C++ 传递一个 Int，错误说它是一个 Double

Question

我目前正在完成一项作业，我们正在使用 C++ 编写汇编程序。我已经尽可能让代码在输入数字时进行区分，并且正在尝试让程序将 16 位数组中的相关位更改为所需的二进制数。

问题是在计算幂后，输入数字的余数和超过它的第一个幂被分配给一个 int，但将它传回 dec_to_bin() 我收到以下错误。

assembler.cpp: In function ‘void dec_to_bin(int, int*)’: assembler.cpp:135:32: error: invalid conversion from ‘double (*)(double, double)throw ()’ to ‘int’ [-fpermissive] dec_to_bin(remainder, pointer);

//*pointer points to an array[16]
void dec_to_bin(int num, int *pointer)
{
    //Base cases
    if(num == 2)
    {
        pointer[14] = 1;
        return;
    }
    else if(num == 1)
    {
        pointer[15] = 1;
        return;
    }
    else
    {
        int power = 0;
        //power function outlined below. The inbuilt pow(a, b) returns a
        //double and gave the same problem, so I tried this instead.
        while(PowerFunc(2, power) < num)
        {
            power++;
        }
        //Problem arises here. remainder is assigned as an int
        //Using typeinfo shows it as an int, yet passing it a few lines
        //down gives the quoted error.
        int remainder = PowerFunc(2, power)-num;
        pointer[16-power] = 1;
    }
    if(remainder == 0)
    {
        return;
    }
    else
    {
        //Line that throws the error. Saying remainder is a double.
        dec_to_bin(remainder, pointer);
    }
}
//Power function to replace pow(a, b). Not ideal but it works.
int PowerFunc(int number, int powernum)
{
    if(powernum == 0)
    {
        return 1;
    }
    else if(powernum == 1)
    {
        return number;
    }
    else
    {
        return number * PowerFunc(number, powernum-1);
    }
}

老实说，这让我很困惑。起初我以为使用 pow(a, b) 并将其分配给一个 int 只会拼接掉小数部分，但即使是我自己的幂函数仍然告诉我我的 int 是一个 double。

Answer 1

线

    int remainder = PowerFunc(2, power)-num;

出现在 else 下的花括号分隔 block 之间, 所以 remainder 的范围变量仅向下延伸一行。 remainder在

    if(remainder == 0)

在

    dec_to_bin(remainder, pointer);

是其他实体，碰巧具有相同的名称。您可以通过重命名您的 remainder 来验证这一点对于其他任何事情，请说 myremainder ...

补充说明:
考虑更换 if(powernum == 0)与 if(powernum <= 0) — 在负指数的情况下，这将为您节省大量计算。

我还建议替换递归

return number * PowerFunc(number, powernum-1);

随着迭代:

int result = 1;
while(powernum-- >= 0)
    result *= number;
return result;