我无法使此功能正常工作。我正在尝试编写一个中位函数，它接受用户输入的数组和大小，验证它是否正确，然后对其进行排序并显示中位数和排序数组。我尝试了几种不同的东西，无论我尝试什么，我都无法使这个程序工作。任何帮助，将不胜感激。非常感谢你。
#include <iostream>
#include <iomanip>
using namespace std;
double median(int n[], int size);
int main(int argc, char** argv) {
cout << "Calculate The Median of an Array" << endl;
cout << "---------------------------------" << endl;
int size, n;
cout << "Array Size (Maximum is Ten)? ";
cin >> size;
if (size > 10 || size < 0) {
cout << "Invalid size. Please Re-enter." << endl;
};
cout << "Array Contents? ";
cin >> n;
if ([n] != size) {
cout << "Invalid Array. Please Re-enter. " << endl;
};
median(n, size);
return 0;
};
double median(int n[], int size) {
// Allocate an array of the same size and sort it.
double* dpSorted = new double[size];
for (int i = 0; i < size; ++i) {
dpSorted[i] = n[i];
};
for (int i = size - 1; i > 0; --i) {
for (int j = 0; j < i; ++j) {
if (dpSorted[j] > dpSorted[j+1]) {
double dTemp = dpSorted[j];
dpSorted[j] = dpSorted[j+1];
dpSorted[j+1] = dTemp;
};
};
};
// Middle or average of middle values in the sorted array.
int median = 0;
if ((size % 2) == 0) {
median = (dpSorted[size/2] + dpSorted[(size/2) - 1])/2.0;
}
else {
median = dpSorted[size/2];
};
cout << "Median of the array " << dpSorted << "is " << median << endl;
};
我收到以下错误，我无法弄清楚如何修复它们。
34 16 C:\Users\ryanw\Desktop\C++\Labs\Lab 6\main.cpp [Error] invalid
conversion from 'int' to 'int*' [-fpermissive]
和
18 8 C:\Users\ryanw\Desktop\C++\Labs\Lab 6\main.cpp [Note] initializing
argument 1 of 'double median(int*, int)'
答案 0 :(得分：0)
您正在将{int}传递给median：
median(n, size);
这条线几乎肯定没有你认为的那样做
if ([n] != size)
我有点担心你的编译器接受它作为有效的语法。
您需要更改main以获取n的整数集合，以将其传递到中位数。
评论者建议使用向量而不是数组是正确的，原始数组永远不会在c ++中使用正确的集合类型。请注意，数组不作为与C互操作的代码中的参数出现，它们被视为指针。
#include <algorithm>
#include <iterator>
#include <vector>
int median(const vector<int> & numbers); // takes a reference to a vector of ints, and ensures the vector isn't changed
int main(int argc, char** argv) {
cout << "Calculate The Median of some integers" << endl;
cout << "-------------------------------------" << endl;
vector<int> numbers;
for(int n; cin >> n;) { numbers.push_back(n); }
int result = median(numbers);
cout << "The Median of the numbers is " << result << endl;
return 0;
};
int median(const vector<int> & numbers)
{
int size = numbers.size(); // vectors know how many elements they have
vector<int> copy = numbers; // we copy the vector into a local, that we will modify later, leaving numbers unchanged
sort(copy.begin(), copy.end()); // sort is a std:: function
return (size % 2) ? // this is a single expression that chooses from two subexpressions. It is called (the) "ternary operator" (ternary being of three parameters)
copy[size/2] : // even
(copy[size/2] + copy[(size/2) - 1])/2; // odd
}
答案 1 :(得分：0)
我将解释内容作为代码中的注释。这是阵列版本。我保持尽可能接近OP的来源。
#include <iostream>
#include <iomanip>
using namespace std;
double median(int n[], int size);
int main() { // the parameters asre not being used. You can safely leave them out.
cout << "Calculate The Median of an Array" << endl;
cout << "---------------------------------" << endl;
unsigned int size; // unsigned disallows negative numbers. Lest testing required
int n[10]; // n is an array of 10 elements
cout << "Array Size (Maximum is Ten)? ";
cin >> size;
while (size > 10) { // repeat until user provides a valid size
cout << "Invalid size. Please Re-enter." << endl;
cin >> size;
};
// the above loop will be an infinite loop if the user types in a value
// that cannot be converted into an integer.
cout << "Array Contents? ";
for (unsigned int index = 0; index < size; index++)
{
cin >> n[index];
}
// don't need to test the size. This is ensured by the for loop. Mostly
// for now we are ignoring the simple problem: "what if the user inputs
// a value that is not an integer?
median(n, size);
return 0;
}// don't need ; after function.
double median(int n[], int size) {
// Don't need an array of doubles. Ignoring it.
// assuming the logic here is correct. If it isn't, that's a different
// topic and another question.
for (int i = size - 1; i > 0; --i) {
for (int j = 0; j < i; ++j) {
if (n[j] > n[j+1]) {
int dTemp = n[j];
n[j] = n[j+1];
n[j+1] = dTemp;
};
};
};
// Middle or average of middle values in the sorted array.
double result = 0; //reusing an identifier is a dangerous business. Avoid it.
if ((size % 2) == 0) {
result = (n[size/2] + n[(size/2) - 1])/2.0;
}
else {
result = n[size/2];
};
cout << "Median of the array is " << result << endl;
// it's harder to print out an array than I think it should be.
// Leaving it out for now
// Other than main, a function with a return type must ALWAYS return.
return result;
}
使用std::vector和其他图书库魔法：
#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
// using namespace std; generally should avoid this
// moving median up here so forward declaration isn't needed
double median(std::vector<int> &n) { //don't need size. Vector knows how big it is
std::sort(n.begin(), n.end()); // use built-in sort function
double result = 0;
auto size = n.size();
if ((size % 2) == 0) {
result = (n[size/2] + n[(size/2) - 1])/2.0;
}
else {
result = n[size/2];
};
// since this function calculates and returns the result, it shouldn't
// also print. A function should only do one thing. It make them easier
// to debug and more re-usable
return result;
}
int main() {
// can chain couts into one
// endl is more than just a line feed and very expensive. Only use it
// when you need the message to get out immediately
std::cout << "Calculate The Median of an Array\n" <<
"---------------------------------\n" <<
"Array Size (Maximum is Ten)? " << std::endl;
unsigned int size;
std::cin >> size;
while (size > 10) {
// here we use endle because we want he user to see the message right away
std::cout << "Invalid size. Please Re-enter:" << std::endl;
std::cin >> size;
};
std::vector<int> n(size);
std::cout << "Array Contents? " << std::endl;
for (int & val: n) // for all elements in n
{
std::cin >> val;
}
std::cout << "Median of the array is " << median(n) << std::endl;
return 0;
}