c++ - cout 函数调试断言失败

Question

在我的代码中，我有几个地方有一个相同的 cout 函数，我决定将其移到 int main() 之外。 然而，当我这样做并重新运行我的代码时，当相关函数被触发时，我收到了“调试断言失败”错误。

我放在主函数之外的代码。

string printresult(double smallest, double largest)
{
    cout << "the smaller value is: " << smallest << "\n";
    cout << "the larger value is: " << largest << "\n";
    return 0;
}

我相信它与我在 cout 函数末尾添加的 return 0 代码有关，但是如果没有它，我的代码现在可以甚至编译。任何人都可以指出我正确的方向，以便我可以更好地识别我的错误吗？

我的完整代码

#include "../../std_lib_facilities.h"

// conversion of units to meters

double funconvert(double x, string unit)
{
    const double cm_m = 1.0 / 100.0, in_cm = 2.54, ft_in = 12;
    if (unit == "m")
        return x = x;
    else if (unit == "cm")
        return x = x * cm_m;
    else if (unit == "in")
        return x = x * in_cm * cm_m;
    else if (unit == "ft")
        return x = x * ft_in * in_cm * cm_m;
    else
        cout << "Unknown unit value.\n";
}

// printing the large and the small variables
string printresult(double smallest, double largest)
{
    cout << "the smaller value is: " << smallest << "\n";
    cout << "the larger value is: " << largest << "\n";
    return 0;
}

int main() 
{
    vector<double>numbers;
    double a, b, smallest, largest;
    string unit;
    cout << "Please enter an intiger followed by a measurment unit [e.g 10cm]:";

    // Take input
    while (cin >> a >> unit)
    {
        // check if this is the first number entered
        if (numbers.size() == 0)
        {
            a = funconvert(a, unit);
            b = a;
            numbers.push_back(a);
            numbers.push_back(b);
        }
        else
        {
            //assign a and b to the last two numbers in a vector
            a = funconvert(a, unit);
            numbers.push_back(a);
            a = numbers[(numbers.size() - 1)];
            b = numbers[(numbers.size() - 2)];
            // numbers.erase(numbers.begin()); // enable if desire to keep the vector empty
        }

        // find out which variable is larger and smaller
        if (a > b && (a/b-1) >= 0.01)
        {
            smallest = b;
            largest = a;
            printresult(smallest, largest);
        }
        else if (a < b && (b / a - 1) >= 0.01)
        {
            smallest = a;
            largest = b;
            printresult(smallest, largest);
        }
        else if ((a / b - 1) < 0.01 && (a / b - 1) != 0.00 || (b / a - 1) < 0.01 && (a / b - 1) != 0.00)
        {
            cout << "the nubers are almost equal\n";
        }
        else
            cout << a << " equals " << b << "\n";
    }
    return 0;
}

Answer 1

当您编写return 0 时，C++ 标准库会尝试使用来自const char* 的std::string 构造函数。但是该构造函数要求您向以 nul 结尾的字符缓冲区提供有效地址，否则程序行为未定义。

要修复，请将函数更改为 void 返回类型:

void printresult(双最小，双最大)

并将 return 语句放入函数体中。

还可以考虑用 return x;

替换特殊的 return x = x;