Answer 1

std::stoi() requires a constant std::string reference as first argument, here you have given it a `char`, due to the use of subscript (index) operator.

You can also play with the digits ASCII value, given you are sure all the characters in the std::string were digit (0 ~ 9).

for( size_t i = 0; i < L; i++)
{
arr[ i ] = Num[ i ] - '0';
}

Answer 2

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Answer 3

Incidental header inclusion

https://www.sololearn.com/post/69831/?ref=app

Answer 4

The error happens because you pass a char to std::stoi() which were expecting a constant std::string reference. It's not because <string> header not explicitly included.

Answer 5

string Num="1234";

In the previous code, you get each character from the string variable <Num> by using subscript operator []; you then pass the character as argument for std::stoi(). But that didn't work because std::stoi() was expecting a constant string reference, whereas a char was given instead.

arr[i] = stoi(Num[i]);

Num[0] = '1'
Num[1] = '2'
Num[2] = '3'
Num[3] = '4'

In the recent code, you use an array of string, thus each element are a string themselves, and passing them as argument to std::stoi() is fine. It's the type of argument that was expected by std::stoi().

(Edited)

Answer 6

Yes that's right Manav Roy,

std::string str {"Hello"};

str[0] = 'H'
str[1] = 'e'
str[2] = 'l'
str[3] = 'l'
str[4] = 'o'

And each of them are char ...

Answer 7

Contd:

string s(1,Num[i]); is creating a string with constructor , argument 1 represents number of length and is filled with Num[i] charecter

ex:

string s(3,'4'); //this creates a string 444 so now you can pass to stoi(), like
cout<<stoi(s);

There are other ways also..
it's equal to
string s ;
s += Num[i];

ex2:
s += 'a'; //this makes s="a"; string from char 'a'

Answer 8

#include <iostream>
using namespace std;

int main() {
string Num="1234abcde";

int arr;
size_t ss;//unsigned integer datatype used to represent the size of objects in bytes,
//here ss used as pointer to store position of argument after the last digit taken for convertion
arr=stoi(Num,&ss,10); //10 base represents base type of number in string, which is going to be extracted from strong to convert to int.

cout<<arr<<endl; //output: 1234
cout<<Num.substr(ss)<<endl; //output: abcde

//in case, you want specify base type but you dont want pointer position then you can use nullptr

cout<<stoi("112",nullptr,2); //output:3 (11 in base 2 => 3 in base 10)

//because cout<<stoi("112",10)

cout<<stoi("112",&ss); valid

cout<<stoi("112",nullptr); valid but equalto cout<<stoi("112");

return 0;

}

edit:

Answer 9

What you want is string.substr().

Num[i] gives you char, stoi needs string.

Num.substr(i,1) gives you string.

Why use char only to convert back to string.

https://code.sololearn.com/cyqC7o57xw45/?ref=app

Answer 10

You can do it without loop just in one line.

Error is because the first argument to
stoi() function is expect a string. But you are giving a character.
Syntax :
stoi( string, position, int base);
position is from where to start convertion
Base is type to convert,default base 10 (decimal)

#include <iostream>
using namespace std;

int main() {
string Num="1234";
//int L=Num.length();
int arr;
//for(int i=0;i<L;i++)
{
arr=stoi(Num);
}
cout<<arr;
return 0;
}

edit:
Manav Roy
'1' is a character (single quotes)
where as
"1" is a string (using of double quotes)

Your first code loop, from string Num[i] extracts character, and 2nd code loop, from string of array, Num[I] extracts a string. so its fine to stoi(), but first not.

Answer 11

Then convert char to string and use it like:

#include <iostream>
using namespace std;

int main() {
string Num="1234";
int L=Num.length();
int arr[L];
for(int i=0;i<L;i++)
{
string s(1,Num[i]);
arr[i]=stoi(s);

cout<<arr[i];
}
return 0;
}

Answer 12

hello guys can anyone help me