Левый операнд должен быть левосторонним значением c как исправить

The message says that you try to assign to an expression which is not an lvalue. For built-in types, you can only assign to lvalues (that’s where the name comes from: lvalue = value that can be on the left hand side of the assignment operator, while rvalue = value that must be on the right hand side of the assignment operator).

So what is an lvalue or an rvalue? Consider the following code:

int a;
a = 3;

In this assignment a is an lvalue (if it weren’t, the compiler would complain). That is, the expression a refers to an object which can be modified. On the other hand, 3 is an rvalue, that is, basically a value. Of course you cannot assign to 3; the compiler would complain about the statement 3=a; with exactly the same message you got in your code.

So as a first approximation, an lvalue designates an object, while an rvalue designates a value. Note that this is also true for assignment of the form

a = b;

where b also is a variable. What happens here is the so-called lvalue to rvalue conversion: What is assigned is not the object b, but its current value.

Now consider the following case:

int f();
f() = 3;

Here you might argue that the function f does return an object (if you use some user-defined type, you even can see its construction/destruction). But the compiler still complains with the message you got. Why?

Well, even if you consider f to return an object, it is a temporary object which will go away immediately. So it does not make much sense to assign a value because you cannot do anything with it anyway afterwards.

Therefore here’s the second rule:

Whenever there’s an expression which produces a temporary object, C++ defines that expression as rvalue.

And now we come to the definition of MyVector::at() which you did not show, but which, according to the error message, probably looks similar to this:

template<typename T>
 T MyVector<T>::at(int i)
{
  return data[i];
}

This has essentially the same form as f above, as it also returns a T (an employee* in your case). This is why the compiler complains.

And that complaint is helpful: Even if the compiler wouldn’t complain, the code would not dio what you almost certainly intended. The return statement returns a copy of the object data[i]. Thus if the statement payment.at(i)=NULL; had compiled, what would actually happen would be the following:

1. The internal object data[i] (or however you called it in your code) is copied and the temporary copy returned.
2. The statement assigned that temporary copy, but leaves the original object in MyVector unchanged.
3. The temporary copy gets destructed, leaving no trace of your assignment.

This is almost certainly not what you wanted. You wanted to change the internal object. To do so, you have to return a reference to that object. A reference refers to the object it was initialized with instead of making a copy. Correspondingly, a reference, even when returned, is an lvalue (since C++11 there’s a second type of reference which behaves differently, but we don’t need to care about that here). Your corrected function then reads

template<typename T>
 T& MyVector<T>::at(int i)
{
  return data[i];
}

and with that definition, payment.at(i)=NULL; not only compiles, but actually does what you want: Change the internally stored i-th pointer in payment to NULL.

The message says that you try to assign to an expression which is not an lvalue. For built-in types, you can only assign to lvalues (that’s where the name comes from: lvalue = value that can be on the left hand side of the assignment operator, while rvalue = value that must be on the right hand side of the assignment operator).

So what is an lvalue or an rvalue? Consider the following code:

int a;
a = 3;

In this assignment a is an lvalue (if it weren’t, the compiler would complain). That is, the expression a refers to an object which can be modified. On the other hand, 3 is an rvalue, that is, basically a value. Of course you cannot assign to 3; the compiler would complain about the statement 3=a; with exactly the same message you got in your code.

So as a first approximation, an lvalue designates an object, while an rvalue designates a value. Note that this is also true for assignment of the form

a = b;

where b also is a variable. What happens here is the so-called lvalue to rvalue conversion: What is assigned is not the object b, but its current value.

Now consider the following case:

int f();
f() = 3;

Here you might argue that the function f does return an object (if you use some user-defined type, you even can see its construction/destruction). But the compiler still complains with the message you got. Why?

Well, even if you consider f to return an object, it is a temporary object which will go away immediately. So it does not make much sense to assign a value because you cannot do anything with it anyway afterwards.

Therefore here’s the second rule:

Whenever there’s an expression which produces a temporary object, C++ defines that expression as rvalue.

And now we come to the definition of MyVector::at() which you did not show, but which, according to the error message, probably looks similar to this:

template<typename T>
 T MyVector<T>::at(int i)
{
  return data[i];
}

This has essentially the same form as f above, as it also returns a T (an employee* in your case). This is why the compiler complains.

