Conversion of char * to bool

[Christopher Benson-Manica]

I have compiled the following program with two different
implementations without any warnings at all. Why is a char *
converted to a bool (apparently) without a cast?

void foo( bool a ) {}

int main() {
foo( "123" );
const char *bar="123";
foo( bar );
return 0;
}

 

[TB]

Christopher Benson-Manica sade:

I have compiled the following program with two different
implementations without any warnings at all. Why is a char *
converted to a bool (apparently) without a cast?

Because the C++ Standard allows it. It's convenient.
A zero arithmetic value or null pointer value is converted to false,
everything else to true.

void foo( bool a ) {}

int main() {
foo( "123" );
const char *bar="123";
foo( bar );
return 0;
}

TB

 

[Christopher Benson-Manica]

Because the C++ Standard allows it. It's convenient.

I disagree wholeheartedly WRT the "convenience" of this behavior. I
spent some time tracking down some bizarre behavior that turned out to
be caused by a change in the prototype of a function from

void foo( const char *, const char * );

to

void foo( const char *, bool, const char * );

A zero arithmetic value or null pointer value is converted to false,
everything else to true.

What would have been so complicated about

foo( "foo", myCharPtr != NULL );

?

 

[Bo Persson]

I disagree wholeheartedly WRT the "convenience" of this behavior.

All too true!

The convenience comes for old C from before the appearance of the bool
type. Traditionally any expression could be used as a condition value,
treating zero as false and non-zero as true.

Unfortunately, there is way too much old code out there, for anyone to
dare to change this rule.


Bo Persson