Question

我试图调用该函数

(breaks <- data.frame(
  start = index(z[c(1, bp$breakpoints+1)]),
  end = c(index(z[bp$breakpoints]), index(z[length(z)]))
))
#                 start                 end
# 1 2015-01-14 00:00:00 2015-01-14 07:00:00
# 2 2015-01-14 08:00:00 2015-01-14 09:00:00
# 3 2015-01-14 10:00:00 2015-01-14 17:00:00
# 4 2015-01-14 18:00:00 2015-01-14 20:00:00
# 5 2015-01-14 21:00:00 2015-01-14 23:00:00
fits <- lapply(seq_len(nrow(breaks)), function(x) {
  idx <- index(z)>=breaks[x, 1] & index(z)<=breaks[x, 2]
  fit <- lm(z[idx]~index(z[idx]))
})
sapply(fits, coefficients)
#                        [,1]          [,2]          [,3]          [,4]          [,5]
# (Intercept)   -4.048094e+02 -1.007876e+05  8.358223e+03  3.750603e+04  1.873346e+04
# index(z[idx])  2.850529e-07  7.091667e-05 -5.880291e-06 -2.638889e-05 -1.318056e-05

在

int Even( int a[][MAX_COLUMNS], int length, int width)

内

将输出打印给用户。

运行程序后，我收到以下错误

`int Even(int,int,int) connot convert argument 1 from int [3][2] to int

这是代码，感谢您抽出宝贵时间帮助我。

main

Answer 1

你应该避免前瞻性声明（特殊情况除外）。在你的Q导致你错误。完整代码：

#include <iostream>
#define MAX_ROWS 3
#define MAX_COLUMNS 2

using namespace std;

int Even( int a[][MAX_COLUMNS], int length, int width){

    int sum = 0;
    int i; 
    int j; 

    for (i = 0; i < length, i++;){

        for (j = 0; j < width, j++;){

            if (a[i][j]  % 2 == 0){

                sum++;
            }
        }

    }

    return sum;

}

int main(){


    int A[MAX_ROWS][MAX_COLUMNS] = { { 3, 2 }, { 4, 5 }, { 2, 2 } };

    cout << "There are " << Even(A,MAX_ROWS,MAX_COLUMNS) << " even numbers in the matrix." << endl;
    cout << "Goodbye :-)" << endl;

    system("PAUSE");
    return 0;

}

如何在数组中调用函数？

1 个答案: