理解C代码编译警告有困难,函数定义有问题

oalqel3c  于 2023-03-07  发布在  其他
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我正在努力理解数组/指针/函数,以及如何在我开始编程c代码时正确地应用它们。
这个程序根据所选的难度显示一串含有隐藏单词的字母几秒钟。然后它给你一个机会看看你是否找到了隐藏的单词。它这样做三次,然后游戏结束。
收到代码块编译器生成警告:(我在代码中包含了行号(在行的末尾),这样就很容易找到警告所指的行)

80 Warning: passing argument 3 of 'verifyanswer' makes pointer from integer without a cast

10 note: expected 'int* ' but argument is of type 'int' In function 'verifyanswer':

114 warning: comparison between pointer and integer

下面是我的代码:

#include <stdio.h> 
#include <stdlib.h> 
#include <time.h> 
#include <string.h>
 #include <ctype.h>

void verifyanswer (char*StrGameArray, char response [], int Strlength []);// 10 10 10 10 10

int main () { 
char *StringGameArray [50] = {"cathoropordscodiscopwnsestropthprds",//disco                "cassormocedimahonahteidbballoonhelspq",//balloon "cheeszpistsrwhystacifitercinotonomwerd",};//why

int Stringlength [3] = {5,7,3};

char Response [80]={0};

int LevelChoice = 0; int DisplayTime = 0; int StartTime = 0; int Displayed = 0; int x = 0; system("clr");

while (LevelChoice !=4 )
{
  printf("\nWELCOME TO THE FIND THE WORD GAME!");
  printf("\nChoose your difficulty level 1-3");
  printf("\n\n 1-Super easy");
  printf("\n 2-Regular");
  printf("\n 3-Hard");
  printf("\n 4-Quit\n");
  scanf("%d",&LevelChoice);

     switch (LevelChoice)
    {
      case 1:
      DisplayTime = 5;
      break;

      case 2:
      DisplayTime = 3;
      break;

      case 3:
      DisplayTime = 1;
      break;

     }//end of switch

StartTime = time(NULL);

  for (x=0; x<3;x++)
     {
       //DISPLAY TEXT FOR A FEW SECONDS DICTATED BY DIFFICULTY CHOICE
       while (StartTime + DisplayTime > time(NULL))
          {
               if (Displayed == 0)
               {
                printf("\nFind a word in :\n\n");
                printf("%s\n\n",StringGameArray [x]);
                Displayed = 1;
               }//end if

          }//end while

       system("clear");

       printf("\nEnter found word\n");
       scanf("%s",Response);

       verifyanswer(StringGameArray[x],Response,Stringlength [x]);//function call  80 80 80 

       Displayed = 0;
       StartTime = time(NULL);

     }//end for loop 3x


}//end of while
return 0;
}//end of function main

void verifyanswer (char*StrGameArray, char response [],int Strlength []) {

int LenghtofResponse = 0;
LenghtofResponse = (strlen(response));

printf("\nResponse is %s",response);

//CONVERT ANSWER TO LOWER CASE TO PERFORM A VALID COMPARISON int x =0; for (x=0; x<= strlen(response);x++){ response[x] = (tolower (response[x]));}

if ( (strstr(StrGameArray,response) != 0)&&  (LenghtofResponse == Strlength) && (response[0] != 0)) //114 114 114 114 114 

printf("\nCongratulations you found the word");

else printf("\nYou did not find the word");

}//end of check answer
kyxcudwk

kyxcudwk1#

您已经声明并在稍后定义了函数以获取int[]

void verifyanswer (char*StrGameArray, char response [], int Strlength []);

这已经很可疑了,因为你会期望“length”是一个整数,而实际上,当你在main()中调用时,你传递给它一个int

int Stringlength [3] = {5,7,3};
//...
verifyanswer(StringGameArray[x],Response,Stringlength [x]);

这就是警告的含义。参数会被转换成一个指向int的指针,但这可能不是你想要的。为了解决这个问题,你应该声明并定义一个函数来接受int:

void verifyanswer (char*StrGameArray, char response [], int Strlength);
pn9klfpd

pn9klfpd2#

这里

verifyanswer(StringGameArray[x],Response,Stringlength [x]);/

你正在传递一个整数(Stringlength [x])到verifyanswer .函数需要一个数组

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