simple casting problem

cout << 'a'+'b';
is c++ ??

when u manimulate the char it work like int retherthan char.

use (char)('a' + 1)

or (char)('a' + 'b')

it will print char.

the first line of your code will output the ASCII value of 'a' + 1 which is the char 'b' which = 98.  The second cout ('a'+ 'b') will out put 195 (ASCII 97 + 98).  You code is not incorrect but you need to state what you are trying to achieve.
-Mel

one other thing... if you were wishing to cast 'a'+1 as a char then use this:
cout << char ('a'+1) << endl;  the method of casting you used in your sample code will not perform the proper cast if you were looking to get the next character.  The output of what I just gave you will be: "b"  

does this help?

I know to cast char('a'+1)
my question is:
where is it written / how could I know in advance
that 'a'+'b' will give me an integer.

Simple--a single character delimited by single quotes is always treated by C (and C++) as an integer, even though it's referred to as a character. If you want a single character to be treated as a string, you have to enclose it in double quotes, so

cout << "a" + "b";

would give "ab", which presumably what you wanted.

if u print directly by cout it behave as int..
if u assigne it to any char variable it will behave like
char..

cout<< 'a' + 1;
output : 66

char ch;
ch = 'a' + 1;
cout << ch;

output: b


 

>Simple--a single character delimited by single quotes is >always treated by C (and C++) as an integer, even though >it's referred to as a character. If you want a single >character to be treated as a string, you have to enclose >it in double quotes, so
>cout << "a" + "b";
>would give "ab", which presumably what you wanted.

so why:
cout << 'a' // prints 'a'

I can't see a simple rule here.
maybe something like
"arithmentic operation with char convert it to int"

we you all agree ?

 

 

if you want the interger value of 'a' + 'b' then use:
cout << 'a' + 'b';
if you want the char value of 'a' + 'b' use the following:
cout << char ('a' + 'b');  // this will give you the cumlative value in ASCII which in the case of 'a' + 'b' will be a non-recognizable character.  if you want the interger value of 'a' + 'b' just do as the first example shows.

if u print directly by cout it behave as int..
if u assigne it to any char variable it will behave like
char..

cout<< 'a' + 1;
output : 66

char ch;
ch = 'a' + 1;
cout << ch;

output: b


 

according to 'avadhesh' your output would be upper case values...  Lower case 'a' starts at 97 decimal (in ASCII) while upper case 'A' starts at 65 decimal (in ASCII)

yes first out will be 98

So you are looking for the integer value of the output?  For each character or just the sum of the chars?

My question is not practical as I can cast every expression the any type.

I what to understand the rule.
I thought that there is a type hirracy char < int < float
and that an expression containing elements of the same type carry the same type as the elements.

can one phrase a different rule ?

no rule is same but at the time of manipulating char behave like int.

char a = 'a' it actually have 97
char b = 'b' it actually have 98

if a + b it actually like 97 + 98 ->> 195.

if u print that value using cout it give 195.

bcoz the cost of (a + b) is int default ratherthan char.
the cout take it as int.

char c = a + b;
   the c store 195.

then u cout << c . cout take cost of c is char and it print char value of 195.

it also apply for short ...



 

you want to know the casting rules... is that what you are asking???  I've tried to provide you with detailed answers to the previous posts but you are now asking a completely different question.

hi
like it has been told many times, when you get ascii value and try to manipulate it, it will return an int.

so the rule could be "arithmetic computation on ASCII values of chars gives int".

Seems similar to what you wrote,
bye