Similar code, yet one version does not return a String when it should
For a class exercise, both my friend and I wrote a similar code. Hers works while mine gives out:
Error: This method must return a result of type java.lang.String
Her code :
public static String alphaString (String a, String b) {
for (int i=0; i<a.length(); i++){
if ((int)(a.charAt(i)) < ((int)(b.charAt(i)))) {
return a;
}
if ((int)(a.charAt(i)) > ((int)(b.charAt(i)))) {
return b;
}
}
return ("these are the same words") ;
}
My code :
// method that takes two Strings and returns the one that comes first in the alphabet
// alphaString("banana", "apple")--> "apple"
// alphaString("snake", "squirrel")--> "snake".
public static String alphaString(String s1, String s2) {
for (int i = 0; i<s1.length(); i++ ) {
if (s1.charAt(i)<s2.charAt(i)) {
return s1;
}
if (s1.charAt(i)>s2.charAt(i)){
return s2;
}
}
Now, I can tell that our variables are named differently and that she uses (int) (though I don't think it is necessary). I don't understand why her code works and mine not. If I try to also add the (int) and make the code even more similar, I still get the error.
Does anyone have any idea why that might be?
java string for-loop types alphabetical
edited Nov 23 '18 at 2:20
camickr
274k15126239
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
|
show 4 more comments
For a class exercise, both my friend and I wrote a similar code. Hers works while mine gives out:
Error: This method must return a result of type java.lang.String
Her code :
public static String alphaString (String a, String b) {
for (int i=0; i<a.length(); i++){
if ((int)(a.charAt(i)) < ((int)(b.charAt(i)))) {
return a;
}
if ((int)(a.charAt(i)) > ((int)(b.charAt(i)))) {
return b;
}
}
return ("these are the same words") ;
}
My code :
// method that takes two Strings and returns the one that comes first in the alphabet
// alphaString("banana", "apple")--> "apple"
// alphaString("snake", "squirrel")--> "snake".
public static String alphaString(String s1, String s2) {
for (int i = 0; i<s1.length(); i++ ) {
if (s1.charAt(i)<s2.charAt(i)) {
return s1;
}
if (s1.charAt(i)>s2.charAt(i)){
return s2;
}
}
Now, I can tell that our variables are named differently and that she uses (int) (though I don't think it is necessary). I don't understand why her code works and mine not. If I try to also add the (int) and make the code even more similar, I still get the error.
Does anyone have any idea why that might be?
java string for-loop types alphabetical
edited Nov 23 '18 at 2:20
camickr
274k15126239
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
What is the last statement of her method? What is the last statement of your method? What is the difference?
– camickr
Nov 23 '18 at 2:21
What would her method return if the first string was empty (length 0)? What would yours return?
– David
Nov 23 '18 at 2:21
@David mine wouldn't return anything at all because of the error, while hers would return "these are the same words"
– Patricia Manarazan
Nov 23 '18 at 2:24
1
@PatriciaManarazan: "mine wouldn't return anything at all" - That is the error. The compiler has to ensure that the method will always return the advertised type. There exists a logical condition where your method wouldn't return anything, which is invalid.
– David
Nov 23 '18 at 2:25
@camickr hers: return ("these are the same words") ; mine: return s2;
– Patricia Manarazan
Nov 23 '18 at 2:26
|
show 4 more comments
For a class exercise, both my friend and I wrote a similar code. Hers works while mine gives out:
Error: This method must return a result of type java.lang.String
Her code :
public static String alphaString (String a, String b) {
for (int i=0; i<a.length(); i++){
if ((int)(a.charAt(i)) < ((int)(b.charAt(i)))) {
return a;
}
if ((int)(a.charAt(i)) > ((int)(b.charAt(i)))) {
return b;
}
}
return ("these are the same words") ;
}
My code :
// method that takes two Strings and returns the one that comes first in the alphabet
// alphaString("banana", "apple")--> "apple"
// alphaString("snake", "squirrel")--> "snake".
public static String alphaString(String s1, String s2) {
for (int i = 0; i<s1.length(); i++ ) {
if (s1.charAt(i)<s2.charAt(i)) {
return s1;
}
if (s1.charAt(i)>s2.charAt(i)){
return s2;
}
}
Now, I can tell that our variables are named differently and that she uses (int) (though I don't think it is necessary). I don't understand why her code works and mine not. If I try to also add the (int) and make the code even more similar, I still get the error.
