unique_ptr Ownership Consume

Question

Problem: unique_ptr Ownership Consume. Solve this hands-on electronics task on EWskills.

Accepted Answer

#include 
#include 
#include 
using namespace std;

// Function clearly consumes ownership
void consumeBuffer(unique_ptr buf, int size) {
    for (int i = 0; i < size; i++) {
        cout << (int)buf[i];
        if (i != size - 1) cout << " ";
    }
    cout << endl;
}

int main() {
    int N;
    cin >> N;

    unique_ptr buffer(new uint8_t[N]);

    for (int i = 0; i < N; i++) {
        int temp;
        cin >> temp;
        buffer[i] = (uint8_t)temp;
    }

    // Ownership transfer is explicit at call site
    consumeBuffer(std::move(buffer), N);

    // Caller no longer owns the buffer
    if (buffer == nullptr)
        cout << "No data" << endl;
    else
        cout << "Has data" << endl;

    return 0;
}

Explanation & Logic Summary:

A function that takes std::unique_ptr by value clearly states that it consumes ownership.
The caller must use std::move to transfer ownership, making the handoff explicit.
After the call, the caller’s pointer becomes empty, enforcing correct ownership semantics through the type system.

Firmware Relevance & Real-World Context:

In systems and firmware software, functions often represent processing stages that fully consume a buffer. Expressing ownership transfer directly in the function signature prevents misuse, avoids accidental sharing, and makes APIs safer and easier to reason about in large embedded codebases.

77. unique_ptr Ownership Consume