Reputation: 9
public boolean isNumber(String t) {
for (int i = 0, i<= 9, i++) {
if t.equals(i) {
return true;
}
}
return false;
}
Copypastad the wrong method originally -_- I have this inside a class compiling with this error:
data_structures/ExpressionEvaluator.java:40: illegal start of type for (int i = 0, i< 10, i++) {
Upvotes: 0
Views: 95
Reputation: 2785
Your method only verifies if the String that you passed is a digit, not a number (a number can have more than one digit). You could verifiy it only using a char and calling the ,Character.isDigit
char c = '1';
boolean isDigit = Character.isDigit(c);
If you really want to create your own method, passing a String param, I suggest you to modify like this:
public boolean isDigit(String t) {
return t.length() == 1 && Character.isDigit(t.charAt(0));
}
Upvotes: 0
Reputation: 66637
You should use semi-colon and your if
should be surrounded with brackets.
public boolean isNumber(String t) {
for (int i = 0; i <= 9; i++) {
if (t.equals(i)) {
return true;
}
}
return false;
}
I would suggest reading Language Basics
Upvotes: 2
Reputation: 1
public boolean isNumber(String t) {
for (int i = 0; i<= 9; i++) {
if( t.equals(i) ){
return true;
}
}
return false;
}
1 . use ";" replace of ","
2 .
if(boolean) {
//do stuff
}
Upvotes: 0
Reputation: 28687
Semicolons separate the qualities of a for loop. Also, the condition of your if block must be surrounded by parentheses.
public boolean isNumber(String t) {
for (int i = 0; i <= 9; i++) {
if (t.equals(i)) {
return true;
}
}
return false;
}
Upvotes: 0
Reputation: 15766
Use semi-colons instead of commas.
for(int i = 0; i < 10; i++) {
//do stuff
}
Upvotes: 3