the operator int() function converts the string to an int
class mystring
{
private:
chat str[20];
public:
operator int() // i'm assuming this converts a string to an int
{
int i=0,l,ss=0,k=1;
l = strlen(str)-1;
while(l>=0)
{
ss=ss+(str[l]-48)*k;
l--;
k*=10;
}
return(ss);
}
}
int main()
{
mystring s2("123");
int i=int(s2);
cout << endl << "i= "<<i;
}
So what's the logic behind operator int() ? What's the 48 in there? Can someone explain to me the algorithm behind the conversion from string to int.
Yes this converts a string to an integer. 48 is the ASCII value for '0'. If you subtract 48 from an ASCII digit you'll get the value of the digit (ex: '0' - 48 = 0, '1' - 48 = 1, ..). For each digit, your code calculates the correct power of 10 by using k (ranges between 1...10^{ log of the number represented by the input string}).
It does indeed convert a string to an integer. The routine assumes that all characters are decimal digits (things like minus sign, space, or comma will mess it up).
It starts with the ones place and moves through the string. For each digit, it subtracts off the ASCII value of '0', and multiplies by the current place value.
This does indeed convert the string to an integer. If you look at an ascii table the numbers start at the value 48. Using this logic (and lets say the string "123") the while loop will do:
l=2
ss=0+(51-48)*1
so in this case ss = 3
next loop we get
l=1
ss=3+(50-48)*10
so ss = 23
next loop
l=0
ss=23+(49-48)*100
so ss= 123
The loop breaks and we return an integer of value 123.
Hope this helps!
Related
I'd like to count number 1 in my input, for example,111 (1+1+1) must return 3and
101must return 2 (1+1)
To achieve this,I developed sample code as follows.
#include <iostream>
using namespace std;
int main(){
string S;
cout<<"input number";
cin>>S;
cout<<"S[0]:"<<S[0]<<endl;
cout<<"S[1]:"<<S[1]<<endl;
cout<<"S[2]:"<<S[2]<<endl;
int T = (int) (S[0]+S[1]+S[2]);
cout<<"T:"<<T<<endl;
return 0;
}
But when I execute this code I input 111 for example and my expected return is 3 but it returned 147.
[ec2-user#ip-10-0-1-187 atcoder]$ ./a.out
input number
111
S[0]:1
S[1]:1
S[2]:1
T:147
What is the wrong point of that ? I am totally novice, so that if someone has opinion,please let me know. Thanks
It's because S[0] is a char. You are adding the character values of these digits, rather than the numerical value. In ASCII, numerical digits start at value 48. In other words, each of your 3 values are exactly 48 too big.
So instead of doing 1+1+1, you're doing 49+49+49.
The simplest way to convert from character value to digit is to subtract 48, which is the value of 0.
e.g, S[0] - '0'.
Since your goal is to count the occurrences of a character, it makes no sense to sum the characters together. I recommend this:
std::cout << std::ranges::count(S, '1');
To explain the output that you get, characters are integers whose values represent various symbols (and non-printable control characters). The value that represents the symbol '1' is not 1. '1'+'1'+'1' is not '3'.
I am asking about the job of minus and plus notation on string , in this situation specifically :
Solver(string s) {
for (unsigned int i = 0; i < s.length(); i++) {
grid[i] = (int) (s[i] - '0'); // the minus here will remove 0's of string or not ?
}
}
int main() {
Solver ss(
(string) "850002400" + // the plus here will combine all strings together like Java or not ?
(string) "720000009" +
(string) "004000000" +
(string) "000107002" +
(string) "305000900" +
(string) "040000000" +
(string) "000080070" +
(string) "017000000" +
(string) "000036040"
);
}
operator+ for string concatenates them – as you discovered already. But there's no operator- for strings!
Have a close look, you are not subtracting from the string (s - '0'), but from the character s[i]. This won't remove the character from the string, but instead calculate a new value based on the character's value minus the value of zero character (which has a value of 48, in ASCII and compatible, at least – not the value null!). As digits are guaranteed to be contiguous by C++ standard (just like in C as well), you can reliably calculate decimal digits from characters that way.
This works for bases smaller than 10, too, but not larger ones, as next characters used for representation don't follow the decimal digits directly (and you might have to distinguish upper and lower case letters).
Side note: You don't need the cast to int: as type char is smaller in size than int, both operands will be promoted to int implicitly, so actually the calculation is done in int anyway and the result remains int...
string - C++ Reference
http://www.cplusplus.com/reference/string/string/
As is said in link above, the operator+ means "Concatenate strings".
If you want to CLIP the string, you can use the s.substr() function.
grid[i] = (int) (s[i] - '0')
the the minus in code means transform the 'char' to 'int'. For example,
string s="425";
char c = s[0]; //c='4';
int value = c-'0'; //value=4 it is a number
It is not the function of string, just a utilization of ASCII.
The function stoi(s[i]) can realize the same thing.
Subtracting '0' from any character of digit will return the integer value of that digit.
char seven = '7';
int value = (int)(seven - '0');
cout<<value<<endl;
Output:
7
In your example - was used to convert a character grid(1D) to integer grid(1D).
On the other hand, + sign between two or more string type data represent concatenation of string.
string s = "abc" + "def";
cout<<s<<endl;
Output:
abcdef
I want to make a number like 77 into a string but I can't use ascii because it's only from 0-9. Is there some way to make numbers become a string? So this is the result at the end: You input a number and it outputs the number but as a string. Example: input:123; output:"123".
Each digit can use ASCII. The number 123 uses a 1, a 2, and a 3, and the string that represents that value uses the characters '1', '2', and '3'.
