Home > OS >  Error in assignment to expression with array type
Error in assignment to expression with array type

Time:10-26

I am trying to code a program that shows the Easter day when the user inputs the year between 1991 and 2099. All are good but when I am trying to run the program, it says there is an error: assignment to expression with array type. How can I fix this? and what is the wrong with my code?

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

int main()
{
char month[50];
int year, day;
printf("Please enter a year between 1990 and 2099: ");
scanf("%d", &year);


int a = year-1900;
int b = a%19;
int c = (7*b 1)/19;
int d = (11*b 4-c)%29;
int e = a/4;
int f = (a e 31-d)%7;
int g = 25 - (d   f);

if (g <= 0 )
{
    month = "March";
    day = 31   g;
}
else
{
    month = "April";
    day = g;
}
    printf("The date is %s %d", month, day);

}

CodePudding user response:

Arrays do not have the assignment operator. Arrays are non-modifiable lvalues.

To change the content of an array use for example the standard string function strcpy.

#include <string.h >

//...

strcpy( month, "March" );

An alternative approach is to declare the variable month as a pointer. For example

char *month;

or (that is better)

const char *month;

In this case you may write

month = "March";
  • Related