C++ Convert Char To Int?

[supereater14]

i am looking to convert a char array to an int array

example:

"4" becomes 4

never mind, instead, tell me what's wrong with this code:

#include <fstream>

#include <iostream>

using namespace std;

int main(){

int x[25];

int y = 0;

char m[100];

int z[100];

ifstream in("input.txt");

/*if (!in){

cout << "read error\n";

return 1;

}*/

in >> m;

/*for(int b = 0; b == 99; b++){

z[c] = m[c];

}*/

//*z << m;

for (int c = 0; c == 99; c++){

switch(m[c]){

case '1':

y--;

break;

case '2':

y++;

break;

case '3':

x[y]--;

break;

case '4':

x[y]++;

break;

case '5':

ofstream out("output.txt");

out << x[y];

out.close();

in.close();

return 0;

break;

}

}

in.close();

return 0;

}

Edited March 30, 2010 by supereater14

[exxon]

You could use atoi which converts strings into an int. If you would like to write your own function for that then something like this would work:

char array[50];
int n[50];
...
for( int i = 0; i < 50; i++ )
    n[i] = (int)array[i] - 48;

That would just convert to single digits. But you get the idea.

EDIT:

instead of that whole switch statement and stuff you can do what I posted. Although depends on how your input file is setup. Assuming in your input it says something like:

1 2 3 4 5 6 7 8

you would have to include something like:

if( array[i] != ' ' )
...

Edited March 30, 2010 by exxon

[supereater14]

that would actually work

[supereater14]

[quote

EDIT:

instead of that whole switch statement and stuff you can do what I posted. Although depends on how your input file is setup. Assuming in your input it says something like:

1 2 3 4 5 6 7 8

you would have to include something like:

if( array[i] != ' ' )
...

wouldn't

wouldn't that stop at the first whitespace?

[requiem]

Put his snippet in a loop.

while (fin)  {
   if ( array[i] != ' ' )
   ....
   }

[supereater14]

i'm trying atoi, it's not working. it replies:

cannot convert parameter 1 from 'char' to 'const char *'

[requiem]

i'm trying atoi, it's not working. it replies:

cannot convert parameter 1 from 'char' to 'const char *'

Can you post a snippet of how you're using it?

[supereater14]

char blahblah[100];

int hi[100];

...

for (int b = 0; b == 99; b++){

hi = atoi(blahblah;

}

[supereater14]

if it helps, i'm using microsoft's visual c++ 2008 (really annoying compiler)

[exxon]

The thing about atoi is that its a C function so you would have to include a C library. Cant remember which.

anyways just did this really fast so sorry if its messy and not that optimal

#include <iostream>
#include <cstring>

using namespace std;

int main(){
    char array[] = "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15";
    int n[15];
    int x = 0;
    int j = 0;
    for( int i = 0; i<=strlen(array); i++ )
    {
        if( array[i] == ' ' || array[i] == '\0')
        {
            n[j] = x;
            j++;
            x=0;
        }
        else
        {
            x *= 10;
            x += (int)array[i]-48;
        }
    }

    for( int i = 0; i < 15; i++ )
    {
        cout << n[i] << endl;
    }
    return 0;
}

[requiem]

After some googling, (and then some self hatred) I found that atoi only works for cstrings, not just a char.

What does your input file look like?

[supereater14]

the entire error is

error C2664: 'atoi' : cannot convert parameter 1 from 'char' to 'const char *'

though, meaning that it recognizes atoi

[supereater14]

After some googling, (and then some self hatred) I found that atoi only works for cstrings, not just a char.

What does your input file look like?

the input file looks like, for example:

4445

[exxon]

the input file looks like, for example:

4445

where 4445 is one number or 4 separate numbers?

by the way with the code I posted you should be able to get it to work. All you have to do is change the char part.

[supereater14]

4 numbers

btw, atoi requires a const char*, not a cstring

Edited March 30, 2010 by supereater14

[supereater14]

well, atoi(&whatever) works (where whatever is a char), but the code still doesn't work as a whole.

[supereater14]

i think the best solution is (int(whatever) - 48)

[exxon]

if the input file is "44459103123" where each digit is its own separate number then the initial code i posted should be all you need for this.

[requiem]

4 numbers

btw, atoi requires a const char*, not a cstring

Well all a cstring is is a character array. :P

[supereater14]

here's the code, but when i run it, it doesn't produce an output file.

#include <cstring>
#include <fstream>
#include <iostream>
using namespace std;
int main(){
    int x[25];
    int y = 0;
    char m[100];
    int z[100];
    //*char h[100];
    ifstream in("input.txt");
    /*if (!in){
        cout << "read error\n";
        return 1;
    }*/
    in >> m;
    /*for(int clap = 0; clap == 99; clap++){
        h[clap] = m[clap];
    }*/
    for(int b = 0; b == 99; b++){
    z[b] = (int(m[b]) - 48);
    }
    //*z << m;
    for (int c = 0; c == 99; c++){
        switch(z[c]){
            case 0:
                break;
            case 1:
                y--;
                break;
            case 2:
                y++;
                break;
            case 3:
                x[y]--;
                break;
            case 4:
                x[y]++;
                break;
            case 5:
                ofstream out("output.txt");
                out << x[y];
                out.close();
                in.close();
                return 0;
                break;
        }
    }
    in.close();
    return 0;
}

[supereater14]

all right, apparently it cannot find the "input.txt" file

[exxon]

dont you have to open the file? Try doing it like this:

ifstream input;
input.open("input.txt", ifstream::in);
or just
input.open("input.txt");

and that should work

[requiem]

It's not generating an output file cause you never call ofstream...

You have an ifstream in, but no ofstream out.

[supereater14]

It's not generating an output file cause you never call ofstream...

You have an ifstream in, but no ofstream out.

sure i do, it's under case 5.

[exxon]

I still believe you might need that file.open("filename") part.