How to compare input with ASCII code?

Question

I am trying to write a console program, which gets a number from the input and puts it into an array. The read in lasts as long as the user does not hit the "space" button. I have tried several things, but my program won't compare the input with the ASCII code of "space". Thanks in advance!

#include <iostream>
using namespace std;

int main()
{
    int fabcd[25],number_of_items = 0;
    cout << "The read in loop lasts while you don't hit space:\n";
    while((char)cin.get() != 32)
    {
            cin >> fabcd[number_of_items];
            number_of_items++;      
    }   
    return 0;
}

Answer 1

There are several problems with your code. The most fundamental is that you do two inputs each time through the loop, the first in the while, and the second using >> in the loop. Note too that the second will skip any spaces, then convert the input to an integer. If the input cannot be legally converted to an integer (e.g. the next character is an 'a'), the input goes into an error state, and cin.get() will always return EOF. And of course, if you are entering numbers, you can easily enter more than 25, and overrun the buffer.

If you just want to enter a sequence of numbers, then:

std::vector<int> fabcd;
int tmp;
while ( std::cin >> tmp ) {
    fabcd.push_back( tmp );
}

is all you need. (The numbers must be separated by white space.) If you want to enter the digits, up until you encounter a white space, something like the following should work:

std::vector<int> digits;
int tmp = std::cin.get();
while ( tmp != EOF && ! std::isspace( tmp ) ) {
    digits.push_back( tmp /* or tmp - '0' */ );
    tmp = std::cin.get();
}

Notice that the results of std::cin.get() are assigned to an int, not a char. This function returns an int intentionally, because it must be able to return the out of band value EOF. There are a couple of variants on this: if you use std::cin::peek(), instead of std::cin::get(), you'll not actually extract the character (which you'll have to do in the loop); and there is also std::get( char& ), which extract the character into the given char, provided it isn't at EOF. Using this function, the loop would be:

std::vector<int> digits;
char tmp;
while ( std::get( tmp ) && ! std::isspace( tmp ) ) {
    digits.push_back( tmp );
}

(You might also want to use std::isdigit, instead of ! std::ispace to control the loop.)

EDIT:

One last point: the actual numeric code for a space or whatever depends on the implementation, and is not necessarily 32. Also, you'll want to handle tabs and newlines similarly. You should always use the isspace, etc. functions. Or if you really do want to check for a space, and only a space, compare with ' ', and not 32.

Answer 2

(char)13 ist the carriage return not the space.

isn't the point that you would do:

((char)cin.Get()) !=' '

Answer 3

The acii code for space is 32. So you should substitute 64 or 13, which is carraiage return to 32:

while((char)cin.get() != 32)

However the problem might be that cin may stop on some characters for example whitespace: You can disable this behaviour by setting: cin.unsetf(ios::skipws); according to C++, user input check for '\0' stops at spaces?

Answer 4

Here is the answer:

int fabcd[25],number_of_items = 0;
cout << "The read in loop lasts while you don't hit space:\n";
char a;
while((a=cin.get()) != ' ')
{

        fabcd[number_of_items]=a;
        number_of_items++;      
}   
return 0;

Answer 5

try this :

 #include <iostream>
#include <conio.h >
using namespace std;

int main()
{
    int fabcd[25],number_of_items = 0;
    cout << "The read in loop lasts while you don't hit space:\n";
    while( _kbhit() != 32  )
    {
            char ch =getche();
            if (ch != 32 )
            {
           cin >> fabcd[number_of_items];
            number_of_items++;
            }
            else
                break;

    }   
    cout << "\nnumber of items:"<<number_of_items;
    cin.get();
    cin.get();
    return 0;
}