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。
最佳答案
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;
关于C++ 传递一个 Int,错误说它是一个 Double,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37069628/