冲突类型，当它们都无效时

Question

我的代码出现以下错误，其中一个找不到解决方法。 我了解函数类型的声明必须与定义匹配，但是可以。

#include <stdio.h>
#include <math.h>

//Global Declarations
int input_win(void);
int input_scores(int,int);
void results(int,int,int,int);

int main()
{
  //Local Declarations
    int winNum;        //The number of points needed to win
    int scores[25];    //The array of scores
    int score_old = 0; //The second to most recent score input
    int score_new;     //The most recent score input
    int cnt_score = 1; //The count for input
    int cnt_won = 0;   //The count for games won
    int cnt_played =0; //The count for games played
  //Executeable Statements
    winNum = input_win();
    do
    {
      score_new = input_scores(cnt_score,cnt_won);
      if(score_new != -1)
      {
        if(score_new <= score_old)
        {
          cnt_played += 1;
        }
        if(score_new == winNum)
        {
          cnt_won += 1;
        }
        scores[cnt_score - 1] = score_new;
        score_old = score_new;
        cnt_score += 1;
      }
    }while(score_new != -1 || score_new != 25);
    results(cnt_score,scores,cnt_won,cnt_played);
return 0;
}

......

void results(int cnt_score,int scores[25],int cnt_won,int cnt_played)
{
  //Local Declarations
    int cnt_loop; //The counter for the for loop

  //Executeable Statements
    printf("\nScores entered (%d): {",cnt_score);
    for (cnt_loop = 1;cnt_loop < cnt_score;cnt_loop++)
    {
      printf("%d, ",scores[cnt_loop - 1]);
    }
    printf("%d}\n",scores[cnt_score - 1]);
    printf("\nNumber of games won: %d\nNumber of games played: %d\n",cnt_won,cnt_played);
}

我得到的错误如下：

hw06.c: In function 'main':
hw06.c:62: warning: passing argument 2 of 'results' makes integer from pointer without a cast
hw06.c:29: note: expected 'int' but argument is of type 'int *'
hw06.c: At top level:
hw06.c:160: error: conflicting types for 'results'
hw06.c:29: note: previous declaration of 'results' was here

Answer 1

您的函数原型与您的函数定义不匹配：

void results(int,int,int,int);
void results(int cnt_score,int scores[25],int cnt_won,int cnt_played) { ... }

请注意，第二个参数是int * ，而不是int 。

Answer 2

将结果函数重新声明为：

void results(int,int[],int,int);

Answer 3

函数的第二个参数声明为int类型，但实现的类型为int[25] （整数数组）。