Suppose the user types:

Michael

in response to the first prompt. The %c format reads the M; the %s format reads ichael without bothering to get any new data.

Also, you should not be passing &firstname or &lastname; you should be passing just firstname and lastname to scanf(). The difference is in the type; with the ampersand, you're passing a char (*)[256] which is not the same as the char * that scanf() expects. You get away with it, but 'get away with it' is the operative term.

Use a %s format (or, better, %255s format) for the two scanf() calls. Then pass firstname[0] and lastname to printf(). You might want to think about using tolower() from <ctype.h> on the first letter, and maybe on the last name too.

This is a reasonable approximation to a good program:

#include <stdio.h>

int main(void)
{
    char firstname[256];
    char lastname[256];
    printf("What's your first name? ");
    if (scanf("%255s", firstname) != 1)
        return 1;
    printf("What's your last  name? ");
    if (scanf("%255s", lastname) != 1)
        return 1;
    printf("Your school.edu e-mail address is: %c%[email protected]\n",
           firstname[0], lastname);
    return 0;
}

It avoids quite a lot of problems, one way or another. It is not completely foolproof, but most people won't run into problems with it.

char source[] = "abcdefthijklmn";
char target[100];

strncpy(target, source, 10);
target[10] = '\0'; // IMPORTANT!

Given what you've provided in the question and from interpreting your meaning I can only assume that argv is a two dimensional array of either int or char.

string a = argv[1][0]; // so this has 1
string b = argv[1][1]; // this has 2 in it

You want c to be 12.

    int[,] t = new int[2, 6] {{1,2,3,4,5,6},
                              {1,2,2,0,8,5}};

    string a = Convert.ToString(t[1,0]);
    string b = Convert.ToString(t[1,1]);

    string c = a+b; // "12"

from here you can convert it to a number if you need it as 12.

Regarding char *temp6=NULL;

Here you're trying to tell the compiler, "Hey ! set up temp6 as a pointer to char but don't allocate memory for it. "

If you do something like strcpy(temp,str); later you will get segmentation fault because you are trying to write to memory you don't own.

In your case you didn't go so far to see the segmentation fault, the compiler caught another error which is mentioned by the other answerer, ie in the line :

strcpy(temp6,str[0]);

where the compiler expected second argument to be char* but you passed char.

You have to allocate memory to the pointer first, or point it to an array.
It is also possible to de-allocate the memory you have allocated for a pointer.

 char* temp= malloc(sizeof(char) * 10) ;  // allocating memory
 temp='\0'; // In essence de-allocating the memory.

Below is a complete example.

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

 main()
 {
  char* temp= malloc(sizeof(char) * 10) ; //allocating memory enough to store 10 chars
  char* str="abcdefgh"; // play safely - always store less than 10 characters. Consider that \0 will be appended to the end.
  strcpy(temp,str);
  printf("Temp : %s\n",temp);
  char* str1="ijklmnop";
  strcpy(temp,str);
  printf("Temp : %s\n",temp);
  temp='\0'; // In essence deallocating the memory.
  printf("Temp : %s\n",temp);
  strcpy(temp,str);
  printf("Temp : %s\n",temp);
 }

will give you the below result.

Temp : abcdefgh
Temp : abcdefgh
Temp : (null)
Segmentation fault (core dumped)

Also make sure that free(temp6) is placed at the end of your code to clean up the memory. Though this doesn't directly answer your question, hope it will be useful.