决方案：在fianlly语句块内关闭。
Remove or correct this useless self-assignment.
说明：缺少this
解决方案：用this.xxx=xxx来替代
举例：
public void setName(String name) {
name = name;
}
换为
public void setName(String name) {
this.name = name;
}
Correct this “&” to “&&”.
说明：错误的使用&和&&
解决方案：根据情况使用&和&&
This branch can not be reached because the condition duplicates a previous condition in the same sequence of “if/else if” statements
说明:if else if 语句判断条件重复
解决方案：去掉多余的判断条件
Make this “XXX” field final.
说明：将某个字段置为final，常见在Exception的参数
解决方案：将字段置为final
Remove this call to “equals”; comparisons between unrelated types always return false.
说明：去掉equals判断语句，因为总为false
解决方案：去掉equals语句
Remove this return statement from this finally block.
说明：在finally语句块中有return语句
解决方案：去掉finally语句块的return语句或者放在finally语句块之外
Remove this continue statement from this finally block.
说明：在finally语句块中有continue语句
解决方案：去掉finally语句块中的continue语句或者放在finally语句块之外
Equality tests should not be made with floating point values.
说明：浮点数之间用 == 来比较大小不准确
解决方案：用Number或者BigDecimal来比较
举例：
float myNumber = 3.146;
if ( myNumber == 3.146f ) { //Noncompliant. Because of floating point imprecision, this will be false
// …
}
if ( myNumber != 3.146f ) { //Noncompliant. Because of floating point imprecision, this will be true
// …
}
Add a type test to this method.
说明：强转前未判断类型
解决方案：强转前先判断类型
举例：
ErpVO ev = (ErpVO) obj;
return this.userCode.equals(ev.getUserCode());
换为
if (obj == null) {
return false;
}
if (obj.getClass() != this.getClass()) {
return false;
}
…
Add an end condition to this loop.
说明：没有退出条件
解决方案：根据情况来决定方法的退出条件
Make “XXX” an instance variable.
说明：有些类不是线程安全的，将变量生命为静态的可能会导致线程安全问题
解决方案：将变量声明为实例的。