And that complaint is helpful: Even if the compiler wouldn’t complain, the code would not dio what you almost certainly intended. The return statement returns a copy of the object data[i]. Thus if the statement payment.at(i)=NULL; had compiled, what would actually happen would be the following:

1. The internal object data[i] (or however you called it in your code) is copied and the temporary copy returned.
2. The statement assigned that temporary copy, but leaves the original object in MyVector unchanged.
3. The temporary copy gets destructed, leaving no trace of your assignment.

This is almost certainly not what you wanted. You wanted to change the internal object. To do so, you have to return a reference to that object. A reference refers to the object it was initialized with instead of making a copy. Correspondingly, a reference, even when returned, is an lvalue (since C++11 there’s a second type of reference which behaves differently, but we don’t need to care about that here). Your corrected function then reads

template<typename T>
 T& MyVector<T>::at(int i)
{
  return data[i];
}

and with that definition, payment.at(i)=NULL; not only compiles, but actually does what you want: Change the internally stored i-th pointer in payment to NULL.

The message says that you try to assign to an expression which is not an lvalue. For built-in types, you can only assign to lvalues (that’s where the name comes from: lvalue = value that can be on the left hand side of the assignment operator, while rvalue = value that must be on the right hand side of the assignment operator).

So what is an lvalue or an rvalue? Consider the following code:

int a;
a = 3;

In this assignment a is an lvalue (if it weren’t, the compiler would complain). That is, the expression a refers to an object which can be modified. On the other hand, 3 is an rvalue, that is, basically a value. Of course you cannot assign to 3; the compiler would complain about the statement 3=a; with exactly the same message you got in your code.

So as a first approximation, an lvalue designates an object, while an rvalue designates a value. Note that this is also true for assignment of the form

a = b;

where b also is a variable. What happens here is the so-called lvalue to rvalue conversion: What is assigned is not the object b, but its current value.

Now consider the following case:

int f();
f() = 3;

Here you might argue that the function f does return an object (if you use some user-defined type, you even can see its construction/destruction). But the compiler still complains with the message you got. Why?

Well, even if you consider f to return an object, it is a temporary object which will go away immediately. So it does not make much sense to assign a value because you cannot do anything with it anyway afterwards.

Therefore here’s the second rule:

Whenever there’s an expression which produces a temporary object, C++ defines that expression as rvalue.

And now we come to the definition of MyVector::at() which you did not show, but which, according to the error message, probably looks similar to this:

template<typename T>
 T MyVector<T>::at(int i)
{
  return data[i];
}

This has essentially the same form as f above, as it also returns a T (an employee* in your case). This is why the compiler complains.

And that complaint is helpful: Even if the compiler wouldn’t complain, the code would not dio what you almost certainly intended. The return statement returns a copy of the object data[i]. Thus if the statement payment.at(i)=NULL; had compiled, what would actually happen would be the following:

1. The internal object data[i] (or however you called it in your code) is copied and the temporary copy returned.
2. The statement assigned that temporary copy, but leaves the original object in MyVector unchanged.
3. The temporary copy gets destructed, leaving no trace of your assignment.

This is almost certainly not what you wanted. You wanted to change the internal object. To do so, you have to return a reference to that object. A reference refers to the object it was initialized with instead of making a copy. Correspondingly, a reference, even when returned, is an lvalue (since C++11 there’s a second type of reference which behaves differently, but we don’t need to care about that here). Your corrected function then reads

template<typename T>
 T& MyVector<T>::at(int i)
{
  return data[i];
}

and with that definition, payment.at(i)=NULL; not only compiles, but actually does what you want: Change the internally stored i-th pointer in payment to NULL.

