How to concatenate a char array case [0] and [1] to a char pointer in C?

Question

I want to concatenate two characters '7' and '9' to form this string "79".

First, I initialized the variables. (Restriction: I have to use char types only and I must not make another charac alike variable.)

char *charac = (char *) malloc(sizeof(char) * 200);
char *combined = (char *) malloc(sizeof(char) * 200);

Then I gave the values for charac[0] and charac[1].

charac[0] = '7';
charac[1] = '9';
printf("charac[0] : %c\n", charac[0] );
printf("charac[1] : %c\n", charac[1] );

I want to concatenate charac[0] and charac[1]

strcpy(combined, (char*)charac[0]);
strcat(combined, (char*)charac[1]);
printf("combined : %s\n", combined);
free(combined);

But the code above doesn't work. I got the error message: "warning: cast to pointer from integer of different size".

After reading all of your comments and suggestions, I obtain the output result I want.

Here is the final code:

char *charac = (char *) malloc(sizeof(char) * 200);
        char *combined = (char *) malloc(sizeof(char) * 200);

        charac[0] = '7';
        charac[1] = '9';
        charac[2] = '\0';
        printf("charac[0] : %c\n", charac[0] );
        printf("charac[1] : %c\n", charac[1] );
        printf("charac[2] : %c\n", charac[2] );

        strcpy(combined, & charac[0]);
        strcat(combined, & charac[1]);
        strcpy(combined, & charac[0]);
        strcat(combined, & charac[2]);
        printf("combined : %s\n", combined);
        free(combined);

Answer 1

They are already concatenated as is. This is because they are right next to each other in memory. You have declared them in the same array, and at adjacent indexes.

+---------------+
|  '7'  |  '9'  |
+---------------+
   [0]     [1]

This is your array in memory. The seven (charac[0]) is right next to the nine (charac[1]). As suggested in the comments, simply null terminate the array to complete the concatenation.

charac[2] = '\0';

The array is now printf() friendly as it is null terminated.

Answer 2

A string in C consists of the individual characters followed by a null (zero byte). So the string "79" is of length 3.

Strings are placed into character array storage. You can allocate it on the heap, but just declaring an array is simpler. So the code would be:

char charac[3];
charac[0] = '7';
charac[1] = '9';
charac[2] = '\0';  // or just 0

Even simpler:

char charac[3] = { '7', '9' };   // the 0 just happens

Even simpler:

char charac[] = "79";

Each of these three pieces of code does exactly the same thing.

---

In response to a comment, the treatment of initialised aggregates in the C standard is defined as follows: n1570 S6.7.9/21:

If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.

In other words, any time you use brace initialisation the remaining items are zero filled.