英文:
Why is C confusing my multiplicative operations with pointer assignment?
问题
int hashedValue = n * ((int)((k * A%1)));
被标记为指针错误expression must have an integral type。但我不理解这个错误，因为*与后缀而不是前缀操作一起使用。请帮忙解释。谢谢！
英文:
int multiplicationMethod(int k, int n){if(n > SIZE || n < SIZE){printf("IndexOutOfBounds Error. Please Try different slot number.");}else{int N = rand();float A = N/RAND_MAX;int hashedValue = n * ((int)((k * A%1)));HashTable[hashedValue] = k;return hashedValue;}}
int hashedValue = n * ((int)((k * A%1)));
is being tagged with a pointer error expression must have an integral type
I don't understand this though because the * is used with the post NOT prefix operation. Please help. Thanks!
答案1
得分: 3
k * A%1：
你不能这样做。A 是一个浮点数；而 % 只接受整数参数。
这对于 C 语言来说是个尴尬之处。对于这个限制，现在已经没有好的理由了。只需在遇到这个运算符时自动调用 fmod。
应该写成 k * fmod(A, 1)。
正如评论中指出的，你需要将 float A = N/RAND_MAX; 改为 float A = N/(float)RAND_MAX; 才能使你的代码正常工作；否则 A 将永远为 0。
英文:
k * A%1
You can't do that. A is a float; and % only takes integer arguments.
This is an embarrassment to C. There's no good for this restriction anymore. Just auto-call fmod when you see that operator.
Should read k * fmod(A, 1)
As pointed out in comments, you need to change float A = N/RAND_MAX; to float A = N/(float)RAND_MAX; for your code to work; otherwise A will always be 0.
答案2
得分: 0
srand((unsigned int)time(NULL));float A = (float)(rand()) / (float)(RAND_MAX);int B = 1;int hashedValue = (int)((n * (k * (modf(A, &B)))));HashTable[hashedValue] = k;
"srand() 生成一个随机种子，以在每次调用方法时生成一个随机数。rand() 方法中的随机整数生成原本产生一个整数，但被强制转换为浮点数，然后除以一个极大的浮点数。这样的结果是一个介于 0 和 1 之间的随机浮点数。
接下来的过程相对容易理解。但需要注意的是，modf 解析了浮点数 A 和整数 B 的内存地址。"
<details><summary>英文:</summary>srand((unsigned int)time(NULL));float A = (float)(rand()) / (float)(RAND_MAX);int B = 1;int hashedValue = (int)((n* (k* (modf(A,&B)))));HashTable[hashedValue] = k;sreand() generates a random seed to generate a random number upon each subsequent call to method. Rand int generation in rand() method originally produces an int but is typecasted to a float and then divided by an incredibly large float. The result of this is a random float between 0 and 1.The next process is relatively self-explanatory. However, it is important to note that modf parses the float a and the memory address to B.</details># 答案3**得分**: -3根据DaveS的说法，当尝试计算浮点数的模数时，需要使用fmod()而不是%取模运算符。以下是您的示例代码，经过一些调整。需要指出的重要事项是，您需要非常小心您的命名约定，使用单个字母字符很容易，但在循环之外使用会影响代码的可读性。此外，方法名称应该具有描述性，"multiplicationMethod" 并没有提供关于方法实际功能的真正信息。```cpp#include <iostream>#include <unordered_map>#include <math.h>const int SIZE = 20;class TestClass{std::unordered_map<int, std::string> hashTable;public:int multiplicationMethod(int value, int slotNum){if (slotNum > SIZE){printf("IndexOutOfBounds 错误。请尝试不同的槽号。");}else{int N = rand();float A = N / RAND_MAX;int hashedValue = slotNum * ((int)(value * fmod(A, 1)));hashTable[hashedValue] = value;printf(std::to_string(hashedValue).c_str());return hashedValue;}}};int main(){TestClass t;t.multiplicationMethod(4, 4);}
希望这有助于您理解代码的内容。
英文:
As DaveS stated you need to use fmod() rather than the % modulo operator when trying to work out the modulus of a float.
Here is your example with some tweaks. Big thing I have to point out is that you need to be very careful with your naming conventions it is easy to use one letter characters but outside of a loop it will affect the readability of your code. Also method names should be descriptive multiplicationMethod gives no real information as to what the method actually does.
#include <iostream>#include <unordered_map>#include <math.h>const int SIZE = 20;class TestClass{std::unordered_map<int, std::string> hashTable;public:int multiplicationMethod(int value,int slotNum){if(slotNum > SIZE){printf("IndexOutOfBounds Error. Please Try different slot number.");}else{int N = rand();float A = N/RAND_MAX;int hashedValue = slotNum * ((int)(value * fmod(A,1)));hashTable[hashedValue] = value;printf(std::to_string(hashedValue).c_str());return hashedValue;}}};int main(){TestClass t;t.multiplicationMethod(4,4);}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论