const char* test(bool i)
{
const char t[] = "aa\n";
const char* p = "bbb\n";
if(i)
return p;
return t;
}
int main(array<System::String ^> ^args)
{
printf(test(true));
printf(test(false));
return 0;
}
That returns something of sort:
bbb
%^&$^$%
It is clear that test(false) returns a pointer to a local variable. The question is that p is also local variable. Why the memory for "bbb\n" is not cleaned after the function returns. I thought const char[] is interpreted same way as const char* but it is not true as it seems.
p is a local variable, which you return by value, but points to a string literal, which resides in read-only memory, not in the automatic memory allocated for the method.
Returning t and the using it indeed results in undefined behavior.
Also, don't think of pointers and arrays to be equivalent.
Although p is a local variable, what it points to is not local - it is a compile-time string constant; it is legal to return that constant's address from a function.
t is different, because the compile-time string constant is copied into an automatic storage area, causing an undefined behavior on dereferencing the returned pointer.
