diff --git a/gdb.py b/gdb.py
index b6d90b429254a63064d17697234259edd27b3b21..b97279c8ef6614574d24a32b9b1f34c9dc13d007 100644
--- a/gdb.py
+++ b/gdb.py
@@ -16,7 +16,7 @@ autobuild = True
debug_file = "resource"
-#klee_out_folder = 'target/x86_64-unknown-linux-gnu/debug/examples/'
+# klee_out_folder = 'target/x86_64-unknown-linux-gnu/debug/examples/'
klee_out_folder = 'target/x86_64-unknown-linux-gnu/release/examples/'
stm_out_folder = 'target/thumbv7em-none-eabihf/release/examples/'
@@ -147,7 +147,7 @@ def stop_event(evt):
try:
ceiling = int(gdb.parse_and_eval(
"ceiling").cast(gdb.lookup_type('u8')))
- except:
+ except gdb.error:
print("No ceiling found, exciting!")
sys.exit(1)
@@ -167,13 +167,15 @@ def stop_event(evt):
elif imm == 3:
if debug:
- print("Debug: found finish bkpt_3 at cycle {}".format(gdb_cyccnt_read()))
+ print("Debug: found finish bkpt_3 at cycle {}"
+ .format(gdb_cyccnt_read()))
gdb.post_event(Executor("si"))
elif imm == 4:
if debug:
- print("Debug: found finish bkpt_4 at cycle {}".format(gdb_cyccnt_read()))
+ print("Debug: found finish bkpt_4 at cycle {}"
+ .format(gdb_cyccnt_read()))
gdb.post_event(posted_event_init)
@@ -204,7 +206,8 @@ def posted_event_init():
else:
if debug:
- print("Debug: Append Finish action at cycle {}".format(gdb_cyccnt_read()))
+ print("Debug: Append Finish action at cycle {}"
+ .format(gdb_cyccnt_read()))
outputdata.append(
[file_name, task_name, gdb_cyccnt_read(), priority, "Finish"])
@@ -411,7 +414,8 @@ def tasklist_get():
for line in fin:
# print(line)
if not line == "// autogenerated file\n":
- return [x.strip().strip("[]\"").split(' ') for x in line.split(',')]
+ return [x.strip().strip("[]\"").split(' ')
+ for x in line.split(',')]
""" Run xargo for building """
@@ -420,8 +424,9 @@ def tasklist_get():
def xargo_run(mode):
if "klee" in mode:
- xargo_cmd = ("xargo build --release --example " + example_name + " --features " +
- "klee_mode --target x86_64-unknown-linux-gnu ")
+ xargo_cmd = ("xargo build --release --example " + example_name
+ + " --features "
+ + "klee_mode --target x86_64-unknown-linux-gnu ")
elif "stm" in mode:
xargo_cmd = ("xargo build --release --example " + example_name +
" --features " +
@@ -447,7 +452,8 @@ def klee_run():
bc_file = (glob.glob(PWD + "/" +
klee_out_folder +
- '*.bc', recursive=False))[-1].split('/')[-1].strip('\'')
+ '*.bc', recursive=False))[-1].split('/')[-1].strip(
+ '\'')
if debug:
print(PWD + "/" + klee_out_folder)
print(bc_file)
@@ -493,8 +499,9 @@ def gdb_cyccnt_write(num):
def gdb_bkpt_read():
# Read imm field of the current bkpt
try:
- return int(gdb.execute("x/i $pc", False, True).split("bkpt")[1].strip("\t").strip("\n"), 0)
- except:
+ return int(gdb.execute("x/i $pc", False, True).
+ split("bkpt")[1].strip("\t").strip("\n"), 0)
+ except gdb.error:
if debug:
print("Debug: It is not a bkpt so return 4")
return 4
@@ -581,9 +588,9 @@ for t in priorities:
""" Subscribe stop_event_ignore to Breakpoint notifications """
gdb.events.stop.connect(stop_event)
-"""
- continue until bkpt 3,
- this will pick the next task (through a posted_event_init event)
+"""
+ continue until bkpt 3,
+ this will pick the next task (through a posted_event_init event)
"""
gdb.execute("continue")
@@ -631,7 +638,7 @@ gdb.execute("continue")
# -'2' the threshold (ceiling 2) of X
# -'Enter' the 'Enter' event
#
-# after 19 cycles we clam Y, raising treshold to 3
+# after 19 cycles we clam Y, raising threshold to 3
# after 28 cycles we exit the Y claim, threshold 3 *before unlock Y*
# after 29 cycles we exit the X claim, threshold 2 *before unlock X*
# and finally we finish at 36 clock cycles
@@ -651,7 +658,7 @@ gdb.execute("continue")
# interarrival = [100, 30, 40]
# should match the arrival time of EXTI1, EXTI2, and EXTI3 respectively
# you may need to change the order depending or your klee/tasks.txt file
-# (in the future interrarrival and deadlines will be in the RTFM model,
+# (in the future interarrival and deadlines will be in the RTFM model,
# but for now we introduce them by hand)
#
# Implement function that takes output data and computes the CPU demand
@@ -670,10 +677,10 @@ gdb.execute("continue")
# Looking up the WCETs from the `output_data`.
# (It may be a good idea to make first pass and extract wcet per task)
##
-# The total utililization bound allows us to discard task sets that are obviously illegal.debug
-# (not the case here though)
+# The total utilisation bound allows us to discard task sets that are
+# obviously illegal.debug (not the case here though)
#
-# Assignment3.
+# Assignment 3.