Does anyone have any idea why that might be?
java string for-loop types alphabetical
edited Nov 23 '18 at 2:20
camickr
274k15126239
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
For a class exercise, both my friend and I wrote a similar code. Hers works while mine gives out:
Error: This method must return a result of type java.lang.String
Her code :
public static String alphaString (String a, String b) {
for (int i=0; i<a.length(); i++){
if ((int)(a.charAt(i)) < ((int)(b.charAt(i)))) {
return a;
}
if ((int)(a.charAt(i)) > ((int)(b.charAt(i)))) {
return b;
}
}
return ("these are the same words") ;
}
My code :
// method that takes two Strings and returns the one that comes first in the alphabet
// alphaString("banana", "apple")--> "apple"
// alphaString("snake", "squirrel")--> "snake".
public static String alphaString(String s1, String s2) {
for (int i = 0; i<s1.length(); i++ ) {
if (s1.charAt(i)<s2.charAt(i)) {
return s1;
}
if (s1.charAt(i)>s2.charAt(i)){
return s2;
}
}
Now, I can tell that our variables are named differently and that she uses (int) (though I don't think it is necessary). I don't understand why her code works and mine not. If I try to also add the (int) and make the code even more similar, I still get the error.
Does anyone have any idea why that might be?
java string for-loop types alphabetical
java string for-loop types alphabetical
edited Nov 23 '18 at 2:20
camickr
274k15126239
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
edited Nov 23 '18 at 2:20
camickr
274k15126239
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
edited Nov 23 '18 at 2:20
camickr
274k15126239
edited Nov 23 '18 at 2:20
camickr
274k15126239
edited Nov 23 '18 at 2:20
camickr
274k15126239
274k15126239
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
asked Nov 23 '18 at 2:18
Patricia Manarazan
206
206
What is the last statement of her method? What is the last statement of your method? What is the difference?
– camickr
Nov 23 '18 at 2:21
What would her method return if the first string was empty (length 0)? What would yours return?
– David
Nov 23 '18 at 2:21
@David mine wouldn't return anything at all because of the error, while hers would return "these are the same words"
– Patricia Manarazan
Nov 23 '18 at 2:24
1
@PatriciaManarazan: "mine wouldn't return anything at all" - That is the error. The compiler has to ensure that the method will always return the advertised type. There exists a logical condition where your method wouldn't return anything, which is invalid.
– David
Nov 23 '18 at 2:25
@camickr hers: return ("these are the same words") ; mine: return s2;
– Patricia Manarazan
Nov 23 '18 at 2:26
|
show 4 more comments
What is the last statement of her method? What is the last statement of your method? What is the difference?
– camickr
Nov 23 '18 at 2:21
What would her method return if the first string was empty (length 0)? What would yours return?
– David
Nov 23 '18 at 2:21
@David mine wouldn't return anything at all because of the error, while hers would return "these are the same words"
– Patricia Manarazan
Nov 23 '18 at 2:24
1
@PatriciaManarazan: "mine wouldn't return anything at all" - That is the error. The compiler has to ensure that the method will always return the advertised type. There exists a logical condition where your method wouldn't return anything, which is invalid.
– David
Nov 23 '18 at 2:25
@camickr hers: return ("these are the same words") ; mine: return s2;
– Patricia Manarazan
Nov 23 '18 at 2:26
What is the last statement of her method? What is the last statement of your method? What is the difference?
– camickr
Nov 23 '18 at 2:21
What is the last statement of her method? What is the last statement of your method? What is the difference?
– camickr
Nov 23 '18 at 2:21
What would her method return if the first string was empty (length 0)? What would yours return?
– David
Nov 23 '18 at 2:21
What would her method return if the first string was empty (length 0)? What would yours return?
– David
Nov 23 '18 at 2:21
@David mine wouldn't return anything at all because of the error, while hers would return "these are the same words"
– Patricia Manarazan
Nov 23 '18 at 2:24
@David mine wouldn't return anything at all because of the error, while hers would return "these are the same words"
– Patricia Manarazan
Nov 23 '18 at 2:24
1
1
@PatriciaManarazan: "mine wouldn't return anything at all" - That is the error. The compiler has to ensure that the method will always return the advertised type. There exists a logical condition where your method wouldn't return anything, which is invalid.
