reading a character string

[cprog]

/*
lotname(void){
char lots[18]={'1', '2', '3', '4', '5', '6', '7', '8', '9', '10E', '10N', '10S', 'BB', 'CC', 'EE', 'GG', 'LL','EOB'};



char num[2];
int number, x, size;

printf("Enter parking permit number (1 through 999 inclusive):\t");
scanf("%s", &num);
number=atoi(num);
if(number <=0 || number >=1000){
printf("\nInvalid number entered\n");
lotname();
}
else
printf("it works");
printf("\"%s\"\n", lots);
size=strlen(lots);
printf("This sting is %i characters long.\n", size);
for(x=0; x<size; x++)
printf("lots[%i]='%c'\n", x, lots[x]);

}
*/

I dont know what im doing wrong ity doesn't output the character string could some1 please help me.

[Vber]

Use code-tags

Also, why you take the number as a string? dude, declare it as an integer and than take the number... you don't need to atoi() and so on... or else, if you still want this way, change the size of the buffer, it's small, put something like 10 or 20, or if you need more exactly 4-5.

[ronin]

char lots[18]={ '10E', '10N', '10S', 'BB', 'CC', 'EE', 'GG', 'LL','EOB'};

This isn't right. A character array as you're using it would be a single char. '1', '0', 'E' etc.

char lots[18] = "10E10N10SBB...";

char lots[][20] = { "1", "2", .... , "10E", "10N" ... };

char *lots[] = { "1", "2", .... "10N" };

Check your string in num[] before using it as you could overflow the array. fgets(num, sizeof(num), stdin);

if(number < 1 || number > 999) ...

printf("\"%s\"\n", lots); isn't right either. That is the format for a single string... not an array of strings.

printf("%s", lots[index]); for an array of strings
or
printf("%c", lots[index]); for an array of chars