How to get the result of EXPECT_EXIT() in gtest?
I want to do something according to the results of EXPECT_EXIT. For example:
for(int i = 0; i < 1000; i++) {
for(int j = 0; j < 10; j++) {
auto &result = EXPECT_EXIT(Function(i, inputs[j]));
if (result) {
do something;
break; //jump out of the loop
}
}
}
From your comment:
But once it failed, I need record something and skip rest of the input
check for this function
Short answer:
To stop execution at first mismatch you need to use ASSERT version instead of EXPECT, in your case ASSERT_EXIT.
To output some extra data for failed check you can just use stream operator of check:
ASSERT_EXIT(Function(i, j), testing::ExitedWithCode(0), "") << "Some custom data: " << i << "; " << j;
If you want to do something more on fail - you can use ::testing::Test::HasFailure() function.
Long suggestion:
One of the properties of good unittest is simplicity. That implies absence of any loops or branches. Setup logic (if any) should be moved to fixtures. Also unittests must be executed in isolation.
By switching your test to parametric one you would cover that. Check this:
void Function(int i, int j)
{
if (j != 3)
exit(0);
}
class FunTest : public testing::TestWithParam<std::tuple<int, int>> { };
TEST_P(FunTest, Should_Exit)
{
const auto i = std::get<0>(GetParam());
const auto j = std::get<1>(GetParam());
EXPECT_EXIT(Function(i, j), testing::ExitedWithCode(0), "");
}
INSTANTIATE_TEST_SUITE_P(,
FunTest,
testing::Combine(
testing::Range(1, 20),
testing::Range(1, 10)),
[](const testing::TestParamInfo<std::tuple<int, int>>& info) -> std::string {
const auto i = std::get<0>(info.param);
const auto j = std::get<1>(info.param);
return std::to_string(i) + '_' + std::to_string(j);
});
This approach would help with your problem: you can stop execution by using --gtest_break_on_failure and you can control naming of test cases by implementing custom printer for specific TestParamInfo (lambda in example).
It would also guarantee isolation and provide certain level of simplicity (all complex stuff provided by framework, imho test code is minimalistic and simple).
Also this would allow execution of specific test using gtest_filter
(e.g. --gtest_filter=FunTest.Should_Exit/4_3).
Related
I am working on a compiler and one aspect currently is how to wait for interpolated variable names to be resolved. So I am wondering how to take a nested interpolated variable string and build some sort of simple data model/schema for unwrapping the evaluated string so to speak. Let me demonstrate.
Say we have a string like this:
foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}
That has 1, 2, and 3 levels of nested interpolations in it. So essentially it should resolve something like this:
wait for x, y, one, two, and c to resolve.
when both x and y resolve, then resolve a{x}-{y} immediately.
when both one and two resolve, resolve baz{one}-{two}.
when a{x}-{y}, baz{one}-{two}, and c all resolve, then finally resolve the whole expression.
I am shaky on my understanding of the logic flow for handling something like this, wondering if you could help solidify/clarify the general algorithm (high level pseudocode or something like that). Mainly just looking for how I would structure the data model and algorithm so as to progressively evaluate when the pieces are ready.
I'm starting out trying and it's not clear what to do next:
{
dependencies: [
{
path: [x]
},
{
path: [y]
}
],
parent: {
dependency: a{x}-{y} // interpolated term
parent: {
dependencies: [
{
}
]
}
}
}
Some sort of tree is probably necessary, but I am having trouble figuring out what it might look like, wondering if you could shed some light on that with some pseudocode (or JavaScript even).
watch the leaf nodes at first
then, when the children of a node are completed, propagate upward to resolving the next parent node. This would mean once x and y are done, it could resolve a{x}-{y}, but then wait until the other nodes are ready before doing the final top-level evaluation.
You can just simulate it by sending "events" to the system theoretically, like:
ready('y')
ready('c')
ready('x')
ready('a{x}-{y}')
function ready(variable) {
if ()
}
...actually that may not work, not sure how to handle the interpolated nodes in a hacky way like that. But even a high level description of how to solve this would be helpful.
export type SiteDependencyObserverParentType = {
observer: SiteDependencyObserverType
remaining: number
}
export type SiteDependencyObserverType = {
children: Array<SiteDependencyObserverType>
node: LinkNodeType
parent?: SiteDependencyObserverParentType
path: Array<string>
}
(What I'm currently thinking, some TypeScript)
Here is an approach in JavaScript:
Parse the input string to create a Node instance for each {} term, and create parent-child dependencies between the nodes.
