pub fn removeByPtr(self: *Self, key_ptr: *K) void self: *Selfkey_ptr: *K1270
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1280 pub fn removeByPtr(self: *Self, key_ptr: *K) void {
// if @sizeOf(K) == 0 then there is at most one item in the hash
// map, which is assumed to exist as key_ptr must be valid. This
// item must be at index 0.
const idx = if (@sizeOf(K) > 0)
@as([*]K, @ptrCast(key_ptr)) - self.keys()
else
0;
self.removeByIndex(idx);
}
Delete the entry with key pointed to by key_ptr from the hash map. key_ptr is assumed to be a valid pointer to a key that is present in the hash map.
TODO: answer the question in these doc comments, does this increase the unused capacity by one?