#
# Under SRP response time can be computed by equation 7.22 from
# https://doc.lagout.org/science/0_Computer%20Science/2_Algorithms/Hard%20Real-Time%20Computing%20Systems_%20Predictable%20Scheduling%20Algorithms%20and%20Applications%20%283rd%20ed.%29%20%5BButtazzo%202011-09-15%5D.pdf
@@ -681,27 +688,31 @@ gdb.execute("continue")
# In general the response time is computed as.
# Ri = Ci + Bi + Ii
# Ci the WCET of task i
-# Bi the blockng time task i is exposed to
+# Bi the blocking time task i is exposed to
# Ii the interference (preemptions) task is exposed to
#
# where
# Pi the priority of task i
-# Ai the interrarval of task i
+# Ai the interarrival of task i
#
-# We assign deadline = interrarival and priorties inverse to deadline
+# We assign deadline = interarrival and priorities inverse to deadline
# (rate monotonic assignment, with fixed/static priorities)
#
-# Lets start by looking at EXTI2 whith the highest priority, so no interference (preemption)
+# Lets start by looking at EXTI2 with the highest priority,
+# so no interference (preemption)
# R_EXTI2 = 11 + B_EXTI2 + 0
#
-# In general Bi is the max time of any lower priority task (EXIT1, EXTI3 in our case)
-# holds a resource with a ceiling > Pi (ceileng >= 3 in this case)
+# In general Bi is the max time of any lower priority task
+# (EXTI1, EXTI3 in our case)
+# holds a resource with a ceiling > Pi (ceiling >= 3 in this case)
# B_EXTI2 = 10 (EXTI1 holding Y for 10 cycles)
#
-# Notice 1, single blocking, we can only be blocked ONCE, so bound priority inversion
+# Notice 1, single blocking, we can only be blocked ONCE,
+# so bound priority inversion
#
-# Notice 2, `output_data` does not hold info on WHAT exect resource is held
-# (merely each 'claim time at a specific ceiling'). However this is sufficient for the analysis.
+# Notice 2, `output_data` does not hold info on WHAT exact resource is held
+# (merely each 'claim time at a specific ceiling').
+# However this is sufficient for the analysis.
#
# so
# R_EXTI2 = 11 + 10 = 21, well below our 30 cycle margin
@@ -711,22 +722,24 @@ gdb.execute("continue")
# where I_EXTI3 is the interference (preemptions)
#
# Here we can undertake a simple approach to start out.
-# Assume a deadline equal to our interarraval (50)
-# I_EXTI3 is the sum of ALL preemptions until its deadlne.
+# Assume a deadline equal to our interarrival (50)
+# I_EXTI3 is the sum of ALL preemptions until its deadline.
# in this case EXTI2 can preempt us 2 times (40/30 *rounded upwards*)
# I_EXTI3 = 2 * 11
#
-# The worst case blocking time is 15 (caused by the lower prio task EXTI1 holding X)
-# R_EXTI3 = 8 + 2 * 11 + 15 = 45, already here we see that EXTI2 may miss its deadline
+# The worst case blocking time is 15
+# (caused by the lower prio task EXTI1 holding X)
+# R_EXTI3 = 8 + 2 * 11 + 15 = 45, already here we see that
+# EXTI2 may miss its deadline
#
# EXTI1 (our lowest prio task)
# R_EXTI1 = C_EXTI1 + B_EXTI1 + I_EXTI1
#
# Here we cannot be blocked (as we have the lowest prio)
# I_EXTI1 is the sum of preemptions from EXTI2 and EXTI3
-# our deadline = interarravial is 100
-# we are exposed to 100/30 = 4 (rounded upwards) preepmtions by EXTI2
-# and 100/40 = 3 (rounded upwards) preempions by EXTI3
+# our deadline = interarrival is 100
+# we are exposed to 100/30 = 4 (rounded upwards) preemptions by EXTI2
+# and 100/40 = 3 (rounded upwards) preemptions by EXTI3
#
# I_EXTI1 = 37 + 4 * 11 + 3 * 8 = 105
#
@@ -735,15 +748,15 @@ gdb.execute("continue")
#
# Assignment 4.
# Looking closer at 7.22 we see that its a recurrent equation.
-# Ri(0) indicating the inital value
+# Ri(0) indicating the initial value
# Ri(0) = Ci + Bi
# while
# Ri(s) = Ci + Bi + sum ..(Ri(s-1))..
# so Ri(1) is computed from Ri(0) and so forth,
-# this requires a recursive or looping implmentation.
+# this requires a recursive or looping implementation.
#
# One can see that as initially setting a "busy period" to Ci+Bi
-# and compute a new (longer) "busy period" by taking into accout preemptions.
+# and compute a new (longer) "busy period" by taking into account preemptions.
#
# Termination:
# Either Ri(s) = Ri(s-1), we have a fixpoint and have the exact response time
@@ -754,8 +767,9 @@ gdb.execute("continue")
# Notice, we have not dealt with the case where tasks have equal priorities
# in theory this is not a problem (no special case needed)
#
-# However, for exactly analysing the taskset as it would run on the real hardware
-# requires some (minor) modifications. *Not part of this assignment*
+# However, for exactly analysing the taskset as it would run on the
+# real hardware requires some (minor) modifications.
+# *Not part of this assignment*
#
# Examination for full score.
# Make a git repo of your solution. (With reasonable comments)
@@ -765,14 +779,14 @@ gdb.execute("continue")
# Print response times according to Assignment 3
# Print response times according to Assignment 4
#
-# It should work with different assignments of the interrarival vector.
+# It should work with different assignments of the interarrival vector.
# test it also for
# [100, 40, 50]
# [80, 30, 40]
-# (Verify that your resoults are correct by hand computations)
+# (Verify that your results are correct by hand computations)
#
# Grading
# For this part 1/3 of the exam 35 points
# Assignment 2, 10 points
-# Assingment 3, 10 points
-# Assingment 4, 15 points
+# Assignment 3, 10 points
+# Assignment 4, 15 points