问题
i have learned that to convert charsequence to integer we can use this statement
String cs="123";
int number = Integer.parseInt(cs.toString());
what if
cs = "++-+--25";
will this statement still run and give answer -25 according to string given??
回答1:
You are end up with a NumberFormatException
since ++-+--25
is not a valid integer.
See the docs of parseInt()
Parses the string argument as a signed decimal integer. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value or an ASCII plus sign '+' ('\u002B') to indicate a positive value. The resulting integer value is returned, exactly as if the argument and the radix 10 were given as arguments to the parseInt(java.lang.String, int) method.
So you are allowed to do
CharSequence cs = "-25"; //gives you -25
and
CharSequence cs = "+25"; //gives you 25
Otherwise ,take necessary steps to face the Exception
:)
So know the char Sequence is a valid string just write a simple method to return true or false and then proceed further
public static boolean {
try {
Integer.parseInt(s);
} catch(NumberFormatException e) {
return false; // no boss you entered a wrong format
}
return true; //valid integer
}
Then your code looks like
if(isInteger(cs.toString())){
int number = Integer.parseInt(cs.toString());
// proceed remaining
}else{
// No, Operation cannot be completed.Give proper input.
}
回答2:
Answer to your question is code will run and throw Exception as "++-+--25" is not a valid int,
java.lang.NumberFormatException: For input string: "++-+--25"
回答3:
You will get
java.lang.NumberFormatException: For input string: "++-+--25"
Tested example :
CharSequence cs = "++-+--25";
System.out.println("" + Integer.parseInt(cs.toString()));
来源:https://stackoverflow.com/questions/19488067/charsequence-to-integer-with-multiple-ve-and-ve-signss