Character arithmetic in C and C++

4

As already known character known character range is between -128 to 127 or 0 to 255. This point has to be kept in mind while doing character arithmetic. To understand better let’s take an example.

// C program to demonstrate character arithmetic.
#include <stdio.h>

int main()
{
    char ch1 = 125, ch2 = 10;
    ch1 = ch1 + ch2;
    printf("%d\n", ch1);
    printf("%c\n", ch1 - ch2 - 4);
    return 0;
}

Output:

-121
y 

So %d specifier causes an integer value to be printed and %c specifier causes a character value to printed. But care has to taken that while using %c specifier the integer value should not exceed 127.
So far so good.

But for c++ it plays out a little different.

Look at this example to understand better.

// A C++ program to demonstrate character
// arithmetic in C++.
#include <bits/stdc++.h>
using namespace std;

int main()
{
    char ch = 65;
    cout << ch << endl;
    cout << ch + 0 << endl;
    cout << char(ch + 32) << endl;
    return 0;
}

Output:

A
65
a 

Without a ‘+’ operator character value is printed. But when used along with ‘+’ operator behaved differently. Use of ‘+’ operator implicitly typecasts it to an ‘int’. So to conclude, in character arithmetic, typecasting of char variable to ‘char’ is explicit and to ‘int’ it is implicit.

This article is contributed by Parveen Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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