緩沖區溢出警告

Question

此腳本將數字的所有數字展開到一個數組中，因此通過執行此操作： spread(number)[position]您可以訪問具有相對位置的數字。 現在編譯器給了我一個Buffer overrun警告 (C6386)，我認為這是由超出數組邊界引起的（如果我錯了，請糾正我），但我編寫了函數腳本，這樣這樣的事情就不會發生，程序仍然出現故障

#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 循環的主體會干擾您的結束條件：

• 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

解決方法是size(num)計算size(num)並在循環條件中使用它：

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;
}

您將存儲數字直到數字為 0。使用此條件，您的代碼將起作用。 但是你有內存泄漏......你必須釋放動態內存！

更新：有一個短代碼，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;
}