– David
Nov 23 '18 at 2:25
@PatriciaManarazan: "mine wouldn't return anything at all" - That is the error. The compiler has to ensure that the method will always return the advertised type. There exists a logical condition where your method wouldn't return anything, which is invalid.
– David
Nov 23 '18 at 2:25
@camickr hers: return ("these are the same words") ; mine: return s2;
– Patricia Manarazan
Nov 23 '18 at 2:26
@camickr hers: return ("these are the same words") ; mine: return s2;
– Patricia Manarazan
Nov 23 '18 at 2:26
|
show 4 more comments
1 Answer
1
active
oldest
votes
You seem to be missing the default return of "these are the same words". It isn't clear why you both chose to implement this with a loop (or why neither of you chose to use an else). Regardless, String is Comparable so I would simply do
public static String alphaString(String s1, String s2) {
int c = s1.compareTo(s2);
if (c < 0) {
return s1;
} else if (c > 0) {
return s2;
}
return "these are the same words";
}
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
We used a loop because that was the direction our teacher gave us and we have not seen s1.compareTo(s2) formally in class, so we were not allowed.
– Patricia Manarazan
Nov 23 '18 at 2:29
When I add : return "these are the same words"; to my original method I can make it compile and It does return a String but I don't understand why. How does that statement affect the rest? why does its absence cause the error?
– Patricia Manarazan
Nov 23 '18 at 2:46
1
The method advertises that it returns a String, therefore, every possible exit out of the method must return a String. In the case where both of the if conditions evaluate to false, there is not a return statement. Therefore, there is a possible exit out of the method that does not return a String, which is the reason for the compile error.
– John Camerin
Nov 23 '18 at 2:56
add a comment |
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oldest
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You seem to be missing the default return of "these are the same words". It isn't clear why you both chose to implement this with a loop (or why neither of you chose to use an else). Regardless, String is Comparable so I would simply do
public static String alphaString(String s1, String s2) {
int c = s1.compareTo(s2);
if (c < 0) {
return s1;
} else if (c > 0) {
return s2;
}
return "these are the same words";
}
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
We used a loop because that was the direction our teacher gave us and we have not seen s1.compareTo(s2) formally in class, so we were not allowed.
– Patricia Manarazan
Nov 23 '18 at 2:29
When I add : return "these are the same words"; to my original method I can make it compile and It does return a String but I don't understand why. How does that statement affect the rest? why does its absence cause the error?
– Patricia Manarazan
Nov 23 '18 at 2:46
1
The method advertises that it returns a String, therefore, every possible exit out of the method must return a String. In the case where both of the if conditions evaluate to false, there is not a return statement. Therefore, there is a possible exit out of the method that does not return a String, which is the reason for the compile error.
– John Camerin
Nov 23 '18 at 2:56
add a comment |
You seem to be missing the default return of "these are the same words". It isn't clear why you both chose to implement this with a loop (or why neither of you chose to use an else). Regardless, String is Comparable so I would simply do
public static String alphaString(String s1, String s2) {
int c = s1.compareTo(s2);
if (c < 0) {
return s1;
} else if (c > 0) {
return s2;
}
return "these are the same words";
}
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
We used a loop because that was the direction our teacher gave us and we have not seen s1.compareTo(s2) formally in class, so we were not allowed.
– Patricia Manarazan
Nov 23 '18 at 2:29
When I add : return "these are the same words"; to my original method I can make it compile and It does return a String but I don't understand why. How does that statement affect the rest? why does its absence cause the error?
– Patricia Manarazan
Nov 23 '18 at 2:46
1
The method advertises that it returns a String, therefore, every possible exit out of the method must return a String. In the case where both of the if conditions evaluate to false, there is not a return statement. Therefore, there is a possible exit out of the method that does not return a String, which is the reason for the compile error.
