C Programming, assigning strings to char

Thanks infinity08 for your quick response... sorry, I am really a novice programming.  when I changed my "char mon[16]" , and then run gcc to compile my program, I get the following messages.

test1.c: In function ‘main’:
test1.c:57: error: incompatible types when assigning to type ‘char[16]’ from type ‘char *’

did i do it wrong?

If you don't mind, let me share my codes with you... it's an assignment of converting the 3 number arguments (year month day), and then outputs them in american date format.  I had worked on it for a day...  although i know the logic and programming flow is very beginner like ...

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

#define PROGRAM_NAME            "printdate"
#define CORRECT_INPUT_COUNT     4


void main(int argc, char *argv[])
{

        int y;
        int m;
        int d;
        char mon,

/* Check # of input arugments*/

        if(argc != CORRECT_INPUT_COUNT)
        {
                printf("Usage: printdate year month day\n");
                exit(-1);
        }
/* Check year, month, day arguments*/

        if(sscanf(argv[1],"%d", &y)<1)
        {
                printf("Error: year must be a number.\n");
                exit(-2);
        }

        else if(sscanf(argv[2],"%d", &m)<1)
        {
                printf("Error: month must be a number.\n");
                exit(-3);
        }

        else if(sscanf(argv[3],"%d", &d)<1)
        {
                printf("Error:  day must be a number.\n");
                exit(-4);
        }
 else

        if(sscanf(argv[2],"%d", &m)==1)
        {
                mon = "Jan";
        }


                printf("%d-",m );
                printf("%d-",d, ",");
                printf("%d\n",y);

        
                printf("%d-",m );
                printf("%d-",d, ",");
                printf("%d\n",y);

/* Checks for leapyear */     
/*      if(y % 400 == 0)
        {
                printf("Leap Year.\n");
        }
        else if(y % 100 == 0)
        {
                printf("Not Leap Year.\n");
        }
  else if(y % 4 == 0)
        {
                printf("Leap Year.\n");
        }
        else
                printf("Not Leap Year.\n");
*/

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"January" is an array of characters and C compiler holds it as a pointer but mon is a character.
In this assinment Compiler tries to convert a pointer to a character so it shows a warning.
you can use below code:

   

char mon[100] = "September";
    ....
    if (m == 1)
    {
        strcpy (mon, "January");
    }

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>> test1.c:57: error: incompatible types when assigning to type ‘char[16]’ from type ‘char *’

String assignment doesn't exist in C. You cannot assign a string to another.

What you can do, is assign a pointer to a string to another pointer to a string. Like in the example code I posted in my previous post.


>> 45:            if(sscanf(argv[2],"%d", &m)==1)

sscanf returns the amount of succesfully read tokens. So, if it returns 1 in this case, it means the sscanf succeeded (because it needed to read one %d token, and it did so). So, it's good to check for success in this way.

However, you will still have to check the actual value in m to know which month it is.


In any case :

>> 15:            char mon,

if you change this to be a const char* instead of a char (see the code snippet in my earlier post), it should work better.

Note also that there's a comma at the end of that line, while it should be a semicolon.

Thanks all for your input and comments.  especially for the breakdown descriptions.   Actually, i am still unfamiliar with pointers, char syntax.... but nevertheness, with infinity08's code, I did get pass with matching the mon(integer) to text for now... let me keep on writing it.  will return back shortly.  thanks.

Thanks again for the input.  Storing the words in pointer solved the problem!  =)