input char array and output it as integer array

Question

I would like to have input as char array in single line like: "12 5 232 65 76 435 2345" and output it as int array

I started writing a code but it works only for single digitals (like "3 6 4 5 6" etc...)

#include <iostream>

int main(){

char A[100];
int intA[100];

std::cout << "Input Array numbers" << std::endl;
std::cin.getline(A,100);
std::cout << std::endl;

for(int i=0; A[i]!= 0; i++){

    if ( A[i] != ' ' ){
        intA[i] = (int) A[i] - '0';
        std::cout << intA[i] << std::endl;
    }
}

return 0;
}

additional question: could someone explain me what actually "- '0'" does (I know without it a char casted to int would be and ASCII representation and you need to add it to get actual number)

-'0' subtracts the ascii value '0' from your character, so if '0' is 48 then '2' is represented by the number 50, than '2'-'0' is equivelant to 50 - 48, which is 2 — Sam I am says Reinstate Monica
– Sam I am says Reinstate Monica, Commented May 29, 2014 at 20:12

Vlad from Moscow · Accepted Answer · 2014-05-29 20:13:46Z

1

This task is better to do using std::stringstream and standard algorithm std::copy

For example (without testing):

#include <iostream>
#include <sstream>
#include <algorithm>
#include <iterator>

int main()
{
   char A[100];
   int intA[100];

   std::cout << "Input Array numbers" << std::endl;
   std::cin.getline(A,100);
   std::cout << std::endl;

   std::istringstream is( A );

   int *p = std::copy( std::istream_iterator<int>( is ),
                       std::istream_iterator<int>(),
                       std::begin( intA ) );

   std::copy( std::begin( intA ), p, std::ostream_iterator<int>( std::cout, " " ) );
   std::cout << std::endl;

   return 0;
}

answered May 29, 2014 at 20:13

Vlad from Moscow

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Comments

Columbo · Accepted Answer · 2014-05-29 20:24:24Z

1

could someone explain me what actually "- '0'" does

This is used if the character miuend is known to be a digit: Subtracting '0' gives the actual number that the Code-point is representing. For instance: '7' - '0' == 7.

I'm not sure whether this helps, but this is an example implementation:

std::string str; // use std::string
std::getline( std::cin, str );
std::istringstream stream(str);

std::vector<int> vec{ std::istream_iterator<int>{stream}, {} };
for( auto i : vec )
    std::cout << i << '\n';

edited May 29, 2014 at 20:24

answered May 29, 2014 at 20:15

Columbo

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4 Comments

vsoftco Over a year ago

A useful trick I learned recently is that you can just use {} (in C++11) instead of the second iterator to specify the end of the stream, otherwise perfectly fine +1

Columbo Over a year ago

I always thought that would lead to an error in template argument deduction since {} is a non-deduced context (never tried it out, but i'm gonna check the standard on that one). Apparently it works out. Editet it, thanks!

vsoftco Over a year ago

I think the template deduction is ok, as it deduce it from the first iterator, then the second one is just a default initialization object, which for istream_iterator's is the end of stream. I did not check the standard but never had a warning/problem with using {}.

vsoftco Over a year ago

for example, in the following you deduce the template argument of the constructor from the first parameter: struct Foo { template<typename T> Foo(T a, T b){}; }; int main(){ Foo f(2.3,{});}

Chewie · Accepted Answer · 2014-05-29 20:27:23Z

0

It only works for single digits because you are working on the chars as letters (42 is not 42 it is '4' and then '2'), so your for loop runs on first '4', and then it runs on '2' - also incrementing i for the index used in intA. So you are not building 42 out of '4' and '2' but just placing 4 in intA[0] then 2 in intA[1].

When you read your input you need to first gather the multiple digit numbers into a number format and not separate letters.

The -'0' is needed because you are working on the letter 4 (ASCII value 52) and not the number 4, so you subtract 48 (ASCII value of '0') to get the numerical value.

answered May 29, 2014 at 20:27

Chewie

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Comments

user1618143 · Accepted Answer · 2014-05-29 20:37:01Z

0

If you want to do it all manually (for whatever reason), your current approach is pretty close to what you need. Try pushing an integer to the output array only when you see a space, and otherwise multiply the previous number by ten and add the new digit. Like this:

int curSum = 0;
int curIndex = 0;
for(int i=0; A[i]!= 0; i++){

    if ( A[i] == ' ' ){
        intA[curIndex] = curSum;
        curIndex++;
        std::cout << curSum << std::endl;
        curSum = 0;
    } else {
        curSum = curSum*10;
        curSum += (int) A[i] - '0';
    }
}

However, unless this is a homework problem or something, using an existing library is probably a better solution.

answered May 29, 2014 at 20:37

user1618143

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Comments

Rainbacon · Accepted Answer · 2014-05-29 20:53:25Z

0

To convert multiple characters to an integer copy those characters into a separate char[] and call atoi on it.

answered May 29, 2014 at 20:53

Rainbacon

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