c++ : Why does assigning true or false to char compiles fine

Question

I am wondering why the following program compiles fine

#include <iostream> int main() { char a = false; if (!a) { std::cout << "kdjk"; } char b = true; if (!b) { std::cout << "ppp\n"; } return 0; }

Output : prints kdjk

Why does assignment of bool value to char compiles without warnings?

It's implicit type conversion. true is 1 equals to SOH. false is 0 equals to NUL. — Louis Go
– Louis Go, Commented Aug 3, 2020 at 5:51
The why is in part that C++ had been developed with backwards compatibility to C, and in part when the bool type was actually added to the language it also used the fairly flexible implicit type conversions prevalent for other types. This is not a behavior that I like, but it would be a major breaking change to change the rules now, and would break C++'s rather impressive effort to maintain backwards compatibility to earlier C++ standards and C. — Eljay
– Eljay, Commented Aug 3, 2020 at 11:20

Keith Thompson · Accepted Answer · 2020-08-03 06:25:15Z

This:

char a = false;

invokes an implicit conversion from bool to char. The language rules for converting a bool value to an integral type (such as char) say that:

A prvalue of type bool can be converted to a prvalue of type int, with false becoming zero and true becoming one. The result is that the value of a is 0 and the value of b is 1.

zar · Accepted Answer · 2020-08-03 05:58:48Z

1

The value of false is 0 and that fits into character and ! is an operator that checks if the operand is 0 or not. That's why it works.

edited Aug 3, 2020 at 5:58

answered Aug 3, 2020 at 5:51

zar

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2 Comments

Keith Thompson Over a year ago

false is not a character.

zar Over a year ago

@KeithThompson Updated the answer, it's 0 and it fits into the character.

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