Str.charAt() ?!

[HearT.Hunt3r]

why String.charat return number of char?!

like this code :
if(a.length()==b.length()){
for(int i=0;i<a.length();i++){
res += a.charAt(i) + b.charAt(i);
}
return res;
}


=========================
return this : "217219221"
instead of this >> mixString("abc", "xyz") → "axbycz"

[FlyNn]

String res;
if(a.length()==b.length()){
for(int i=0;i<a.length();i++){
res += a.charAt(i) +""+ b.charAt(i);
}
return res;
}

I assume that when you do a.charAt() + b.charAt() you actually adding decimal representations of ascii characters. Maybe if you add string the whole thing should be represented as a String, so the a.charAt() + b.charAt() wont be added together.

[DarrylBurke]

char is an integral type.
Types, Values, and Variables

db

[stchman]

why String.charat return number of char?!

like this code :
if(a.length()==b.length()){
for(int i=0;i<a.length();i++){
res += a.charAt(i) + b.charAt(i);
}
return res;
}


=========================
return this : "217219221"
instead of this >> mixString("abc", "xyz") → "axbycz"

Assumption - res is of type String.

You are attempting to concatenate a char to a String. You will need to convert those char to String. Check out the Character class.

[Junky]

Once again you exhibit your lack of understanding.

You are attempting to concatenate a char to a String.

Which is perfectly fine.

String str = "hell";
char c = 'o';
str += c;
System.out.println(str);

Output is "hello".

However the problem the OP is facing is that they have 2 chars on the righthand side and their ascii values are added together instead of the two chars being concatenated together. The result of the addition is then concatenated to the String.

String str = "hel";
char c1 = 'l';
char c2 = 'o';
str += c1 + c2;
System.out.println(str);

Ouput is "hel219" because 'l' has a value of 108 and 'o' has the value of 111. 108 + 111 = 219 which is concatenated to the end of the String.

[stchman]

Once again you exhibit your lack of understanding.


Which is perfectly fine.

String str = "hell";
char c = 'o';
str += c;
System.out.println(str);

Output is "hello".

However the problem the OP is facing is that they have 2 chars on the righthand side and their ascii values are added together instead of the two chars being concatenated together. The result of the addition is then concatenated to the String.

String str = "hel";
char c1 = 'l';
char c2 = 'o';
str += c1 + c2;
System.out.println(str);

Ouput is "hel219" because 'l' has a value of 108 and 'o' has the value of 111. 108 + 111 = 219 which is concatenated to the end of the String.

Once again, for someone who is ignoring me you are doing a really bad job.

I maintain that the OP is trying to create a string from a series of chars.

In your opinion, what is the OP trying to do?

[Junky]

I explained in another post why I need to keep my eye on you, to ammend all the incorrect replies you make.

I undeerstand what the OP is trying to do. What I have an issue with is you proclaiming that they need to convert the chars to Strings first. To correct the "bad" code above I would use a StringBuilder.

String str = "hel";
char c1 = 'l';
char c2 = 'o';
StringBuilder buffer = new StringBuilder();
buffer.append(str);
buffer.append(c1);
buffer.append(c2);
System.out.println(buffer);

// or

String str = "hel";
char c1 = 'l';
char c2 = 'o';
str += c1;
str += c2;
System.out.println(str);

Saying that someone MUST do something a particular way is misleading. There is usually various ways to achieve the same result.

[stchman]

I explained in another post why I need to keep my eye on you, to ammend all the incorrect replies you make.

So you are now the java-forums.org police?

Concatenating a char to a String is bad form. The proper way to do it is to convert the char to a String. If you wish to engage in sloppy programming, that is your wish.

I would do it this way.

String str = "hel";
char c1 = 'l';
char c2 = 'o';
str += Character.toString( c1 );
str += Character.toString( c2 );
System.out.println(str);

See, this way everything is of the same type....

[Junky]

Concatenating a char to a String is bad form. The proper way to do it is to convert the char to a String.

Yet again your lack of understanding shows.

