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feat: Update esptool to v5

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This commit is contained in:
Peter Dragun
2025-08-11 16:11:49 +02:00
parent 053fb47e78
commit e3198fff3c
129 changed files with 1237 additions and 1318 deletions
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25
examples/security/flash_encryption/pytest_flash_encryption.py
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@@ -1,7 +1,6 @@
# SPDX-FileCopyrightText: 2018-2025 Espressif Systems (Shanghai) CO LTD
# SPDX-License-Identifier: Apache-2.0
import binascii
from collections import namedtuple
from io import BytesIO
import espsecure
@@ -13,9 +12,9 @@ from pytest_embedded_idf.utils import idf_parametrize
# 1. Generate zero flash encryption key:
# dd if=/dev/zero of=key.bin bs=1 count=32
# 2.Burn Efuses:
# espefuse.py --do-not-confirm -p $ESPPORT burn_efuse FLASH_CRYPT_CONFIG 0xf
# espefuse.py --do-not-confirm -p $ESPPORT burn_efuse FLASH_CRYPT_CNT 0x1
# espefuse.py --do-not-confirm -p $ESPPORT burn_key flash_encryption key.bin
# espefuse --do-not-confirm -p $ESPPORT burn-efuse FLASH_CRYPT_CONFIG 0xf
# espefuse --do-not-confirm -p $ESPPORT burn-efuse FLASH_CRYPT_CNT 0x1
# espefuse --do-not-confirm -p $ESPPORT burn-key flash_encryption key.bin
def _test_flash_encryption(dut: Dut) -> None:
@@ -36,17 +35,17 @@ def _test_flash_encryption(dut: Dut) -> None:
key_bytes = b'\xff' + b'\x00' * 31
aes_xts = True
EncryptFlashDataArgs = namedtuple(
'EncryptFlashDataArgs', ['output', 'plaintext_file', 'address', 'keyfile', 'flash_crypt_conf', 'aes_xts']
output = BytesIO()
espsecure.encrypt_flash_data(
output=output,
plaintext_file=BytesIO(plain_data),
address=flash_addr,
keyfile=BytesIO(key_bytes),
flash_crypt_conf=0xF,
aes_xts=aes_xts,
)
args = EncryptFlashDataArgs(BytesIO(), BytesIO(plain_data), flash_addr, BytesIO(key_bytes), 0xF, aes_xts)
try:
# espsecure 5.0 arguments are passed one by one; the following will convert tuple to dict and unwrap it
espsecure.encrypt_flash_data(**args._asdict())
except TypeError:
espsecure.encrypt_flash_data(args)
expected_ciphertext = args.output.getvalue()
expected_ciphertext = output.getvalue()
hex_ciphertext = binascii.hexlify(expected_ciphertext).decode('ascii')
expected_str = (
' '.join(hex_ciphertext[i : i + 2] for i in range(0, 16, 2))