Question

While reading “C++ Primer Plus 5th edition”, I saw this piece of code:

    cin.get(ch);
    ++ch;
    cout << ch;

So, this will lead to display the following character after ch. But, If I did it that way:

    cin.get(ch);
    cout << ch+1;

Now, cout will think ch is an int(try typecasting). So, why cout does so?
And why if I added 1 to a char it will produce a number?. And why there’s a difference between:
ch++, and ch + 1.

Answer 1

The reason this occurs is the type of the literal 1 is int. When you add an int and a char you get an int, but when you increment a char, it remains a char.

Try this:

#include <iostream>

void print_type(char)
{
    std::cout << "char\n";
}
void print_type(int)
{
    std::cout << "int\n";
}
void print_type(long)
{
    std::cout << "long\n";
}

int main()
{
    char c = 1;
    int i = 1;
    long l = 1;

    print_type(c); // prints "char"
    print_type(i); // prints "int"
    print_type(l); // prints "long"

    print_type(c+i); // prints "int"
    print_type(l+i); // prints "long"
    print_type(c+l); // prints "long"

    print_type(c++); // prints "char"

    return 0;
}