– John Camerin
Nov 23 '18 at 2:56
add a comment |
You seem to be missing the default return of "these are the same words". It isn't clear why you both chose to implement this with a loop (or why neither of you chose to use an else). Regardless, String is Comparable so I would simply do
public static String alphaString(String s1, String s2) {
int c = s1.compareTo(s2);
if (c < 0) {
return s1;
} else if (c > 0) {
return s2;
}
return "these are the same words";
}
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
You seem to be missing the default return of "these are the same words". It isn't clear why you both chose to implement this with a loop (or why neither of you chose to use an else). Regardless, String is Comparable so I would simply do
public static String alphaString(String s1, String s2) {
int c = s1.compareTo(s2);
if (c < 0) {
return s1;
} else if (c > 0) {
return s2;
}
return "these are the same words";
}
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
answered Nov 23 '18 at 2:25
Elliott Frisch
152k1389178
152k1389178
We used a loop because that was the direction our teacher gave us and we have not seen s1.compareTo(s2) formally in class, so we were not allowed.
– Patricia Manarazan
Nov 23 '18 at 2:29
When I add : return "these are the same words"; to my original method I can make it compile and It does return a String but I don't understand why. How does that statement affect the rest? why does its absence cause the error?
– Patricia Manarazan
Nov 23 '18 at 2:46
1
The method advertises that it returns a String, therefore, every possible exit out of the method must return a String. In the case where both of the if conditions evaluate to false, there is not a return statement. Therefore, there is a possible exit out of the method that does not return a String, which is the reason for the compile error.
– John Camerin
Nov 23 '18 at 2:56
add a comment |
We used a loop because that was the direction our teacher gave us and we have not seen s1.compareTo(s2) formally in class, so we were not allowed.
– Patricia Manarazan
Nov 23 '18 at 2:29
When I add : return "these are the same words"; to my original method I can make it compile and It does return a String but I don't understand why. How does that statement affect the rest? why does its absence cause the error?
– Patricia Manarazan
Nov 23 '18 at 2:46
1
The method advertises that it returns a String, therefore, every possible exit out of the method must return a String. In the case where both of the if conditions evaluate to false, there is not a return statement. Therefore, there is a possible exit out of the method that does not return a String, which is the reason for the compile error.
– John Camerin
Nov 23 '18 at 2:56
We used a loop because that was the direction our teacher gave us and we have not seen s1.compareTo(s2) formally in class, so we were not allowed.
– Patricia Manarazan
Nov 23 '18 at 2:29
We used a loop because that was the direction our teacher gave us and we have not seen s1.compareTo(s2) formally in class, so we were not allowed.
– Patricia Manarazan
Nov 23 '18 at 2:29
When I add : return "these are the same words"; to my original method I can make it compile and It does return a String but I don't understand why. How does that statement affect the rest? why does its absence cause the error?
– Patricia Manarazan
Nov 23 '18 at 2:46
When I add : return "these are the same words"; to my original method I can make it compile and It does return a String but I don't understand why. How does that statement affect the rest? why does its absence cause the error?
– Patricia Manarazan
Nov 23 '18 at 2:46
1
1
The method advertises that it returns a String, therefore, every possible exit out of the method must return a String. In the case where both of the if conditions evaluate to false, there is not a return statement. Therefore, there is a possible exit out of the method that does not return a String, which is the reason for the compile error.
– John Camerin
Nov 23 '18 at 2:56
The method advertises that it returns a String, therefore, every possible exit out of the method must return a String. In the case where both of the if conditions evaluate to false, there is not a return statement. Therefore, there is a possible exit out of the method that does not return a String, which is the reason for the compile error.
– John Camerin
Nov 23 '18 at 2:56
add a comment |
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What is the last statement of her method? What is the last statement of your method? What is the difference?
– camickr
Nov 23 '18 at 2:21
What would her method return if the first string was empty (length 0)? What would yours return?
– David
Nov 23 '18 at 2:21
@David mine wouldn't return anything at all because of the error, while hers would return "these are the same words"
– Patricia Manarazan
Nov 23 '18 at 2:24
1
@PatriciaManarazan: "mine wouldn't return anything at all" - That is the error. The compiler has to ensure that the method will always return the advertised type. There exists a logical condition where your method wouldn't return anything, which is invalid.
– David
Nov 23 '18 at 2:25
@camickr hers: return ("these are the same words") ; mine: return s2;
– Patricia Manarazan
Nov 23 '18 at 2:26