Adding one character to string

Question

In C++:

If I want to add 0x01 to the string text I would do: text += (char)0x01;

If I want to add 0x02 to the string text I would do: text += (char)0x02;

If I want to add 0x0i (were i is an unsinged int between 0 and 9), what could I do?

EDIT: I probably wasn't quite clear. So by 0x01, I mean the character given in Hex as 01. So in the above if i is the integer (in decimal) say 3, then I would want to add 0x03 (so this is not the character given in decimal as 48 + 3).

What is the type of text? What are you trying to accomplish? — dirkgently
– dirkgently, Commented Jun 10, 2012 at 0:33
@dirkgently: text is a string. So I want to add too the string (i.e. make it longer by one character) the character given in hex by 0i. — Thomas
– Thomas, Commented Jun 10, 2012 at 0:36

Luchian Grigore · Accepted Answer · 2012-06-10 00:22:51Z

4

You can directly do

text += (char)i;

Because 0x0i == i if i is between 0 and 9.

answered Jun 10, 2012 at 0:22

Luchian Grigore

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Comments

Jerry Coffin · Accepted Answer · 2012-06-10 00:41:51Z

1

One other possibility -- you could use push_back instead. Since it takes the string's char_type as its parameter type, you don't need an explicit cast:

text.push_back(i);

In fairness, I should add that you don't really need an explicit cast with += either. Simply text += i; will work fine. For example:

std::string text; 
for (int i=0; i<9; i++)
     text += i;

With either text += i; or text.push_back(i);, this will produce a string that contains: "\x00\x01\x02\x03".

edited Jun 10, 2012 at 0:41

answered Jun 10, 2012 at 0:35

Jerry Coffin

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Comments

Kamran Amini · Accepted Answer · 2012-06-10 00:23:43Z

0

You should generate its ASCII value. '0' is 48 and '9' is 9 + 48

text += (char)( i + 48 ) ;

if i was a multi digit number you could parse it to its digits and use the same technique to generate its equal string.

answered Jun 10, 2012 at 0:23

Kamran Amini

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2 Comments

Luchian Grigore Over a year ago

He doesn't want 'i', he wants i, so there's no need to add to 48. That's why his samples are text += (char)0x01, and not text += (char)0x01+48

Kamran Amini Over a year ago

If you are right, so what's the point of this question ? I think his first 2 lines are not correct. He wanted to concatenate characters but his hexadecimal numbers says you are right.

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