Collect the leaf Nodes of this tree as the tree is being constructed: group these leaf nodes by their identifier. Note that the same identifier could occur multiple times in the input string, leading to multiple Nodes. If a variable x is resolved, then all Nodes with that name (the group) will be resolved.
Each node has a resolve method to set its final value
Each node has a notify method that any of its child nodes can call in order to notify it that the child has been resolved with a value. This may (or may not yet) lead to a cascading call of resolve.
In a demo, a timer is set up that at every tick will resolve a randomly picked variable to some number
I think that in your example, foo, and a might be functions that need to be called, but I didn't elaborate on that, and just considered them as literal text that does not need further treatment. It should not be difficult to extend the algorithm with such function-calling features.
class Node {
constructor(parent) {
this.source = ""; // The slice of the input string that maps to this node
this.texts = []; // Literal text that's not part of interpolation
this.children = []; // Node instances corresponding to interpolation
this.parent = parent; // Link to parent that should get notified when this node resolves
this.value = undefined; // Not yet resolved
}
isResolved() {
return this.value !== undefined;
}
resolve(value) {
if (this.isResolved()) return; // A node is not allowed to resolve twice: ignore
console.log(`Resolving "${this.source}" to "${value}"`);
this.value = value;
if (this.parent) this.parent.notify();
}
notify() {
// Check if all dependencies have been resolved
let value = "";
for (let i = 0; i < this.children.length; i++) {
const child = this.children[i];
if (!child.isResolved()) { // Not ready yet
console.log(`"${this.source}" is getting notified, but not all dependecies are ready yet`);
return;
}
value += this.texts[i] + child.value;
}
console.log(`"${this.source}" is getting notified, and all dependecies are ready:`);
this.resolve(value + this.texts.at(-1));
}
}
function makeTree(s) {
const leaves = {}; // nodes keyed by atomic names (like "x" "y" in the example)
const tokens = s.split(/([{}])/);
let i = 0; // Index in s
function dfs(parent=null) {
const node = new Node(parent);
const start = i;
while (tokens.length) {
const token = tokens.shift();
i += token.length;
if (token == "}") break;
if (token == "{") {
node.children.push(dfs(node));
} else {
node.texts.push(token);
}
}
node.source = s.slice(start, i - (tokens.length ? 1 : 0));
if (node.children.length == 0) { // It's a leaf
const label = node.texts[0];
leaves[label] ??= []; // Define as empty array if not yet defined
leaves[label].push(node);
}
return node;
}
dfs();
return leaves;
}
// ------------------- DEMO --------------------
let s = "foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}";
const leaves = makeTree(s);
// Create a random order in which to resolve the atomic variables:
function shuffle(array) {
for (var i = array.length - 1; i > 0; i--) {
var j = Math.floor(Math.random() * (i + 1));
[array[j], array[i]] = [array[i], array[j]];
}
return array;
}
const names = shuffle(Object.keys(leaves));
// Use a timer to resolve the variables one by one in the given random order
let index = 0;
function resolveRandomVariable() {
if (index >= names.length) return; // all done
console.log("\n---------------- timer tick --------------");
const name = names[index++];
console.log(`Variable ${name} gets a value: "${index}". Calling resolve() on the connected node instance(s):`);
for (const node of leaves[name]) node.resolve(index);
setTimeout(resolveRandomVariable, 1000);
}
setTimeout(resolveRandomVariable, 1000);
your idea of building a dependency tree it's really likeable.
Anyway I tryed to find a solution as simplest possible.
Even if it already works, there are many optimizations possible, take this just as proof of concept.
The background idea it's produce a List of Strings which you can read in order where each element it's what you need to solve progressively. Each element might be mandatory to solve something that come next in the List, hence for the overall expression. Once you solved all the chunks you have all pieces to solve your original expression.
It's written in Java, I hope it's understandable.