When you concatenate anything to a String what happens under the hood is that the code is changed to use a StringBuilder and amends the content regardless of its type. So converting the char to a String beforehand is actually causing the code to do extra work.

[Junky]

By the way I realised the code I posted above can be improved somewhat (depending upon your point of view).

String str = "hel";
char c1 = 'l';
char c2 = 'o';
StringBuilder buffer = new StringBuilder();
str = buffer.append(str).buffer.append(c1).buffer.append(c2).toString();
System.out.println(str);

[stchman]

Yet again your lack of understanding shows.

When you concatenate anything to a String what happens under the hood is that the code is changed to use a StringBuilder and amends the content regardless of its type. So converting the char to a String beforehand is actually causing the code to do extra work.

Let's agree that neither of us is going to change the other's mind. Just go ahead and ignore me again, at best you are annoying.

[Tolls]

Concatenating a char to a String is bad form. The proper way to do it is to convert the char to a String. If you wish to engage in sloppy programming, that is your wish.

I'm afraid Junky's correct.
The standard way to do it is to use the StringBuilder, since one is created under the hood anyway. Using the Character.toString() you're adding an unecessary step.

[JosAH]

Once again, for someone who is ignoring me you are doing a really bad job.

Junky is doing a fine job; if he didn't do this, I or another moderator had to do this because in general your 'advice' (mind the quotes) is incorrect.

kind regards,

Jos

[stchman]

Hmmmmmm.

I don't see how people think this

Junky's code

String str = "hel";
char c1 = 'l';
char c2 = 'o';
StringBuilder buffer = new StringBuilder();
str = buffer.append(str).buffer.append(c1).buffer.append(c2).toString();
System.out.println(str);

Is more clear or correct than this:

String str = "hel";
char c1 = 'l';
char c2 = 'o';
str += Character.toString( c1 ) + Character.toString( c2 );
System.out.println(str);

I accomplished the SAME thing with LESS code. I gather to some that is a bad thing.

[Norm]

Which method uses more computer resources?
String concatenation can be expensive.

[JosAH]

Hmmmmmm.

I accomplished the SAME thing with LESS code. I gather to some that is a bad thing.

Run javap -c on your class and see for yourself what the compiler has generated for you behind your back; please stop being stubborn and please stop spreading bad advice.

kind regards,

Jos

[stchman]

Junky is doing a fine job; if he didn't do this, I or another moderator had to do this because in general your 'advice' (mind the quotes) is incorrect.

kind regards,

Jos

If Junky is, then how come his code is incorrect?

String str = "hel";
char c1 = 'l';
char c2 = 'o';
StringBuilder buffer = new StringBuilder();
str = buffer.append(str).buffer.append(c1).buffer.append(c2).toString();
System.out.println(str);

When the correct code should be:

String str = "hel";
char c1 = 'l';
char c2 = 'o';
StringBuilder buffer = new StringBuilder();
str = buffer.append(str).append(c1).append(c2).toString();
System.out.println(str);

I guess when anyone else posts bad code, it's OK.

[JosAH]

[QUOTE=stchman;230477]If Junky is, then how come his code is incorrect?

It may be a slip of the finger but at least I got the intention of the code; but you are right: the code should've been tested before being posted.

I guess when anyone else posts bad code, it's OK.

The same counts for you: test your code before you claim non-quantitative qualifiers such as 'better'. Your latest conclusion is incorrect.

kind regards,

Jos

[stchman]

[QUOTE=JosAH;230478]

If Junky is, then how come his code is incorrect?

It may be a slip of the finger but at least I got the intention of the code; but you are right: the code should've been tested before being posted.



The same counts for you: test your code before you claim non-quantitative qualifiers such as 'better'. Your latest conclusion is incorrect.

kind regards,

Jos

Fair enough. I bring up Geany, write a small program, then compile and debug the methods I've wrote. They may or may not be the most compiler "efficient", but they are easily readable.

[JosAH]

Fair enough. I bring up Geany, write a small program, then compile and debug the methods I've wrote. They may or may not be the most compiler "efficient", but they are easily readable.

That isn't a quantifiable statement either. case closed.

kind regards,

Jos