I'm trying to build an LLVM pass that splits the BasicBlock and make a decision using "SplitBlockAndInsertIfThenElse" every time a binary operation is encountered, however this only allows me to split once (split at the 1st binop). could you please help me make it iterate through all the binop instructions?
Knowing that changing the position of "break;" gives me errors when running the pass. same thing happens when I put the "SplitBlockAndInsertIfThenElse" in a nested loop.
Here is my code:
for (inst_iterator I = inst_begin(F), E = inst_end(F); I != E; ++I)
{
if (auto *op = dyn_cast<BinaryOperator>(&*I))
{
IRBuilder<> Builder(op);
Value *lhs = op->getOperand(0);
Value *rhs = op->getOperand(1);
Value *xpv = Builder.CreateAlloca(llvm::Type::getInt32Ty(llvm::getGlobalContext()), nullptr, "x");
Value *xpv2 = Builder.CreateAlloca(llvm::Type::getInt32Ty(llvm::getGlobalContext()), nullptr, "x2");
Value *add1 = Builder.CreateAdd(lhs, rhs);
Value *add2 = Builder.CreateAdd(lhs, rhs);
Value *icmp1 = Builder.CreateICmpEQ(add1, add2);
TerminatorInst *ThenTerm , *ElseTerm ;
SplitBlockAndInsertIfThenElse(icmp1, op, &ThenTerm, &ElseTerm,nullptr);
Builder.SetInsertPoint(ThenTerm);
Value *xp1 = Builder.CreateStore(add1, xpv);
Builder.SetInsertPoint(ElseTerm);
break ;
}
}
Don't perform concurrent iteration and modification of the instruction list. Iterate until you find the first instruction you care about, then break out of the loop, perform the modification, and restart the loop, starting from the next instruction after the split-before one (so the next instruction after op, in your case).
Related
vector& vector::operator = (const vector& a)
//make this vector a copy of a
{
double* p = new double [ a.sz ]; // allocate new space
copy(a.elem, a.elem+a.sz, elem); // copy elements
delete[] elem; // deallocate old space
elem = p; // now we can reset elem
sz = a.sz;
return *this; // return a self-reference
}
I thought that the third argument of std::copy() should be the pointer p, but the book (Programming principles and practice using C++ - 2nd edition) says:
"When implementing the assignment, you could consider simplifying the code by freeing the memory for the old elements before creating the copy, but it is usually a very good idea not to throw away information before you know that you can replace it. Also, if you did that, strange things would happen if you assigned a vector to itself" - Page 635 and 636.
So, the pointer elem must be third argument of std::copy() to not let the pointer be invalid for a moment. I think...
But from where does p gets the information to be put in the array it points to, to be able to do: elem = p ?
I already know copy and swap strategy exist, you don't have to explain that.
I want to comprehend what is above.
No, that is a typo.
std::copy(a.elem, a.elem+a.sz, p);
is what the code should read.
i have declared a map below using stl and inserted some values in it.
#include<bits/stdc++.h>
int main()
{
map<int,int> m;
m[1]=1;
m[2]=1;
m[3]=1;
m[4]=1;
m[5]=1;
m[6]=1;
for(auto it=m.begin();it!=m.end();)
{
cout<<it->first<<" "<<it->second<<endl;
it=it++;
}
return 0;
}
When i executed the above written code it ended up in an infinite loop. Can someone tell me why it does so?
I am incrementing the value of iterator it and then it gets stored in it which should get incremented next time the loop is executed and eventually it should terminate normally.Am i wrong?
The bad line is it = it++;. It is undefined behavior! Because it is not defined, when it is increased, in your case it is increased before the assingment to itsself again, that the value of it before it is increased is assigned to it again and so it keeps at the first position. The correct line would be it = ++it; or only ++it;/it++;, because it changes itsself.
Edit
That is only undefined with the builtin types, but in here that is defined by the source-code of the map in the stl.
If you try doing something similar with an int, you'll get a warning:
int nums[] = { 1, 2, 3, 4, 5 };
for (int i = 0; i < sizeof nums / sizeof *nums; ) {
cout << nums[i] << '\n';
i = i++;
}
warning: operation on 'i' may be undefined [-Wsequence-point]
However, when you're using a class (std::map::iterator) which has operator overloading, the compiler probably isn't smart enought to detect this.
In other words, what you're doing is a sequence point violation, so the behavior is undefined behavior.
The post-increment operation would behave like this:
iterator operator ++ (int) {
auto copy = *this;
++*this;
return copy;
}
So, what happens to your increment step is that iterator it would get overwritten by the copy of its original value. If the map isn't empty, your loop would remain stuck on the first element.
I am debugging a kernel crash dump. There seems to be a problem with one process was trying to memory map a new region. The problem is that it was not able to hold the memory map semaphore.
When I looked into process's mm_struct and printed its content. I saw that the struct rw_semaphore mmap_sem were as seen below. Now, does he value of count seem suspicious? It has a negative value, as if there was a race condition where it was decremented twice by two different threads after checking for zero.
mmap_sem = {
count = -4294967295,
wait_lock = {
{
rlock = {
raw_lock = {
slock = 262148
}
}
}
},
wait_list = {
next = 0xffff8801f0113e48,
prev = 0xffff8801f0113e48
}
},
Sorry for the confusion. I thought crash pulls the correct data types and uses that properly when printing out the all the values ...
