指针：表达式不可赋值

Question

我正在编写一个函数，其目的是更改结构变量的值。我收到错误消息，指出第一个函数末尾的表达式不可赋值。那么，如何让函数更改 daate 结构的值？

这是我的代码：

#include <stdio.h>
#include <stdbool.h>

struct date
{
    int month;
    int day;
    int year;
};

struct date dateUpdate (struct date *dait) {
    struct date tomorrow;
    int numberOfDays (struct date d);

    if (dait->day != numberOfDays (*dait) ) {
        tomorrow.day = dait->day + 1;
        tomorrow.month = dait->month;
        tomorrow.year = dait->year;
    }
    else if ( dait->month == 12 ) { //end of year
        tomorrow.day = 1;
        tomorrow.month = 1;
        tomorrow.year = dait->year + 1;
    }
    else { //end of month
        tomorrow.day = 1;
        tomorrow.month = dait->month + 1;
        tomorrow.year = dait->year;
    }

    &dait->day = tomorrow.day;
    &dait->month = tomorrow.month;
    &dait->year = tomorrow.year;

    return *dait;
}

//Function to find the number of days in a month;

int numberOfDays (struct date d)
{
    int days;
    bool isLeapYear (struct date d);
    const int daysPerMonth[12] = 
    { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

    if ( isLeapYear (d) && d.month == 2 )
        days = 29;
    else
        days = daysPerMonth[d.month - 1];

    return days;
}

}

int main (void) {
    struct date daate;
    daate.day = 21;
    daate.month = 6;
    daate.year = 2013;

    dateUpdate(&daate);
    printf("%d", daate.day);
}

Answer 1

正确的语法（您已经在函数中使用过）如下

dait->day = tomorrow.day;
dait->month = tomorrow.month;
dait->year = tomorrow.year;

请注意，您可以简单地编写

而不是这三个语句

*dait = tomorrow;

Answer 2

dait 已经是指针，所以 &dait 是指向指针的指针，因此它没有要取消引用的字段。

解决办法：把dait前的&去掉。

指针：表达式不可赋值

Pointers: Expression is not assignable

c

function

pointers

structure