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我正在为这个程序的最后一点苦苦挣扎，我需要将数组传递给两个不同的函数，但我不知道该怎么做。

我得到的唯一错误发生：这里：

 input(array[20]);
 calculate(array[20],&pairs);

和这里：

//exit,
exit;

除此之外，它应该按照我需要的方式工作，我想出了如何在普通变量上使用指针，但是数组的行为不同，我不知道该怎么做......

文档已经完成了一半，我仍然需要在描述概述的末尾添加循环，但我只需要传递数组的帮助。

涉及我的退出行的错误也与问题无关，但如果您知道一个很好的修复程序！

/*
Description: Do not use global variables. Pass your arguments by value and
             by reference. Using arrays and modular programming techniques,
             write a C program that will allow a user to populate an array
             with integers, and then compute and print out the number of
             adjacent pairs in the array (i.e. the number of occurrences where
             an array element is the same as its neighbors). For example, if
             the array contained [2,3,3,4,52,52,4,4,4,4,7,7,1], the number of
             adjacent pairs is 6. The program should give the user the option
             of examining more than one array (i.e. loop). Assume the array has
             a max. of 20 elements. The main() function should primarily be
             responsible to direct the flow of logic.  That is: use one
             function to obtain input (pass by reference), another function to
             do processing (pass by value), and perhaps a third function to do
             output (pass by value).  The main() function should call the input
             function and the processing function. 
*/

//Include statements.
#include <cstdlib>
#include <iostream>
#include <math.h>

//Standard namespace.
using namespace std;

void input (int array[20]); //Used when user inputs the numbers.
void calculate(int array[20], int *pairs); //Used to calculate the matches.
void output(int *pairs); //Used to output the number of pairs.

int main(void) 
{ 
     int array[20] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};
     char quit;

     start:
     int pairs = 0;
     input(array[20]);
     calculate(array[20],&pairs);
     output(&pairs);

     //Ask the user if they want to exit
     printf("\nWould you like to continue testing my project, or exit?");
     printf("\nTo exit input: 'N' or 'n'. To continue testing input anything else.");
     //store their input in variable: exit
     scanf("%s",&quit);
     //If they want to exit...
     if (quit == 'N' || quit == 'n')
     {
         //exit,
         exit;
     }
     //otherwise,
     else
     {
         //clear the screen
         system("cls");
         //and go back to the start.
         goto start;
     }
}

void input(int array[20])
{
     int count = 0;
     for (count;count<20;count++)
     {
         printf("Enter values . . . \n");
         scanf("%i", &array[count]);
     }
}

void calculate(int array[20], int *pairs)
{
     int counter = 0;
     for (counter;counter<19;counter++)
     {
         if (array[counter] == array[counter+1])
            *pairs+=1;
     }
}

void output(int *pairs) 
{
     printf("Number of pairs: [%i]\n", *pairs);
}
4

1 回答 1

0

调用函数时不传递大小。

input(array);

这应该足够了。

另一种解决方案是使函数采用int*. 您将添加一个额外的参数来传递数组的大小。

void func(int* my_array, int size);

你会像这样声明你的数组：

int i[20];

并像这样调用你的函数：

func(i, 20);

目前，通过传入array[20]，您将返回 an int，如果您想知道，您将超出范围，因为最大索引是19. 表达式的不适当的返回类型不允许您编译程序。

于 2013-03-23T23:43:27.170 回答