为什么我在写入“ptr”时收到“C6386”缓冲区溢出警告？

Question

`Variable 'size' is 10 in this example. If I were to hardcode 10 in place of 'int(size)' the overrun warning goes away. Any suggestions/reasoning why this is occurring? I want to allocate 80 bytes for my pointer, each of the allocated values being a timestep from the given timespan.

Thank you!

int main() {
    const double h = 0.1;
    const double tspan[2] = { 0.0, 1.0 };
    const double size =round(tspan[1] / h);
    double *ptr = (double*)malloc(int(size) * sizeof(double));

    if (!ptr) {
        cout << "Memory Allocation Failed";
        exit(1);
    }

    double j = 0;
    for (int i = 0; i < size; i++) {
        ptr[i] = j;
            
        //cout << j << '\n';
        j++;
    }

    cout << '\n';
    for (int i = 0; i < size; i++) {
        cout << *(ptr + i) << endl;
        //cout << i << '\n';
    }
    
    
    free(ptr);
    return 0;
}

I have tried dereferencing the pointer and making sure it isn't NULL. I have also printed out the result. The result being a pointer that counts 0-9. `

Answer 1

double size can be 10.1 , the condition i < 10.1 does not terminate the loop if i is 10, the allocated buffer size is int(10.1) , that is 10, ptr[10] causes the buffer overrun.

Answer 2

As an addedum to what 273K has answered, you csn solve this problem by having size as a std::size_t variable.

Let's floor the result of that division and cast to int .

int main() {
    const double h = 0.1;
    const double tspan[2] = { 0.0, 1.0 };
    const int size = static_cast<int>(std::floor(tspan[1] / h));
    double *ptr = (double*)malloc(size * sizeof(double));

    if (!ptr) {
        cout << "Memory Allocation Failed";
        exit(1);
    }

    double j = 0;
    for (int i = 0; i < size; i++) {
        ptr[i] = j;
            
        //cout << j << '\n';
        j++;
    }

    cout << '\n';
    for (int i = 0; i < size; i++) {
        cout << *(ptr + i) << endl;
        //cout << i << '\n';
    }
    
    free(ptr);
    return 0;
}

Somewhat better, let's use new and delete instead of malloc and free.

int main() {
    const double h = 0.1;
    const double tspan[2] = { 0.0, 1.0 };
    const int size = static_cast<int>(std::floor(tspan[1] / h));
    double *ptr = new double[size];

    if (!ptr) {
        std::cout << "Memory Allocation Failed";
        exit(1);
    }

    double j = 0;
    for (int i = 0; i < size; i++) {
        ptr[i] = j;
            
        //std::cout << j << '\n';
        j++;
    }

    std::cout << '\n';
    for (int i = 0; i < size; i++) {
        std::cout << *(ptr + i) << std::endl;
        //cout << i << '\n';
    }
    
    delete[] ptr;
    return 0;
}

Better still, use std::vector .

int main() {
    const double h = 0.1;
    const double tspan[2] = { 0.0, 1.0 };
    const int size = static_cast<int>(std::floor(tspan[1] / h));
    std::vector<double> vec(size);

    double j = 0;
    for (auto &x : vec) {
        x = j;
            
        //std::cout << j << '\n';
        j++;
    }

    std::cout << '\n';
    for (auto x : vec) {
        std::cout << x << std::endl;
        //std::cout << i << '\n';
    }

    return 0;
}

Answer 3

round() rounds to the nearest integer number. If anything, you should use ceil() to round up .