I got a question one of my friends asked me to make a program that you type in the number 1-27 on the command line, and get the corresponding letter for it. Like if you enter 1 as the arguement you would get an A as an output, and if you type in 2, and get a B for the output and so on.

I got a question one of my friends asked me to make a program that you type in the number 1-27 on the command line, and get the corresponding letter for it. Like if you enter 1 as the arguement you would get an A as an output, and if you type in 2, and get a B for the output and so on.

It would help if you supplied your code to see where you are having problems.

I don't have any code for it, he was just wanted me to do some code for him, sense I am out fo work for a couple of weeks, with a broke leg, and have nothing to do, but I still do work for job at home, and don't have the time to do this for him.

so tell him you don't have the time to do his homework for him, just like we don't have the time either...

I think this code will help you out

# include<stdio.h>
# include<conio.h>
char change(int); /* Function prototype */
void main()
{   int num;
    clrscr();
    printf("Enter any number : ");
    scanf("%d",&num); /* getting the no from user */
    printf("%c",change(num));
    getch();

}

char change(int n)
{   char k;
    k = (char)n;
    k = k+64;
    return k;
}

for command line arguement

# include<stdio.h>
# include<conio.h>
# include<stdlib.h>
char change(int); /* Function prototype */
int atoi(char);
void main(int count,char *argv[])
{    if(count != 2)
      {  puts("INVALID COMMAND");
          exit(0);
      }
      int num;
     num = atoi(argv[1]);
      printf("%c",change(num));
     getch();
}

char change(int n)
{   char k;
     k = (char)n;
     k = k+64;
    return k;
}
int atoi(char c)
{  return c - '0';
}
Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.