It does though? Just compiled a small program that creates a vector, and GDB is perfectly happy accessing it using this syntax. It will even print std::string’s correctly if you cast them to const char* by hand. (Linux x86-64, GDB 14.2.)
(gdb) p unrolls_[0]
Could not find operator[].
(gdb) p unrolls_[(long)0]
Could not find operator[].
(gdb) p unrolls_.data_.mem[0]
$2 = {
I'm on Linux, gdb 15.1.
That might be it. If that operator isn’t actually ever emitted out of line, then GDB will (naturally) have nothing to call. If it helps, with the following program
template<typename T>
struct Foo {
int operator[](long i) { return i * 3; }
};
Foo<bool> bar;
template int Foo<bool>::operator[](long); // [*]
int main(void) {
Foo<int> foo;
__asm__("int3");
return foo[19];
}
What does that mean? I think `print str.c_str()` has worked for me in GDB before, but sounds like you did something different.
Some implementations even put char[0] at the first byte in the optimized form.
It doesn't seem possible in actual C++ that the cast from non-pointer to pointer would work at all (even if a small string happens to be inlined at offset 0.) Like GP, I looked for a conversion operator, and I don't think it's there. Maybe it is a feature of the gdb parser.
https://devblogs.microsoft.com/oldnewthing/20240510-00/?p=10...