Interesting things of ! operation in bool value
Following are one code snippet
#define T 0xFF
using namespace std;
int main(void) {
char c = T;
bool *pc = (bool *)(&c);
bool nc = !(*pc);
cout << "print: " << hex << nc <开发者_JS百科< endl;
nc = T;
cout << "print: " << hex << nc << endl;
nc = c;
cout << "print: " << hex << nc << endl;
}
The results is
print: fe
print: 1
print: 1
If type cast a char to a bool using the value 0xFF, bool value is 1.
But when type cast a char pointer to a bool pointer, 0xFF becoming 0xFE, only the last bit was flipped by ! operation.
Seems that gcc assumes a bool to be either 0 or 1 and if constructor of bool object is not called, it will just interpret the memory to contain a bool and flip the least significant bit.
But when bool is set by a char, does it trigger the copy contructor? But why different?
A bool
can only take the values true
and false
.
When you perform !(*pc)
you are getting undefined behavior because you telling the compiler to perform !
on a bool
object when in actual fact pc
is pointing at a char
object. (It is the C style cast from char*
to bool*
that is dangerous, even though the undefined behavior only occurs when you apply the !
operator.)
When you convert a char
or an int
(e.g. 0xFF
) to a bool
all non-zero values are converted to true and zero values are converted to false.
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