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Sum of two chars in C/C++
(7 answers)
Closed 4 years ago.
I'm coming from high language, PHP js and things. So this seem strange to me.
I'm using either local or online interpreter but I always get this result.
I suppose this result is because '2' is 50 in ASCII and 98 is 'b' but I'm not sure. Also I don't really understand how the conversion work.
The code is here:
#include <iostream>
#include <string>
int main()
{
std::cout << '1' + 1 << '\n';
std::cout << '1' + '1' << '\n';
}
Type char is integral type. Each character maps to an integer value. The value depends on the encoding used which in your case is probably ASCII. So the character '1' probably has an integer value of 49 thus the '1' + '1' expression is equivalent to 49 + 49 and results in 98. Adding integer value of 1 to 49 results in 50. Which is the same as adding integer value of 1 to (a value represented by the) character '1'.
In a nutshell, values are values, whether represented via character literals or integer literals.
'1' is a char constant with a specific value determined by the encoding used on your system. That encoding might be ASCII, but it might not. When used as an argument to +, it is promoted to an int. So decltype('1') is a char, but decltype('1' + '1') is an int.
On your system, it's clear that '1' has the value 49. That's why'1' + '1' is 98. And therefore '1' + 1 is 50.
Note that in C, '1' is an int type. Arguably that's less confusing than the way C++ has it.
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I'm new to programming and sometimes see expressions like this
...
for (int i=0; i<str1.length(); i++)
{
int sum = ((str1[i]-'0')+(str2[i]-'0'));
str.push_back(sum%10 + '0');
}
...
So that is '0' here? Is it some kind of converting or something?
It's literally the character zero '0'
The operation str[i] - '0' is used to convert the representation of a digit into its numeric value.
Since the characters from '0' to '9' are following in the ascii table, having respectively, the values 48 to 57, when you perform the operation '3' - '0', the interpreter will use the ascii values, 51 - 48 == 3, so you can convert '3' to 3
ASCII Table
(source of the picture : Wikipedia)
'0' is the character zero. Since characters are sequential, adding a digit to 0 will produce the character representing that digit (once cast back to a char). E.g., (char)(2 + '0') (i.e., the integer two plus the character zero) will produce '2' (i.e., the character two).
Well
str2[i] - '0'
Converts character representation of a digit ('3', '7', '9') into its integer value (3, 7, 9). More accurate (and more wordy) construction is
(int)char.GetNumericValue(str2[i])
What is implied, but not really stated, in the other answers is that int and char can be treated as fairly equivalent in c#. You can do math on chars just like you can on ints, and you can convert back and forth between int and char with ease. The number you get is based on the position in the utf8 character tables for the char.
'0' as a char has an int value of 48, so you could do:
int x = 7;
char c = x+48; //c would be '7' as a char, or 55 as an int
Other examples:
char c = 'a';
c++;
Console.Write(c); //prints 'b', because 'a' + 1 is b
It's quite logical and reasonably helpful sometimes* but the main reason you might see '0' is that it's easier to remember '0' than it is to remember 48 (it's slightly easier to remember the hex version 0x30)
All these give you char 5 from int 5:
char five = 5 + 48;
char five = 5 + 0x30;
char five = 5 + '0';
Which one would you find easiest to remember? :)
*for example, say you wanted to count the chars in an ascii string, you could do:
var counts = new int[256];
foreach(char c in string)
counts[c]++;
You can use the char to index the array just like you can an int. At the end of the operation "hello world" would have put a 1 in index 104 (the h), a 3 in index 108(the l) etc..
Sure these days you might use a Dictionary<char, int> but appreciating that intrinsic char/int equivalence and how it can be used has its merits..
What is '0' means?
In C++, it is a character literal.
The fundamental reason why/how str1[i] - '0' works is through promotion.
In particular when you wrote:
str1[i]-'0'
This means
both str1[i] and '0' will be promoted to an int. And so the result will be an int.
Let's looks at some example for more clarifications:
char c1 = 'E';
char c2 = 'F';
int result = c1 + c2;
Here both c1 and c2 will be promoted to an int. And the result will be an int. In particular, c1 will become(promoted to) 69 and c2 will become(promoted to) 70. And so result will be 69 + 70 which is the integer value 139.
In a C++ class I've the following code/while loop:
uint8_t len = 0;
while (*s != ',') {
len = (uint8_t)(len + 1u);
++s;
}
return (len);
The outcome should be a value between 0 and max 20.
As I receive a strange outcome, and started debugging. When I step through this
I get the following values for the variable Len:
‘\01’, ‘\02’, ‘\03’, ‘\04’, ‘\05’, ‘\06’, ‘\a’, ‘\b’, ‘\t’
I don’t understand the change from ‘\06’ to ‘\a’!
Can somebody explain this? I expect that the Len value is simply increased by 1 until character array pointer s hits the ',' char.
The values are correct, but your debugger interprets them as char type, not an integer type.
You can see escape sequences used in C++ here (and the corresponding values in ASCII).