Looks like crash utility is not read the count member as an int ....
When I print it as int, I get the correct value.
crash> p (int) (((struct mm_struct *) 0xffff8801f15fa540)->mmap_sem).count
$13 = 1
I'm new to LLVM and I was wondering if you could help me building a pass to duplicate instructions in LLVM IR, the problem I'm facing is that the cloned instructions couldn't be returned using (user class), is this the correct way to do it ? are there any other ways (excluding this http://llvm.org/docs/ExtendingLLVM.html)
My pass:
BasicBlock *B = I->getParent();
if (auto *op = dyn_cast<BinaryOperator>(&*I))
{
auto temp = op->clone();
B->getInstList().insert(op, temp);
temp->setName(op->getName());
if (temp->getOpcode() == Instruction::Add)
{
IRBuilder<> builder(temp); //building the cloned instruction
Value *lhs = temp->getOperand(0);
Value *rhs = temp->getOperand(1);
Value *add1 = builder.CreateAdd(lhs, rhs);
for (auto &v : temp->uses()) {
User *user = v.getUser(); // A User is anything with operands.
user->setOperand(v.getOperandNo(), add1);
}
}
}
BasicBlock *B = I->getParent();
if (auto *op = dyn_cast<BinaryOperator>(&*I))
{
auto temp = op->clone();
B->getInstList().insert(op, temp);
temp->setName(op->getName());
At this point you have successfully cloned your instruction and inserted it in the BasicBlock where the original instruction lives.
if (temp->getOpcode() == Instruction::Add)
{
IRBuilder<> builder(temp); //building the cloned instruction
Value *lhs = temp->getOperand(0);
Value *rhs = temp->getOperand(1);
Value *add1 = builder.CreateAdd(lhs, rhs);
Now you are building an IRBuilder. An IRBuilder as a helper class that allows you easy insertion of instructions in your code. But it does not build your temp instruction. The temp instruction is already there from calling clone an inserting it in the BasicBlock.
You create another copy of your original instruction (add1).
for (auto &v : temp->uses()) {
User *user = v.getUser(); // A User is anything with operands.
user->setOperand(v.getOperandNo(), add1);
}
Here you are updating all users of temp. But at this point temp has no users. temp is just a clone of your original instruction. You have created two copies of your original instruction that are not used and will be removed by dead code elimination.
What you want to do is to replace all uses of op by one of your copies.
An easier way to achieve this is to use RAUW ReplaceAllUsesWith.
BasicBlock *B = I->getParent();
if (auto *op = dyn_cast<BinaryOperator>(&*I))
{
auto temp = op->clone();
B->getInstList().insert(op, temp);
temp->setName(op->getName());
op->replaceAllUsesWith(temp);
}
With RAUW now op is dead (i.e., has no users) and your cloned instruction is alive.
Why does the following code throw an exception when getting to the second scanf_s after entering an number to put into the struct.
This by no means represents a complete linked list implementation.
Not sure how to get onto the next scanf_s when having entered the value? Any ideas?
EDIT: Updated code with suggested solution, but still get an AccessViolationException after first scanf_s
Code:
struct node
{
char name[20];
int age;
float height;
node *nxt;
};
int FillInLinkedList(node* temp)
{
int result;
temp = new node;
printf("Please enter name of the person");
result = scanf_s("%s", temp->name);
printf("Please enter persons age");
result = scanf_s("%d", &temp->age); // Exception here...
printf("Please enter persons height");
result = scanf_s("%f", &temp->height);
temp->nxt = NULL;
if (result >0)
return 1;
else return 0;
}
// calling code
int main(array<System::String ^> ^args)
{
node temp;
FillInLinkedList(&temp);
...
You are using scanf_s with incorrect parameters. Take a look at the examples in the MSDN documentation for the function. It requires that you pass in the size of the buffer after the buffer for all string or character parameters. So
result = scanf_s("%s", temp->name);
should be:
result = scanf_s("%s", temp->name, 20);
The first call to scanf_s is reading garbage off the stack because it is looking for another parameter and possibly corrupting memory.
There is no compiler error because scanf_s uses a variable argument list - the function doesn't have a fixed number of parameters so the compiler has no idea what scanf_s is expecting.
You need
result = scanf_s("%d", &temp->age);
and
result = scanf_s("%f", &temp->height);
Reason is that sscanf (and friends) requires a pointer to the output variable so it can store the result there.
BTW, you have a similar problem with the parameter temp of your function. Since you're changing the pointer (and not just the contents of what it points to), you need to pass a double pointer so that the changes will be visible outside your function:
int FillInLinkedList(node** temp)
And then of course you'll have to make the necessary changes inside the function.
scanf() stores data into variables, so you need to pass the address of the variable (or its pointer)Example:
char string[10];
int n;
scanf("%s", string); //string actually points to address of
//first element of string array
scanf("%d", &n); // &n is the address of the variable 'n'
%19c should be %s
temp->age should be &temp-age
temp->height should be &temp->height
Your compiler should be warning you
about these errors
I believe you need to pass parameters to scanf() functions by address. i.e. &temp->age
otherwise temp-age will be interpreted as a pointer, which will most likely crash your program.