include path problem when building a simple library with bazel - c++

Question

The demo structure like this:
.
├── demo
│ ├── include
│ │ └── func.h
│ │
│ └── src
│ ├── BUILD
│ │
│ └── func.cc
└── WORKSPACE
func.h:
#pragma once
int square(int);
func.cc: include from the root
#include "demo/include/func.h"
int square(int i) { return i * i; }
BUILD:
cc_library(
name = "simple_demo",
srcs = ["func.cc"],
visibility = ["//visibility:public"],
)
Bazel build command in the dir of WORKSPACE:
bazel build //demo/src:simple_demo
The error occurs.
demo/src/func.cc:1:10: fatal error: 'demo/include/func.h' file not found
And I saw this in https://docs.bazel.build/versions/1.1.0/be/c-cpp.html#hdrs
All header files that are used in the build must be declared in the hdrs or srcs of cc_* rules. This is enforced.
However, if I add hdrs = "../include/func.h" to the BUILD, another error will occur
segment '..' not permitted

Answer 1

By default in Bazel labels are not allowed to cross the package boundaries. In your case func.h lives outside the package defined by the BUILD file, that is limited to demo/src (and sub-folders).
Assuming you're creating a library where func.h is the public header visible from outside, I would change the folder structure to:
.
├── demo
│ ├── include
│ │ └── func.h
│ ├── src
│ │ └── func.cc
│ └── BUILD
└── WORKSPACE
and the BUILD file to also include func.h:
cc_library(
name = "simple_demo",
hdrs = ["include/func.h"],
srcs = ["src/func.cc"],
visibility = ["//visibility:public"],
)
Notice that even with the suggested folder structure, if you don't include the hdrs = ... line (or don't include func.h in srcs = ..., if the header is not meant to be public), you will get an error for the reason you mentioned.

Answer 2

I am not sure if this might work, but can you try giving
'../include/func.h'
instead of 'demo/include/func.h' because I believe its not able to find a demo folder inside your src folder since it starts searching from there.