#include<string>
#include<iostream>
using namespace std;

int main()
{

bool flag = false;
int month, day, year;
void header();

class monthClassException
{
public:
monthClassException()
{
message = "Invalid Month";
}
monthClassException(string str)
{
message = str;
}
string what()
{
return message;
}

private:
string message;
};
class dayClassException
{
};
class yearClassException
{
};header();

do
{
try
{
cout << "Please enter your date of birth (MM-DD-YYYY): " << endl;
cin >> month;

cin.ignore(10,'-');
cin >> day;

cin.ignore(10,'-');
cin >> year;if (month > 12 || month < 1)
throw monthClassException("Invalid Month Entry");

if( ((month == 4 || month == 6 || month == 9 || month == 11) && day > 30) || day < 1)
throw dayClassException();

else if ( ((month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12 ) && day > 31) || day < 1)
throw dayClassException();

else if (month == 2 && year % 4 != 0 && day > 28)
throw dayClassException();

else if((month == 2 && year % 4 = 0) && day > 29)
throw dayClassException();
}
catch(monthClassException mCEO)
{
cout << mCEO.what() << endl;
system("pause");
}
catch(dayClassException)
{
cout << "Invalid Day Entered for Selected Month" << endl;
system("pause");
}
catch(yearClassException yCEO)
{
}
}while(!flag);return 0;
}

            else if((month == 2 && year % 4 = 0) && day > 29)
throw dayClassException();

Решение

   year % 4 = 0

Другие решения

year % 4 = 0

year % 4 == 0

...
else if ((month == 2) && (year % 4 == 0) && (day > 29)) {
throw dayClassException();
}

    if year == 0

    if year = 0

     if 0 = year

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

struct autoservis      
{
    int num;
    char marka[50];
    int probeg;
    struct autoservis *next;
};
 
typedef struct autoservis St;
St *last = NULL;
 
int i, k, n, pr;
char m[50];

void vvod()
{
    St *p;
    p = new St;
 
    printf("Vvedite kol-vo avtomobilej: ");
    scanf_s("%d", &k);
 
    if (p != 0)
    {
        for (i = 0; i < k; i++)
        {
            puts("Vvedite registrazionniy nomer: ");
            scanf_s("%d", &n);
            p->num = n;
 
            puts("Vvedite marku: ");
            fflush(stdin);
            gets_s(m);
            p->marka = m;
 
            printf("Vvedite probeg: ");
            scanf_s("%d", &pr);
            p->probeg = pr;
 
            p->next = last;
            last = p;
        }
    }
    else printf("Oshibka pri vidilenii pamjatin");
}

void poisk()   
{
    St *psk = last;
    int nomer;
 
    if (psk == NULL){
        printf("Net avtomobilejn");
    }
    else
    {
        printf("Vvedite reg. nomer: n");
        scanf_s("%d", &nomer);
 
        while (psk != NULL)
        if (psk->num == nomer)
        {
            printf("Registrazionniy nomer: %d n", psk->num);
            printf("Marka: %s n", psk->marka);
            printf("Probeg: %d n", psk->probeg);
        }
        else
            psk = psk->next;
    }
    
}

void udalenie()  
{
    St *po;
    po = new St;
 
    po = last;
    last = last->next;
    free(po);
    printf("Udalenije vipolnenon");
}

void vivod()  
{
    St *o = last;
 
    if (o == NULL) printf("Stek pustn");
    else
    {
        printf("Stek: n");
        while (o != NULL)
        {
            printf("Reg. nomer: %d, Marka: %s, Probeg: %dn", o->num, o->marka, o->probeg);
            o = o->next;
        }
        printf("n");       
    }
}

int check()
{
    return (last == NULL);
}

int menu()  
{
    int ch;
    printf("n   Menu: n");
    printf("1. Vvod massiva structur n");
    printf("2. Poisk v massive structyr po zadannomy parametry n");
    printf("3. Ydalenie structuri iz massiva n");
    printf("4. Vivod structuri n");
    printf("5. Vikhodn");
    printf("t Vash vibor: ");
    scanf_s("%d", &ch);
    return ch;
}
 
int main()
{
    while (1)
    {
        switch (menu())
        {
        case 1:
            vvod();
            break;
        case 2:
            poisk();
            break;
        case 3:
            if (!(check())){
                udalenie();
                break;
            }
            else{
                printf("Stek pustn");
                break;
            }   
        case 4:
            vivod();
            break;
        case 5:
            return 0;
        default:
            printf("n");
            break;
        }
    }
}