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Objects;
public class StackOverflow {
public static void main(String[] args) {
String exp = "foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}";
List<String> chunks = expToChunks(exp);
//it just reverse the order of the list
Collections.reverse(chunks);
System.out.println(chunks);
//output -> [c, two, one, baz{one}-{two}, y, x, a{x}-{y}]
}
public static List<String> expToChunks(String exp) {
List<String> chunks = new ArrayList<>();
//this first piece just find the first inner open parenthesys and its relative close parenthesys
int begin = exp.indexOf("{") + 1;
int numberOfParenthesys = 1;
int end = -1;
for(int i = begin; i < exp.length(); i++) {
char c = exp.charAt(i);
if (c == '{') numberOfParenthesys ++;
if (c == '}') numberOfParenthesys --;
if (numberOfParenthesys == 0) {
end = i;
break;
}
}
//this if put an end to recursive calls
if(begin > 0 && begin < exp.length() && end > 0) {
//add the chunk to the final list
String substring = exp.substring(begin, end);
chunks.add(substring);
//remove from the starting expression the already considered chunk
String newExp = exp.replace("{" + substring + "}", "");
//recursive call for inner element on the chunk found
chunks.addAll(Objects.requireNonNull(expToChunks(substring)));
//calculate other chunks on the remained expression
chunks.addAll(Objects.requireNonNull(expToChunks(newExp)));
}
return chunks;
}
}
Some details on the code:
The following piece find the begin and the end index of the first outer chunk of expression. The background idea is: in a valid expression the number of open parenthesys must be equal to the number of closing parenthesys. The count of open(+1) and close(-1) parenthesys can't ever be negative.
So using that simple loop once I find the count of parenthesys to be 0, I also found the first chunk of the expression.
int begin = exp.indexOf("{") + 1;
int numberOfParenthesys = 1;
int end = -1;
for(int i = begin; i < exp.length(); i++) {
char c = exp.charAt(i);
if (c == '{') numberOfParenthesys ++;
if (c == '}') numberOfParenthesys --;
if (numberOfParenthesys == 0) {
end = i;
break;
}
}
The if condition provide validation on the begin and end indexes and stop the recursive call when no more chunks can be found on the remained expression.
if(begin > 0 && begin < exp.length() && end > 0) {
...
}
I am using the type std::valarray<std::valarray<double>> and wish to sum each of the contained valarrays element wise, to leave a std::valarray<double>.
The C++ documentation states that the operator .sum() can be applied to std::valarray<T> so long as the operator += is defined for type T. My code below (method1) tries to apply this to std::valarray<std::valarray<double>>, but the result appears to be nonsense.
However if I perform this manually, using the += operator (method2), I get the result I want. But the fact that method2 works seems to imply that the operator += is defined for the type std::valarray<double>, and hence that method1, using .sum(). should work. I really can't understand what is happening here...
My code:
#include <iostream>
#include <valarray>
// Attempt to use .sum() operator
std::valarray<double> method1(const std::valarray<std::valarray<double>>& data) {
return data.sum();
}
// Manual summation using += operator
std::valarray<double> method2(const std::valarray<std::valarray<double>>& data) {
std::valarray<double> sum(data[0].size());
for (size_t i{0}; i < data.size(); i++) {
sum += data[i];
}
return sum;
}
// Display size and elements
void showData(const std::valarray<double> data) {
std::cout << "Size = " << data.size() << "\n";
std::cout << "Data = ";
for (size_t i{0}; i < data.size(); i++) {
std::cout << data[i] << " ";
}
std::cout << "\n\n";
}
int main() {
std::valarray<std::valarray<double>> data{{1,2},{3,4}};
showData(method1(data));
showData(method2(data));
}
My output:
Size = 0
Data =
Size = 2
Data = 4 6
The sum method of std::valarray requires operator+= to be defined for its value type (in your case, std::valarray), but std::valarray also requires it to be default-constructible (from the "Numeric" concept requirement).
This allows the sum method to work without operator+, by first default-constructing an element, and then adding each contained element with operator+=.
Although it isn't defined anywhere, as far as I know, it probably works something like this.
T sum() const {
T result;
for (auto& it : elements) {
result += it;
}
return result;
}
The problem with a valarray of valarrays (std::valarray<std::valarray>) is that a default-constructed valarray is empty. And when operator+= is applied with an empty valarray and a non-empty one, it results in undefined behavior ("The behavior is undefined if size() != v.size()"). What you are likely to get is an empty valarray as a result (but you could potentially get anything).
