Unexpected result from ternary operator

Question

I am trying to understand how the ternary operator works in C++.

I expect to see the same output for both print statements, yet the second print statement outputs 49.

Why is this?

#include <iostream> using namespace std; int main() { int test = 0; cout << "First character " << '1' << endl; cout << "Second character " << (test ? 3 : '1') << endl; return 0; }

Output:

First character 1
Second character 49

shargors · Accepted Answer · 2020-03-13 16:23:54Z

5

'1' got converted to an integer which represented the ASCII code for '1'. The ternary operator is supposed to have two values of the same type. You can not have 3 (an integer) and '1' (a char). That's why the conversion took place. If the implicit conversion could not have happened then a compiler error would have been generated.

answered Mar 13, 2020 at 16:23

shargors

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4 Comments

Mooing Duck Over a year ago

More clearly: Every expression has a type at compile time: The type of (test ? 3 : '1') is the common type of the last two parts: int and char. And the common type of those is an int. And (on this machine) '1' is represented by the value 49 in hardware.

alter_igel Over a year ago

This effect can have other nasty surprises. See cppquiz.org/quiz/question/217

shargors Over a year ago

@alter igel Whoa, never new that about the ternary operator or cppquiz.org. Thanks for both!

alter_igel Over a year ago

@shargors I just found out about it recently. I think it could be very useful to help settle arguments about what code does between developers.

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