c - C 中的函数问题 : Expected declaration before 'int' /'char' , 函数参数太少以及其他一些问题

Question

好吧，首先，我应该让大家知道我是一个相对较新的程序员，而且我似乎无法获得正确的功能，所以我深表歉意如果这些确实是愚蠢/明显的问题，请提前。

无论如何，现在开始实际编程。这是一个学校项目，有点像刽子手。我已经花了好几个星期的时间，我即将完成，但这些讨厌的错误妨碍了我，我就是无法修复它们!

如果有人可以帮助我消除最后几个错误，我将非常感激!再说一次，这里是初学者程序员，我可能犯了一些令人畏缩的错误。也很抱歉代码太长..

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

typedef struct {
    char title[50];
    char hidden[50];
}Title;


void Film(Title* pT)
{
    int i=0;
    int number;
    char movies[44][50];

    FILE *fp = NULL;
    fp = fopen("Film.txt", "r");

    for(i=0; i<44; ++i)
    {
        fgets(movies[i], sizeof(movies[i]), fp);
    }


    fclose(fp);


    number = (rand() % 44);
    strcpy(pT->title, movies[number]);
}

char Star(Title* pT, char lowerc, char higherc, char character)
{
    int val; 
    char c;
    int lenMovie;

    lenMovie = strlen(pT->title);
    strcpy(pT->hidden, pT->title);

    for(val=0; val <= lenMovie; val++)
    {

        c = pT->hidden[val]; 



        if(c == lowerc || c == higherc)
        {
            pT->hidden[val] = character;
        }


        else if(c >= 'a' && c<= 'z')
        {

            pT->hidden[val] = '*';
        }


        else if(c >= 'A' && c<= 'Z')
        {

            pT->hidden[val] = '*';
        }


        else
        {

            pT->hidden[val] = c;
        }

    }

    printf("%s", pT->hidden);
}


char Film_Guess(Title* pT, int attempt)
{

    int guess[50], answer, size;

    printf("What movie do you think it is: ");
    scanf("%s", &guess);

    size = strlen(pT->title);
    answer = strncmp(pT->title, guess[50], size);


    if(answer = 0)
    {
        printf("You beat the Film Genie, nce work!");
        return 0;
    }


    else
    {
        attempt++;
        return main();
    }

}


int main(void)
{

    char option; 
    int attempt = 0;
    char lowerc, higherc, character, reply;

    Title t;
    srand(time(NULL));

    printf("\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\t\t\t Welcome Player\n\n\n");




    while(attempt <= 5)
    {


        printf("\nWould you like to try: ");
        scanf("%c", &reply);


        if(reply == 'y' || reply == 'Y')
        {

        Film (&t);
        Star(&t, lowerc, higherc, character);

        printf("\n Would you like to guess a character(c) or the whole film(f):");
        scanf("%c",&option);



            if(option =='c' || option =='C')
            {

                printf("\nPlease enter a character: ");
                scanf("%c", &character);

                lowerc = tolower(character);
                higherc = toupper(character);

                Star(&t, lowerc, higherc, character);


            }


            else if(option =='f' || option =='F')
            {

                Film_Guess(&t, attempt);

            }

            else
            {

                printf("\nInvalid response");
                return main();

            }


        }




    else{
            break;
        }

    }




    if(reply == 'n' || reply == 'N')
    {

        printf("\nLoser");
        return 0;

    }



    else
    {

        printf("\nInvalid response");
        return main();

    }

}

Answer 1

我发现的一些错误(可能还有更多):

1. strncmp() 接受两个 char* 参数，您提供一个 int* 作为其中之一。检查联机帮助页:http://www.manpagez.com/man/3/strncmp/

2. 第 117 行:Star(&t, char lowerc, char highc, char character); -- 在传递参数时不应提供类型，仅在声明参数时(*下面的注释)。

3. 第 129 行:Star (&t); -- 您的函数声明有 4 个参数。您不能只提供一项，您必须提供全部四项。

4. 第 135 行:Film_Guess(&t, int attempts); -- 与第 117 行相同的问题

*注意:如果您对“参数”和“参数”感到困惑，请参阅 this question/answer .