Question

我试图在我的程序中使用结构，并且我收到一堆错误消息，指出time1，time2和hourDif是未声明的标识符。我以为我按照你应该的方式宣布这些结构，所以我对问题是什么感到困惑。

以下是代码：

#include <stdio.h>

struct time1 {
    int seconds1;
    int minutes1;
    int hours1;
};

struct time2 {
    int seconds2;
    int minutes2;
    int hours2;
};

    int timeDif(time1, time2) {
        struct dif {
            int secDif = time2.seconds2 - time1.seconds1;
            int minDif = time2.minutes2 - time1.minutes1;
            int hourDif = time2.hours2 - time1.hours1;
        };

        printf("The difference is %d:%d:%d\n", hourDif, minDif, secDif);
    }



    int main(void) {

        printf("Enter seconds1 (ss)\n");
        scanf("%d", &time1.seconds1);

        printf("Enter minutes1 (mm)\n");
        scanf("%d", &time1.minutes1);

        printf("Enter hours1 (hh)\n");
        scanf("%d", &time1.hours1);

        printf("Enter seconds2 (ss)\n");
        scanf("%d", &time2.seconds2);

        printf("Enter minutes2 (mm)\n");
        scanf("%d", &time2.minutes2);

        printf("Enter hours2 (hh)\n");
        scanf("%d", &time2.hours2);

        timeDif(time1, time2);

    return 0;
}

Answer 1

您无需定义三个结构，因为它们的类型相同 - 时间。您只能定义一个结构time：

#include <stdio.h>

struct time {
    int seconds;
    int minutes;
    int hours;
};

void timeDif(struct time time1,struct time time2) {
    struct time diff= {
         time2.seconds - time1.seconds,
         time2.minutes - time1.minutes,
         time2.hours - time1.hours
    };

    printf("The difference is %d:%d:%d\n", diff.hours, diff.minutes, diff.seconds);
}

int main(void) {
    struct time time1,time2,diff;
    printf("Enter seconds1 (ss)\n");
    scanf("%d", &time1.seconds);

    printf("Enter minutes1 (mm)\n");
    scanf("%d", &time1.minutes);

    printf("Enter hours1 (hh)\n");
    scanf("%d", &time1.hours);

    printf("Enter seconds2 (ss)\n");
    scanf("%d", &time2.seconds);

    printf("Enter minutes2 (mm)\n");
    scanf("%d", &time2.minutes);

    printf("Enter hours2 (hh)\n");
    scanf("%d", &time2.hours);

    timeDif(time1, time2);

    return 0;
}

Answer 2

在此功能定义中

int timeDif(time1, time2) {
    struct dif {
        int secDif = time2.seconds2 - time1.seconds1;
        int minDif = time2.minutes2 - time1.minutes1;
        int hourDif = time2.hours2 - time1.hours1;
    };

    printf("The difference is %d:%d:%d\n", hourDif, minDif, secDif);
}

编译器将标识符time1和time2视为对象的名称。但它们没有定义，因为你没有指定它们的类型。我认为您忘记指定对象的名称并仅指定其类型。函数声明可能看起来像

int timeDif(struct time1 t1, struct time2 t2);

考虑到必须在结构类型的对象声明中使用结构名称的结构标记。

此函数虽然返回类型为int，但不返回任何内容。它没有任何意义。并且您可能无法在C中定义其结构中的结构的数据成员。

因此这个结构定义

    struct dif {
        int secDif = time2.seconds2 - time1.seconds1;
        int minDif = time2.minutes2 - time1.minutes1;
        int hourDif = time2.hours2 - time1.hours1;
    };

无效。

函数调用

存在同样的问题

timeDif(time1, time2);

我认为你的意思是以下

#include <stdio.h>

struct time 
{
    int seconds;
    int minutes;
    int hours;
};


struct time_dif 
{
    int secDif;
    int minDif;
    int hourDif;
};

struct time_dif timeDif( struct time t1, struct time t2 ) 
{
    struct time_dif dif = 
    { 
        t2.seconds - t1.seconds,
        t2.minutes - t1.minutes,
        t2.hours   - t1.hours
    };

    return dif;
}



int main(void) 
{
    struct time t1, t2;

    printf( "Enter hours1 (hh): " );
    scanf( "%d", &t1.hours );

    printf( "Enter minutes1 (mm): " );
    scanf( "%d", &t1.minutes );

    printf( "Enter seconds1 (ss): " );
    scanf( "%d", &t1.seconds );

    printf( "Enter hours2 (hh): " );
    scanf( "%d", &t2.hours );

    printf( "Enter minutes2 (mm): " );
    scanf( "%d", &t2.minutes );

    printf( "Enter seconds2 (ss): " );
    scanf( "%d", &t2.seconds );


    struct time_dif dif = timeDif( t1, t2 );

    printf( "The difference is %d:%d:%d\n", dif.hourDif, dif.minDif, dif.secDif );

    return 0;
}

考虑到例如分钟的差异可以是负值。你应该重写函数timeDid，以便它能提供两次正确的差异。

Answer 3

您的代码中存在大量错误。我在你的代码中做了一些修改。

#include <stdio.h>

struct time1 {
    int seconds1;
    int minutes1;
    int hours1;
};

struct time2 {
    int seconds2;
    int minutes2;
    int hours2;
 };

int timeDif(struct time1 time1, struct time2 time2) { 
    int secDif = time2.seconds2 - time1.seconds1;
    int minDif = time2.minutes2 - time1.minutes1;
    int hourDif = time2.hours2 - time1.hours1;

    printf("The difference is %d:%d:%d\n", hourDif, minDif, secDif);
}

int main(void) {
    //declare the variable to use.
    struct time1 time1;
    struct time2 time2;
    printf("Enter seconds1 (ss)\n");
    scanf("%d", &time1.seconds1);

    printf("Enter minutes1 (mm)\n");
    scanf("%d", &time1.minutes1);

    printf("Enter hours1 (hh)\n");
    scanf("%d", &time1.hours1);

    printf("Enter seconds2 (ss)\n");
    scanf("%d", &time2.seconds2);

    printf("Enter minutes2 (mm)\n");
    scanf("%d", &time2.minutes2);

    printf("Enter hours2 (hh)\n");
    scanf("%d", &time2.hours2);

    timeDif(time1, time2);

    return 0;
}

Answer 4

每当您提及任何struct或class时，您都必须创建objects。您无法直接使用它们，例如scanf("%d", &time1.seconds1);。相反，首先创建一个time1对象，然后使用该对象。这很简单：

struct time1 t1;    // creates an object t1 of the type time1.
scanf("%d",&t1.seconds1);

主要原因是：当您定义struct时，实际上是在定义type，就像int或char数据类型一样。你永远不会像scanf("%d",&int);那样直接使用它们。您始终创建int类型的变量，然后进一步使用它。同样，您为struct创建objects，然后进一步使用它们。

此外，函数timeDif几乎没有问题，例如它没有返回任何值。因为看起来你的函数在它已经打开结果后需要返回任何内容，所以将它声明为void。此外，您没有理由在此函数中包含struct，因为您在同一个函数中打印结果，并且没有在任何地方使用该结构，并且此结构也容易出错。。

以下是您的代码的修改版本：http://ideone.com/uzBnVS

结构：未声明的标识符

4 个答案: