JuniorTheoryCommon

Where, if anywhere, does the call `add(-1)` unlock the mutex `m`?

C++

#include <mutex>

std::mutex m;
int total = 0;

void add(int x) {
    std::lock_guard<std::mutex> guard(m);
    if (x < 0) {
        return;
    }
    total += x;
}

At return;, where guard goes out of scope and its destructor runs

Expecting return; to skip the unlock: the destructor of guard runs on every exit from its scope, an early return included.
Expecting to have to call m.unlock() by hand: the destructor of lock_guard unlocks m, and that is the reason to use it.

Why is lock_guard safer than calling m.lock() and m.unlock() by hand?
How would you keep m locked for only part of a function?

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JuniorTheoryCommon

Where, if anywhere, does the call `add(-1)` unlock the mutex `m`?

C++

#include <mutex>

std::mutex m;
int total = 0;

void add(int x) {
    std::lock_guard<std::mutex> guard(m);
    if (x < 0) {
        return;
    }
    total += x;
}

At return;, where guard goes out of scope and its destructor runs

Expecting return; to skip the unlock: the destructor of guard runs on every exit from its scope, an early return included.
Expecting to have to call m.unlock() by hand: the destructor of lock_guard unlocks m, and that is the reason to use it.

Why is lock_guard safer than calling m.lock() and m.unlock() by hand?
How would you keep m locked for only part of a function?

Back to questions Open topic in roadmap →Answer it yourself — Quiz

Where, if anywhere, does the call add(-1) unlock the mutex m?

Where, if anywhere, does the call add(-1) unlock the mutex m?

Where, if anywhere, does the call `add(-1)` unlock the mutex `m`?

Where, if anywhere, does the call `add(-1)` unlock the mutex `m`?