tests/cases/compute_robust_buffer_access_ssbo.amber

#!amber
# Copyright 2019 The Amber Authors.
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
#     https://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.

SHADER compute buffer_overrun GLSL
#version 430

// Only one invocation per workgroup.
layout(local_size_x = 1, local_size_y = 1, local_size_z = 1) in;

layout(set = 0, binding = 0) buffer BlockUint {
  uint data[];
} ssbo_uint;

layout(set = 0, binding = 1) buffer BlockInt {
  int data[];
} ssbo_int;

void main() {
  uint upper_bound = ssbo_uint.data[0];
  uint ui;
  for (ui = 0; ui < upper_bound; ++ui) {
    ssbo_uint.data[ui]++;
  }

  int int_lower = ssbo_int.data[0];
  int int_upper = ssbo_int.data[1];
  int i;
  for (i = int_lower; i < int_upper; ++i) {
    ssbo_int.data[i]++;
  }
}
END

BUFFER buf_uint DATA_TYPE uint32 DATA
10000 0 0 0
END

BUFFER buf_int DATA_TYPE int32 DATA
-5 10 0 0 0
END

PIPELINE compute pipeline
  ATTACH buffer_overrun
  SHADER_OPTIMIZATION buffer_overrun
    --graphics-robust-access
  END

  BIND BUFFER buf_uint AS storage DESCRIPTOR_SET 0 BINDING 0
  BIND BUFFER buf_int AS storage DESCRIPTOR_SET 0 BINDING 1
END

# Only one workgroup.  Having only one invocation execute ensures
# there are no race conditions.
RUN pipeline 1 1 1

# In the unsigned int index case:
#   The first 4 entries are incremented, as expected.
#   Beyond that, the remaining indices are clamped to the upper
#   bound of the length of the array, minus 1.
EXPECT buf_uint IDX 0 EQ 10001 1 1 9997

# In the signed int index case:
#
#   Index 0 is incremented 6 times, for i values -5, -4, -3, -2, -1, 0
#   since negative indices are clamped to 0.
#     So initial value -5 + 6 --> 1
#   Indices 1, 2, 3 are incremented once each.
#   Index 4 is incremented 6 times, for i values 4, 5, 6, 7, 8, 9
EXPECT buf_int IDX 0 EQ 1 11 1 1 6