C++ – understanding of pthread_cond_wait() and pthread_cond_signal()

cmultithreadingmutex

Generally speaking, pthread_cond_wait() and pthread_cond_signal() are called as below:

//thread 1:
pthread_mutex_lock(&mutex);
pthread_cond_wait(&cond, &mutex);
do_something()
pthread_mutex_unlock(&mutex);

//thread 2:
pthread_mutex_lock(&mutex);
pthread_cond_signal(&cond);  
pthread_mutex_unlock(&mutex);

The steps are

pthread_cond_wait(&cond, &mutex); is called, it unlocks the mutex
Thread 2 locks the mutex and calls pthread_cond_signal(), which unlocks the mutex
In thread 1, pthread_cond_wait() is called and locks the mutex again

Now in thread 2, after pthread_cond_signal() is called, pthread_mutex_unlock(&mutex) is going to run, it seems to me that it wants to unlock a the mutex which is now locked by thread 1. Is there anything wrong in my understanding?

Besides, it also seems to me that pthread_cond_wait() can be called by only 1 thread for the same cond-mutex pair. But there is a saying "The pthread_cond_signal() function shall unblock at least one of the threads that are blocked on the specified condition variable cond (if any threads are blocked on cond)." So, it means pthread_cond_wait() can be called by many threads for the same cond-mutex pair?

Best Answer

pthread_cond_signal does not unlock the mutex (it can't as it has no reference to the mutex, so how could it know what to unlock?) In fact, the signal need not have any connection to the mutex; the signalling thread does not need to hold the mutex, though for most algorithms based on condition variables it will.

pthread_cond_wait unlocks the mutex just before it sleeps (as you note), but then it reaquires the mutex (which may require waiting) when it is signalled, before it wakes up. So if the signalling thread holds the mutex (the usual case), the waiting thread will not proceed until the signalling thread also unlocks the mutex.

The common use of condition vars is something like:

thread 1:
    pthread_mutex_lock(&mutex);
    while (!condition)
        pthread_cond_wait(&cond, &mutex);
    /* do something that requires holding the mutex and condition is true */
    pthread_mutex_unlock(&mutex);

thread2:
    pthread_mutex_lock(&mutex);
    /* do something that might make condition true */
    pthread_cond_signal(&cond);
    pthread_mutex_unlock(&mutex);

The two threads have some shared data structure that the mutex is protecting access to. The first thread wants to wait until some condition is true then immediately do some operation (with no race condition opportunity for some other thread to come in between the condition check and action and make the condition false.) The second thread is doing something that might make the condition true, so it needs to wake up anyone that might be waiting for it.

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

C++ – What are the differences between a pointer variable and a reference variable in C++

A pointer can be re-assigned:

int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);

A reference cannot be re-bound, and must be bound at initialization:

int x = 5;
int y = 6;
int &q; // error
int &r = x;

A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary & operator and a certain amount of space that can be measured with the sizeof operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
```
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;

assert(p == p2); // &x == &r
assert(&p != &p2);
```

You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.

int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;

**pp = 2;
pp = &q; // *pp is now q
**pp = 4;

assert(y == 4);
assert(x == 2);

A pointer can be assigned nullptr, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr, but this is undefined and will not behave consistently.

/* the code below is undefined; your compiler may optimise it
 * differently, emit warnings, or outright refuse to compile it */

int &r = *static_cast<int *>(nullptr);

// prints "null" under GCC 10
std::cout
    << (&r != nullptr
        ? "not null" : "null")
    << std::endl;

bool f(int &r) { return &r != nullptr; }

// prints "not null" under GCC 10
std::cout
    << (f(*static_cast<int *>(nullptr))
        ? "not null" : "null")
    << std::endl;

You can, however, have a reference to a pointer whose value is nullptr.

Pointers can iterate over an array; you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access its members whereas a reference uses a ..
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
```
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
```
This makes const & more convenient to use in argument lists and so forth.