#Copyright (c) 2025 Huawei Technologies Co., Ltd.
#This program is free software, you can redistribute it and/or modify it under the terms and conditions of
#CANN Open Software License Agreement Version 2.0 (the "License").
#Please refer to the License for details. You may not use this file except in compliance with the License.
#THIS SOFTWARE IS PROVIDED ON AN "AS IS" BASIS, WITHOUT WARRANTIES OF ANY KIND, EITHER EXPRESS OR IMPLIED,
#INCLUDING BUT NOT LIMITED TO NON-INFRINGEMENT, MERCHANTABILITY, OR FITNESS FOR A PARTICULAR PURPOSE.
#See LICENSE in the root of the software repository for the full text of the License.

import sys
sys.stdout.reconfigure(encoding='utf-8')

p = r'D:\AI\Vector_error_testing_angent\skills\ascendc-fuzz-cpp-gen\scripts\gen_error_cpp.py'

with open(p, 'rb') as f:
    data = f.read()

# Problem: emit_raw lines contain Python source like:
#   emit_raw('LOG_PRINT("...\\n");')
# where \\n in Python source = 2 actual chars: backslash + n
# which becomes C++ output: LOG_PRINT("...\n")  <- CORRECT!
#
# BUT some lines have FOUR chars: backslash backslash n
# emit_raw('LOG_PRINT("...\\\\n");') 
# which becomes C++ output: LOG_PRINT("...\\n") <- WRONG, literal backslash-n
#
# We need to find and fix lines where emit_raw has too many backslashes before n

lines = data.split(b'\r\n')
fixed = 0
for i, line in enumerate(lines):
    if b'emit_raw' not in line:
        continue
    # Count consecutive backslashes before 'n' in string arguments
    # Replace occurrences of \\n (two backslashes + n) with single backslash + n
    # but ONLY when it's an even number > 1
    # Simple approach: replace all \\\\n with \\n in emit_raw lines
    old = line
    # Replace 4-backslash+latin-n with 2-backslash+latin-n
    # In raw bytes: find the 2-byte sequence 0x5C 0x5C 0x6E and replace with 0x5C 0x6E
    line = line.replace(b'\\forward slash', b'xxx')  # test first
    
# OK let me just do it character by character
lines = data.split(b'\r\n')
new_lines = []
fixed = 0

for line in lines:
    if b'emit_raw' in line or (b'emit(' in line and b'n' in line):
        # Process: find all backslash sequences before 'n' and reduce by half
        result = bytearray()
        i = 0
        while i < len(line):
            if line[i:i+1] == b'\\' and i + 1 < len(line) and line[i+1:i+2] == b'\\' and i + 2 < len(line) and line[i+2:i+3] == b'n':
                # Found \\\\n -> reduce to \\n
                result.extend(b'\\n')
                i += 3
                fixed += 1
            else:
                result.append(line[i])
                i += 1
        new_lines.append(bytes(result))
    else:
        new_lines.append(line)

new_data = b'\r\n'.join(new_lines)
print(f'Fixed {fixed} occurrences')

# Verify
count = new_data.count(b'\\forward')
if count > 0:
    print('ERROR: test replace still present')
    sys.exit(1)

# Check remaining \\n patterns
# We want: single \\n (1 backslash + n) in C++ output lines
# NOT: double \\n (2 backslashes + n)
remaining = 0
for line in new_data.split(b'\r\n'):
    if b'emit_raw' in line:
        # After fix, should have single \x5c\x6e not double \x5c\x5c\x6e
        idx = 0
        while True:
            idx = line.find(b'\x5c\x6e', idx)  # backslash-n
            if idx < 0:
                break
            # Check if preceded by another backslash
            if idx > 0 and line[idx-1:idx] == b'\x5c':
                remaining += 1
            idx += 2

print(f'Remaining double-backslash+n in emit_raw: {remaining}')

with open(p, 'wb') as f:
    f.write(new_data)
print('Written')