c++ - 如何解决 Visual Studio C++ 问题， "Arithmetic overflow: Using operator ' *' on a 4 byte value and then casting the result to a 8 byte value."？

Question

我已经标记了警告位置。如果我理解正确，将两个 32 位数字相乘可以得到 64 位，所以我得到了一个错误。

如果 C++ int 是 32 位那么存储 64 位不是也溢出了吗？请解释我做错了什么。警告消息显示在 Visual Studio 2019 中。

转换为更大的数据类型解决了这个问题。为什么？

#include <iostream>
using namespace std;


unsigned int m;
unsigned int n;


// Symbols 
int symbols[] = { -1, 0, 1 };
unsigned int symbols_size = sizeof(symbols) / sizeof(symbols[0]);



// Returns false to signify that element exiting in array
bool check_symbol_validity(int val) {
    for (unsigned int i = 0; i < symbols_size; ++i) {
        if (symbols[i] == val) {
            return false;
        }
    }
    return true;
}


int main()
{

    cout << "Place enter graph node count: ";
    cin >> m;
    n = m;


    // Just a separator
    cout << string(30, '#') << "\n";
    cout << "Enter Transition Symbol Between Node i and Node j. \nIf none then enter -1.\n";



    // Transition table, take node and symbol, return next node
    //int* transition_table = new int[m * symbols_size]; // WARNING
    int* transition_table = new int[(int64_t)m * symbols_size]; // OK


    // Declare C++ dynamic array
    //int* weight_matrix = new int[m * n]; // WARNING
    //int* weight_matrix = new int[(long)m * n]; // WARNING
    int* weight_matrix = new int[(long long)m * n]; // OK


    // Store positional values for direct access rather than using multiplication
    int* width_val_arr = new int[m];


    // TAKE VALID INPUT
    int user_input_value;
    for (unsigned int i = 0; i < m; ++i) {
        int tmp1 = i * m;
        width_val_arr[i] = tmp1;
        for (unsigned int j = 0; j < n; j++){
            cout << "Enter connection between " << i << " and " << j << "\n";
            cin >> user_input_value;
            while (check_symbol_validity(user_input_value)) {
                cout << "WRONG! ENTER CORRECT VALUE." << "\n";
                cout << "Enter connection between " << i << " and " << j << "\n";
                cin >> user_input_value;
            }
            weight_matrix[tmp1 + j] = user_input_value;
        }
    }


    cout << string(30, '#') << "\n";


    // SHOW THE WEIGHT MATRIX
    for (unsigned int i = 0; i < m; ++i) {
        //int tmp1 = i * m;
        for (unsigned int j = 0; j < n; j++) {
            cout << weight_matrix[width_val_arr[i] + j] << " ";
        }
        cout << "\n";
    }



    // Delete the dynamic arrays
    delete[] weight_matrix;
    delete[] transition_table;
    delete[] width_val_arr;
}

警告信息:

"Arithmetic overflow: Using operator '*' on a 4 byte value and then casting the result to a 8 byte value. Cast the value to the wider type before calling operator '*' to avoid overflow".

Answer 1

C26451 RESULT_OF_ARITHMETIC_OPERATION_CAST_TO_LARGER_SIZE Using operator

Cast the value to the wider type before calling operator [operator] to avoid overflow

This warning indicates incorrect behavior that results from integral promotion rules and types larger than those in which arithmetic is typically performed. We detect when a narrow type integral value was shifted left, multiplied, added, or subtracted and the result of that arithmetic operation was cast to a wider type value. If the operation overflowed the narrow type value, then data is lost. You can prevent this loss by casting the value to a wider type before the arithmetic operation.

Arithmetic overflow checks in C++ Core Check