The way to do the conversion yourself is to get each digit by itself and add the digit to '0'. Like this:
int value = 123
std::string result;
while (value != 0) {
int digit = value % 10;
char digit_as_character = digit + '0';
result.insert(0, 1, digit_as_character);
value = value / 10;
}
This is pretty much what you'd do if you were doing it by hand:
start with the value get the last digit of the value by dividing by
10 and looking at the remainder
write down a digit for the remainder
divide the value by 10 to remove the last digit, since you don't need
it any more.
Who said anything about ascii? The C++ standard doesn't.
Use the platform-independent std::to_string(77) instead.
Reference: http://en.cppreference.com/w/cpp/string/basic_string/to_string
I am new to C so I do not understand what is happening in this line:
out[counter++] = recurring_count + '0';
What does +'0' mean?
Additionally, can you please help me by writing comments for most of the code? I don't understand it well, so I hope you can help me. Thank you.
#include "stdafx.h"
#include "stdafx.h"
#include<iostream>
void encode(char mass[], char* out, int size)
{
int counter = 0;
int recurring_count = 0;
for (int i = 0; i < size - 1; i++)
{
if (mass[i] != mass[i + 1])
{
recurring_count++;
out[counter++] = mass[i];
out[counter++] = recurring_count + '0';
recurring_count = 0;
}
else
{
recurring_count++;
}
}
}
int main()
{
char data[] = "yyyyyyttttt";
int size = sizeof(data) / sizeof(data[0]);
char * out = new char[size + 1]();
encode(data, out, size);
std::cout << out;
delete[] out;
std::cin.get();
return 0;
}
It adds the character encoding value of '0' to the value in recurring_count. If we assume ASCII encoded characters, that means adding 48.
This is common practice for making a "readable" digit from a integer value in the range 0..9 - in other words, convert a single digit number to an actual digit representation in a character form. And as long as all digits are "in sequence" (only digits between 0 and 9), it works for any encoding, not just ASCII - so a computer using EBCDIC encoding would still have the same effect.
recurring_count + '0' is a simple way of converting the int recurring_count value into an ascii character.
As you can see over on wikipedia the ascii character code of 0 is 48. Adding the value to that takes you to the corresponding character code for that value.
You see, computers may not really know about letters, digits, symbols; like the letter a, or the digit 1, or the symbol ?. All they know is zeroes and ones. True or not. To exist or not.
Here's one bit: 1
Here's another one: 0
These two are only things that a bit can be, existence or absence.
Yet computers can know about, say, 5. How? Well, 5 is 5 only in base 10; in base 4, it would be a 11, and in base 2, it would be 101. You don't have to know about the base 4, but let's examine the base 2 one, to make sure you know about that:
How would you represent 0 if you had only 0s and 1s? 0, right? You probably would also represent the 1 as 1. Then for 2? Well, you'd write 2 if you could, but you can't... So you write 10 instead.
This is exactly analogous to what you do while advancing from 9 to 10 in base 10. You cannot write 10 inside a single digit, so you rather reset the last digit to zero, and increase the next digit by one. Same thing while advancing from 19 to 20, you attempt to increase 9 by one, but you can't, because there is no single digit representation of 10 in base 10, so you rather reset that digit, and increase the next digit.
This is how you represent numbers with just 0s and 1s.
Now that you have numbers, how would you represent letters and symbols and character-digits, like the 4 and 3 inside the silly string L4M3 for example? You could map them; map them so, for example, that the number 1 would from then on represent the character A, and then 2 would represent B.
Of course, it would be a little problematic; because when you do that the number 1 would represent both the number 1 and the character A. This is exactly the reason why if you write...
printf( "%d %c", 65, 65 );
You will have the output "65 A", provided that the environment you're on is using ASCII encoding, because in ASCII 65 has been mapped to represent A when interpreted as a character. A full list can be found over there.
In short
'A' with single quotes around delivers the message that, "Hey, this A over here is to receive whatever the representative integer value of A is", and in most environments it will just be 65. Same for '0', which evaluates to 48 with ASCII encoding.
I am working on a little c++ project that receives a char array input from the user. Depending on the value, I am converting it to an int. I understand there are better ways of doing this but I thought I'd try to convert it through ASCII to allow other uses later on. My current code for the conversion is:-
int ctoi(char *item){
int ascii, num = 0;
ascii = static_cast<int>(item[0]);
if(ascii >= 49 && ascii <=57){
num = ascii - 48;
}else{
return 0;
}
ascii = static_cast<int>(item[1]);
if(ascii >= 48 && ascii <=57){
num = num * 10;
num = num + (ascii - 48);
}else{
return 0;
}
return num;
}
It receives a input into the char array item[2] in the main function and passes this to the conversion function above. The function converts the first char to ASCII then the decimal value of the ASCII to num if its between 1 and 9, then it converts the second char to ASCII, if it is between 0 and 9, it times the value in num by 10 (move along one unit) and adds the decimal value of the ASCII value. At any point it may fail, it returns the value 0 instead.
When I cout the function after receiving a value and run this code in a console, it works fine for single digit numbers (1 - 9), however when I try to use a double digit number, it repeats digits such as for 23, it will output 2233.
Thanks for any help.
I wonder how you're reading the input into a two-character array. Note that it's customary to terminate such strings with a null character, which leaves just one for the actual input. In order to read a string in C++, use this code:
std::string s;
std::cin >> s;
Alternatively, for a whole line, use this:
std::string line;
getline(std::cin, line);
In any case, these are basics explained in any C++ text. Go and read one, it's inevitable!