缓冲区溢出警告

Question

此脚本将数字的所有数字展开到一个数组中，因此通过执行此操作： spread(number)[position]您可以访问具有相对位置的数字。 现在编译器给了我一个Buffer overrun警告 (C6386)，我认为这是由超出数组边界引起的（如果我错了，请纠正我），但我编写了函数脚本，这样这样的事情就不会发生，程序仍然出现故障

#include <iostream>
#include <math.h>
using namespace std;

unsigned int size(long long int num)
{
    unsigned int size = 1;
    while (num >= pow(10, size)) size++;
    return size;
}

int* spread(int num)
{
    unsigned int digit;
    int* nums = new int[size(num)];
    for (unsigned int P = 0; P <= size(num) - 1; P++)
    {
        digit = num - num / 10 * 10;
        num /= 10;
        nums[P] = digit; //Right in this line the program doesn't seem to behave correctly
    }
    return nums;
}

int main()
{
    cout << split(377)[0] << endl;
    cout << split(377)[1] << endl;
    cout << split(377)[2] << endl;
    system("PAUSE");
    return 0x0;
}
/*
Output of the program:
7
7
-842150451 <-- there should be a 3 here
Press any key to continue . . .
*/

Answer 1

for 循环的主体会干扰您的结束条件：

• P=0, num=377, size(num) = 3, P <= 2 = true
• P=1, num=37, size(num) = 2, P <= 1 = true
• P=2, num=3, size(num) = 1, P <= 0 = false

解决方法是size(num)计算size(num)并在循环条件中使用它：

int numdigits = size(num);
for (int P=0; P < numdigits; P++) { ... }

Answer 2

{
    unsigned int digit;
    int* nums = new int[size(num)];
    int P = 0;
    while(num!=0)
    {
        digit = num - num / 10 * 10;
        num /= 10;
        nums[P++] = digit; //Right in this line the program doesn't seem to behave correctly
    }
    return nums;
}

您将存储数字直到数字为 0。使用此条件，您的代码将起作用。 但是你有内存泄漏......你必须释放动态内存！

更新：有一个短代码，可以工作;-)

#include <iostream>
#include <vector>
#include <string>
using namespace std;

vector<size_t> spread(int const num)
{
    string number = to_string(num);
    vector<size_t> digits;
    for(size_t i = 0; i < number.size(); ++i)
    {
        digits.push_back(number[i] - '0');
    }
    return digits;
}

int main()
{
    auto vec = spread(12345000);
    for (auto elem : vec)
    {
        cout << elem << endl;
    }
    system("PAUSE");
    return 0x0;
}