为什么在整数位数后加“ 0”可以转换为字符？

Question

I've seen examples of this all over the place:

int i = 2;
char c = i + '0';
string s;
s += char(i + '0');

However, I have not yet seen an explanation for why adding the zero allows for the conversion.

Answer 1

If you look at the ASCII table, asciitable , you'll see that the digits start at 48 (being '0') and go up to 57 (for '9'). So in order to get the character code for a digit, you can add that digit to the character code of '0'.

Answer 2

When ASCII encoding is used, the integer value of '0' is 48 .

'0' + 1 = 49 = '1'
'0' + 2 = 50 = '2'

...

'0' + 9 = 57 = '9'

So, if you wanted convert a digit to its corresponding character, just add '0' to it.

Even if the platfrom uses non-ASCII encoding, the lanuage still guarantees that the characters '0' - '9' must be encoded such that:

'1' - '0' = 1
'2' - '0' = 2
'3' - '0' = 3
'4' - '0' = 4
'5' - '0' = 5
'6' - '0' = 6
'7' - '0' = 7
'8' - '0' = 8
'9' - '0' = 9

When ASCII encoding is used, that becomes:

'1' - '0' = 49 - 48 = 1
'2' - '0' = 50 - 48 = 2
'3' - '0' = 51 - 48 = 3
'4' - '0' = 52 - 48 = 4
'5' - '0' = 53 - 48 = 5
'6' - '0' = 54 - 48 = 6
'7' - '0' = 55 - 48 = 7
'8' - '0' = 56 - 48 = 8
'9' - '0' = 57 - 48 = 9

Hence, regardless of the character encoding used by a platform, the lines

int i = 2;
char c = i + '0';

will always result in the value of c being equal to the character '2' .

Answer 3

The C++ standard says, in its [lex.charset] section “the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous” (quoted from draft N4659).

Thus, the value of '1' is ' 0'+1 , and the value of '2' is one more than that, which is '0'+2 , and so on. If n has a value from 0 to 9, then '0'+n is the value for the corresponding character from '0' to '9' .

(Unlike the earlier answers, this answer does not assume ASCII and shows how the property derives from the C++ standard.)

Answer 4

It's based on ASCII values. Adding the ASCII value of 0 which is 48 means that 48 + 5 will be 53 or ASCII 53 which is 5.

Google ASCII and find a good chart and study it. It should make sense once you look at the values for each char (character).

Answer 5

(char)(i+c) where c is a char gives you a new char with an ascii value equal to (ascii value of c ) + i . Since the ascii values of the digits 0-9 are sequential, i+'0' gives you the char corresponding to i , as long as i lies in the appropriate range.

EDIT : (i+c) is an int, as Jarod42 pointed out. Added a cast to maintain correctness.

Answer 6

字符将根据ASCII值将整数转换为相应的值，因此，如果将48分配给字符，则它将存储0，因为48是ASCII值0.相应地，将ASCII值0添加到任何整数您需要转换为char