Why does the result of an expression depend on where the expression is placed?

Question

I am currently working on a more complex program, and I came across a very weird syntax error best demonstrated with the following minimal example:

#include <iostream>

int main(int argc, char *argv[]){
        char c = 1 + '0';
        std::cout << 1 + '0' << std::endl;
        std::cout << c << std::endl;
        std::cout << '0' + 1 << std::endl;
        return 1;
}

This example produces the following output:

$ ./program
49
1
49

What appears to be happening here is that when the conversion from a single digit integer expression to a character happens outside a stream statement, it succeeds, but when it happens inside such a statement, it produces a garbage answer.

I tried to find someone else asking something similar on Google, but I can't find anything relevant.

I am using g++ (Ubuntu 5.4.0-6ubuntu1~16.04.9) 5.4.0 20160609 on Ubuntu 16.04 LTS x64, but the issue occurs in clang version 3.8.0-2ubuntu4 (tags/RELEASE_380/final) as well, which rules out a compiler bug.

With `auto i = 1 + '0'; std::cout << i << std::endl;`. You will get `49` too. — Jarod42, Apr 04 '18 at 17:54

NathanOliver · Accepted Answer · 2018-04-04T15:35:25.030

7

The result of a int + char is an int. So when you do

char c = 1 + '0';

You actually get

char c = 49;

And now c has the value of 49. When yo go to print that though, cout's operator << is overloaded to take a char and instead of displaying 49 it will display the character that 49 represents.

When you do

std::cout << 1 + '0' << std::endl;

Since 1 + '0' is an int you get the int overload of operator << which just outputs the integer 49.

It should be noted that anything smaller than an int will be promoted to an int when doing arithmetic operations. That means that, char + char, short + short, char + short, ..., will all be an int so even doing something like

std::cout << static_cast<char>(1) + '0' << std::endl;

Will still give you a integer output. You have to cast the result back to a char to get a char like

std::cout << static_cast<char>(1 + '0') << std::endl;

edited Apr 04 '18 at 15:35

answered Apr 04 '18 at 15:28

NathanOliver

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When I change the first cout statement to `std::cout << static_cast(1) + '0' << std::endl;` the output remains unchanged, despite the fact it is now a char + char = char statement, so it should use the overload of `operator<<` for a char. Why? – john01dav Apr 04 '18 at 15:29
@john01dav Change it to `static_cast(1 + '0')`. – Hatted Rooster Apr 04 '18 at 15:30
@john01dav Anything that is smaller than an `int` is promoted to an `int` when doing operations to it. So `char + char`, `short + short`, `char + short`, ... will all be an `int` – NathanOliver Apr 04 '18 at 15:30
@NathanOliver Thanks! Although can I recommend that as a good addition to the answer for future readers? – john01dav Apr 04 '18 at 15:31

score 0 · Answer 2 · answered Apr 04 '18 at 15:29

but when it happens inside such a statement, it produces a garbage answer.

No.

char c = 1 + '0';
std::cout << c << std::endl;

In this case the rvalue int is implicitly converted to char. Then the char overload is used with std::cout, resulting in '1' as ASCII.

std::cout << 1 + '0' << std::endl;

In this case you're printing out the integer right away, so the integer overload is picked, this prints the numeric value of the '1' which is 49.

Why does the result of an expression depend on where the expression is placed?

2 Answers2