What you could use instead is std::accumulate. It requires an initial value as third parameter, which takes care of the problem.
std::accumulate(std::begin(data), std::end(data), std::valarray<double>(data[0].size()))
Live on Coliru.
PS: don't ask me why std::valarray has no method begin and end.
I populate a very large array using a stxxl::VECTOR_GENERATOR<MyData>::result::bufwriter_type (something like 100M entries) which I need to sort in parallel.
I use the stxxl::sort(vector->begin(), vector->end(), cmp(), memoryAmount) method, which in theory should do what I need: sort the elements very efficiently.
However, during the execution of this method I noticed that only one processor is fully utilised, and all the other cores are quite idle (I suspect there is little activity to fetch the input, but in practice they don't do anything).
This is my question: is it possible to exploit more cores during the sorting phase, or is the parallelism used only to fetch the input asynchronously? If so, are there documents that explain how to enable it? (I looked extensively the documentation on the website, but I couldn't find anything).
Thanks very much!
EDIT
Thanks for the suggestion. I provide below some more information.
First of all I use MacOs for my experiments. What I do is that I launch the following program and I study its behaviour.
typedef struct Triple {
long t1, t2, t3;
Triple(long s, long p, long o) {
this->t1 = s;
this->t2 = p;
this->t3 = o;
}
Triple() {
t1 = t2 = t3 = 0;
}
} Triple;
const Triple minv(std::numeric_limits<long>::min(),
std::numeric_limits<long>::min(), std::numeric_limits<long>::min());
const Triple maxv(std::numeric_limits<long>::max(),
std::numeric_limits<long>::max(), std::numeric_limits<long>::max());
struct cmp: std::less<Triple> {
bool operator ()(const Triple& a, const Triple& b) const {
if (a.t1 < b.t1) {
return true;
} else if (a.t1 == b.t1) {
if (a.t2 < b.t2) {
return true;
} else if (a.t2 == b.t2) {
return a.t3 < b.t3;
}
}
return false;
}
Triple min_value() const {
return minv;
}
Triple max_value() const {
return maxv;
}
};
typedef stxxl::VECTOR_GENERATOR<Triple>::result vector_type;
int main(int argc, const char** argv) {
vector_type vector;
vector_type::bufwriter_type writer(vector);
for (int i = 0; i < 1000000000; ++i) {
if (i % 10000000 == 0)
std::cout << "Inserting element " << i << std::endl;
Triple t;
t.t1 = rand();
t.t2 = rand();
t.t3 = rand();
writer << t;
}
writer.finish();
//Sort the vector
stxxl::sort(vector.begin(), vector.end(), cmp(), 1024*1024*1024);
std::cout << vector.size() << std::endl;
}
Indeed there seems to be only one or maximum two threads working during the execution of this program. Notice that the machine has only a single disk.
Can you please confirm me whether the parallelism work on macos? If not, then I will try to use linux to see what happens. Or is perhaps because there is only one disk?
In principle what you are doing should work out-of-the-box. With everything working you should see all cores doing processing.
Since it doesnt work, we'll have to find the error, and debugging why we see no parallel speedups is still tricky business these days.
The main idea is to go from small to large examples:
what platform is this? There is no parallelism on MSVC, only on Linux/gcc.
By default STXXL builds on Linux/gcc with USE_GNU_PARALLEL. you can turn it off to see if it has an effect.
Try reproducing the example values shown in http://stxxl.sourceforge.net/tags/master/stxxl_tool.html - with and without USE_GNU_PARALLEL
See if just in memory parallel sorting scales on your processor/system.
I can't seem to find a great way to express the following in Xtend without resorting to a while loop:
for(int i = 0; i < 3; i++){
println("row ");
}
println("your boat");
So, I guess my question has two parts:
Is there a better way to do the above? I didn't see anything promising in their documenation
A large portion of the features the language has are just Xtend library extensions (and they're great!). Is there range() functionality à la Python that I don't know about?