\01 - 1 in octal, 1 in decimal
\02 - 2 in octal, 2 in decimal
...
\06 - 6 in octal, 6 in decimal
\a - equivalent to \07, the ASCII code to use the computer bell
\b - equivalent to \010 (10 octal, 8 decimal), the ASCII code for "backspace" character
\t - equivalent to \011 (11 octal, 9 decimal), the ASCII code for tabulator
etc.
I don't know if you can change the way your debugger interprets the data. Worst case, you can always print the value after casting it to int.
(gdb)p static_cast<int>(len)
This question already has answers here:
C++- Adding or subtracting '0' from a value
(4 answers)
Closed 3 years ago.
class Complex
{
public:
int a,b;
void input(string s)
{
int v1=0;
int i=0;
while(s[i]!='+')
{
v1=v1*10+s[i]-'0'; // <<---------------------------here
i++;
}
while(s[i]==' ' || s[i]=='+'||s[i]=='i')
{
i++;
}
int v2=0;
while(i<s.length())
{
v2=v2*10+s[i]-'0';
i++;
}
a=v1;
b=v2;
}
};
This is a class complex and the function input inputs string and convert it into integers a and b of class complex.
what is the requirement of subtracting '0' in this code
The characters representing the digits, '0' thru '9' have values that are (and must be) sequential. For example, in the ASCII character set the '0' character is encoded with the value 48 (decimal), '1' is 49, '2' is 50 and so on, until '9', which is 57. Other encoding systems may use different actual values for the digits (for example, in EBCDIC, '0' is 240 and '9' is 249), but the C standard requires that they are sequentially congruent. From §5.2.1 of the C11 (ISO/IEC 9899:201x) Draft:
In both the source and execution basic character sets, the value of
each character after 0 in the above list of decimal digits shall be
one greater than the value of the previous.
Thus, when you subtract the '0' character from another character that represents a digit, you get the numerical value of that digit (rather than its encoded value).
So, in the code:
int a = '6' - '0';
the value of the a will be 6 (and similarly for other digits).
The reason for not just using a value of (say) 48, rather than writing '0' is that the former would only work on systems that use that particular (i.e. ASCII) character encoding, whereas the latter will work on any compliant system.
"What does '0' means in c++" - The symbol '0' designates a single character (constant) with the value 0, which, when interpreted as an ASCII character (which it will be) has the numerical value 0x30 (or 48 in decimal). So, you are basically just subtracting 48.
I dont quite understand the logic of this function but I hope this will help:
'0' is a character literal for 0 in ASCII. The [] operator of string returns a character. So most likely s[i] - '0' is supposed to get you the digit stored in s[i] as a character. Example: '3' -'0' = 3. Note lack of ' around the 3.
The C and C++ standards require that the characters '0'..'9' be
contiguous and increasing. So to convert one of those characters to
the digit that it represents you subtract '0' and to convert a digit
to the character that represents it you add '0'.
In this case the goal is to convert the character in the integer digit that represent.
While watching a tutorial the speaker used
int deca['f' + '9' + 2 ] = {0};
I've never seen this on any other C++ tutorial and they didn't explain what it meant, and when I tried to implement it on my computer an error showed up.
As reference they were in a Linux environment.
It simply declares an integer array of N elements and initializes it to zero. What N evaluates to is determined by the 'f' + '9' + 2 expression. It evaluates to 161 if you are using ASCII code page or something else if you are using different code page. Every character literal has its corresponding integral value depending on the encoding used. In ASCII code page the character 'f' is represented by a number of 102 and the character '9' has a value of 57. The expression becomes 102 + 57 + 2 which equals 161. In other code pages those characters might have other values. Equivalent of:
int deca[161] = { 0 }; // If ASCII code page is used
My main goal is to convert int to char type. I used (char)1 to type cast, but it doesn't seem to work due to the following result:
When I compare '1' and (char)1 in c++ in the following code
if ('1' == (char)1)
{
return 1;
}
However, it seems that the comparison is either invalid due to different variable type or they are actually not the same thing. I always thought converting integer 1 to character is (char)1. Can anyone tell me how I can convert integer 1 to char '1'?
'1' is equal to (char)49 according to http://www.asciitable.com/
(char)1 is equal to SOH (start of heading) which is a non-printable character.
Because the ASCII equivalent of '1' is 49, not 1.
'1' == The character CODE value for the printable 1, traditionally ASCII value, but today, the code point value in whatever charset is used.
The old trick is (ch - '0') to get the numeric value.
Depending on the language you should use a conversion function for a full string.
C++ - stoi, stol or strol or stringstream
C - atoi or atol (these work in C++ too)
As ibiza said, char(49) is in fact what 1 is. This is because char draws from the ASCII library.
Because when you do (char)X with X a number, you are just converting X into the range of a char, either -128 to 127 or 0 to 255 (like a modulo).
For example, (char)300 gives 44 (because 300 % 256 = 44) and (char)1 gives 1. As said in the others comments, 1 is the ASCII equivalent of SOH (Start of Heading), and not of the character '1'.