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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.
Related
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.
This question already has answers here:
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.
I have an example:
int var = 5;
char ch = (char)var;
char ch2 = var+48;
cout << ch << endl;
cout << ch2 << endl;
I had some other code. (char) returned wrong answer, but +48 didn't. When I changed ONLY (char) to +48, then my code got corrected.
What is the difference between converting int to char by using (char) and +48 (ASCII) in C++?
char ch=(char)var; has the same effect as char ch=var; and assigns the numeric value 5 to ch. You're using ASCII (supported by all modern systems) and ASCII character code 5 represents Enquiry 'ENQ' an old terminal control code. Perhaps some old timer has a clue what it did!
char ch2 = var+48; assigns the numeric value 53 to ch2 which happens to represent the ASCII character for the digit '5'. ASCII 48 is zero (0) and the digits all appear in the ASCII table in order after that. So 48+5 lands on 53 (which represents the character '5').
In C++ char is a integer type. The value is interpreted as representing an ASCII character but it should be thought of as holding a number.
Its numeric range is either [-128,127] or [0,255]. That's because C++ requires sizeof(char)==1 and all modern platforms have 8 bit bytes.
NB: C++ doesn't actually mandate ASCII, but again that will be the case on all modern platforms.
PS: I think its an unfortunate artifact of C (inherited by C++) that sizeof(char)==1 and there isn't a separate fundamental type called byte.
A char is simply the base integral denomination in c++. Output statements, like cout and printf map char integers to the corresponding character mapping. On Windows computers this is typically ASCII.
Note that the 5th in ASCII maps to the Enquiry character which has no printable character, while the 53rd character maps to the printable character 5.
A generally accepted hack to store a number 0-9 in a char is to do: const char ch = var + '0' It's important to note the shortcomings here:
If your code is running on some non-ASCII character mapping then characters 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 may not be laid out in order in which case this wouldn't work
If var is outside the 0 - 9 range this var + '0' will map to something other than a numeric character mapping
A guaranteed way to get the most significant digit of a number independent of 1 or 2 is to use:
const auto ch = to_string(var).front()
Generally char represents a number as int does. Casting an int value to char doesn't provide it's ASCII representation.
The ASCII codes as numbers for digits range from 48 (== '0') to 58 (== '9'). So to get the printable digit you have to add '0' (or 48).
The difference is that casting to char (char) explicitly converts the digit to a char and adding 48 do not.
Its important to note that an int is typically 32 bit and char is typically 8 bit. This means that the number you can store in a char is from -127 to +127(or 0 to 255-(2^8-1) if you use unsigned char) and in an int from −2,147,483,648 (−231) to 2,147,483,647 (231 − 1)(or 0 to 2^32 -1 for unsigned).
Adding 48 to a value is not changing the type to char.
This question already has answers here:
How does subtracting the character '0' from a char change it into an int?
(4 answers)
Closed 8 years ago.
I was trying to go in a loop and each time convert a character in a string to it's integer value and I don't mean the ASCII value. Tried to use atoi() with no luck but then I stumbled upon this question Convert single char to int and my code worked. The code is as follows:
std::string tmp = "87532621";
for(i=0;i<tmp.length();i++)
{
**int num = tmp[i] - '0';**
//do some processing
}
I fail to understand why the following line of code works. My question is how is it converting the char value to integer type?
int num = tmp[i] - '0';
Each char in your string is an ascii value. The ascii values are just 7 bit numbers.
The numerical values for the character digits lies in a sequence 0123456789 which is very convenient because it makes it possible to write
int zero = '0' - '0'; // 0 (zilch)
int one = '1' - '0'; // one (1)
int nine = '9' - '0'; // 9 (three times three)
And so on.
The actual numerical values are not important for this to work. The fact that the are next to each other in the character set is.
See wikipedia - ascii for the actual numerical values.
Is the code below converting a character into its ASCII value?.
I faced a piece of code while studying evaluation of postfix operation,where it says "the expression converts a single digit character in C to its numerical value".?
int x=getch();
int c=x-'0'; /*does c have the ASCII value of x?*/
printf("%d",c);
No, it's converting the ASCII value to a number >= 0.
Let's say you type '1'. getch() will return 49 which is the ASCII value of '1'. 49 - '0' is the same as 49 - 48 (48 being the ASCII value for '0'). Which gives you 1.
Note that this code is broken if you enter a character that is not a number.
E.g. if you type 'r' it will print 'r' - '0' = 114 - 48 = 66
(Ref.)
No, it's giving the numeric value of a digit. So '5' (53 in ASCII) becomes 5.
Is the code below converting a character into its ASCII value?
It isn't. It's doing the opposite (converting an ASCII value to the numerical value) and it only works for decimal digits.
To print the ascii value all you need to do is :
int x=getch();
printf("%d",x);
If you are sure that you only want to accept integers as input then you need to put some constraints to the input before proceeding to process them.
int x = getch();
if (x >='0' || x <= '9') {
int c = x - '0'; // c will have the value of the digit in the range 0-9
. . .
}
Any character(in your case numbers) enclosed within single quotes is compiled to its ASCII value. The following line in the snippet above translates to,
int c=x-'0'; ---> int c= x-48; //48 is the ASCII value of '0'
When the user inputs any character to your program, it gets translated to integer as follow,
If x = '1', ASCII of '1' = 49, so c= 49-48 = 1
If x = '9', ASCII of '9' = 57, so c= 57-48 = 9 and so on.