I ended up rolling my own and got something like the following:
class LanguageUtil {
def static Iterable<Integer> range(int stop) {
range(0, stop)
}
def static Iterable<Integer> range(int start, int stop) {
new RangeIterable(start, stop, 1)
}
def static Iterable<Integer> range(int start, int stop, int step) {
new RangeIterable(start, stop, step)
}
}
// implements Iterator and Iterable which is bad form.
class RangeIterable implements Iterator<Integer>, Iterable<Integer> {
val int start
val int stop
val int step
var int current
new(int start, int stop, int step) {
this.start = start;
this.stop = stop;
this.step = step
this.current = start
}
override hasNext() {
current < stop
}
override next() {
val ret = current
current = current + step
ret
}
override remove() {
throw new UnsupportedOperationException("Auto-generated function stub")
}
/**
* This is bad form. We could return a
* new RangeIterable here, but that seems worse.
*/
override iterator() {
this
}
}
The exact replacement for
for(int i = 0; i < 3; i++){
println("row ");
}
is
for (i : 0 ..< 3) {
println("row ")
}
Notice the exclusive range operator: ..<
Also you can doing it more idiomatically with
(1..3).forEach[println("row")]
Very new to Xtend but man it makes programming on the jvm awesome.
To me a range-based forEach implies the range is somehow meaningful. For looping a specific number of times with no iteration variable, I find Ruby's times loop expresses the intent more clearly:
3.times [|println("row")]
Sadly it's not a part of IntegerExtensions, but the implementation is trivial:
def static times(int iterations, Runnable runnable) {
if (iterations < 0) throw new IllegalArgumentException(
'''Can't iterate negative («iterations») times.''')
for (i: 0 ..< iterations) runnable.run()
}
Heh, I found the answer a little while later:
for(i: 1..3) {
println("row ")
}
Since Xtend 2.6, we also support the "traditional" for-loop, just like in Java.
There is actually a version of forEach() that accepts a lambda with two parameters.
It is useful if you need to access the iteration index within the loop.
(10..12).forEach[ x, i | println('''element=«x» index=«i»''')]
prints:
element=10 index=0
element=11 index=1
element=12 index=2
const char IsPressed = 1; // 1
const char WasHeldDown = 2; // 10
const char IsFirstPress = 4; // 100
char* keystates[256];
Class::CalculateKeyStates()
{
for(int i = 0; i < 256; ++i)
{
if(this->IsDown(i))
{
keystates[i] |= IsPressed; // turn on
if(keystates[i] & WasHeldDown)
{
//keystates[i] |= IsFirstPress;
keystates[i] &= ~IsFirstPress; // turn off
}
else
{
keystates[i] |= WasHeldDown + IsFirstPress; // Turn on
}
}
else
{
keystates[i] = 0; // Turn ALL off
}
}
}
This function would be a member function of a class, Class. The other member function, IsDown, will return a true if the key in question is down and false if not.
Can you see any way of further improving this function?
Thanks
EDIT:
I will expand a bit as to what is done why. This is a modification of an bit of code that works through an array keyStates (which was a struct of three bools) setting IsPressed to false for all of the keys. then again setting Ispressed to the value of this->IsDown and then a third time looping through checking if the key had been held, if it has then its no longer the first press so set that to false. if it was not held down, then set first press to true and was held to true as well, so next time it is flagged as having been held.
EDIT2:
Added some comments to code and corrected one line
Personally, I would define the key-states as disjoint states and write a simple state-machine, thus:
enum keystate
{
inactive,
firstPress,
active
};
keystate keystates[256];
Class::CalculateKeyStates()
{
for (int i = 0; i < 256; ++i)
{
keystate &k = keystates[i];
switch (k)
{
inactive:
k = (isDown(i)) ? firstPress : inactive;
break;
firstPress:
k = (isDown(i)) ? active : inactive;
break;
active:
k = (isDown(i)) ? active : inactive;
break;
}
}
}
This is easier to extend, and easier to read if it gets any more complex.
You are always setting IsFirstPress if the key is down, which might not be what you want.
I'm not sure what you want to achieve with IsFirstPress, as the keystate cannot remember any previous presses anyways. If you want to mark with this bit, that it's the first time you recognized the key being down, then your logic is wrong in the corresponding if statement.
keystates[i] & WasHeldDown evaluates to true if you already set the bit WasHeldDown earlier for this keystate.
In that case, what you may want to do is actually remove the IsFirstPress bit by xor-ing it: keystates[i] ^= IsFirstPress