CODEWITHSUNDEEP
c++pointers
Domata Jemisti
Domata Jemisti

Reputation: 83

Printing an incremented character using cout

Why do I have to typecast the pointer in the following example inside the code function in order to return values of type char since I already have declared it as a constant pointer to char in the function definition?

In particular cout << (*p + 1) the result is in integer format, while when changing that to cout << (char) (*p + 1) then the result is displayed in char format, since I am typecasting it.

Does the << operator have some default arguments as to what type to display?

#include <iostream>

using namespace std;

void code(const char* p);

int main()
{
    code("This is a test");

    return 0;
}

void code(const char* p)
{
    while(*p)
    {
        cout << (*p + 1);
        p++;
    }
}

Upvotes: 0

Views: 71

Answers (2)

LogicStuff
LogicStuff

Reputation: 19617

*p is const char and adding 1 (integer literal) promotes it to int (See numeric promotions in this link). You have to cast it back to char:

cout << static_cast<char>(*p + 1); // abandon C-style cast

And yes, std::ostream::operator<< is overloaded to treat different types differently.

Upvotes: 5

SevenBits
SevenBits

Reputation: 2874

Change

cout << (*p+1);

to

cout << *(p+1);

This works for the reasons stated by @LogicStuff.

Upvotes: 0

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