CODEWITHSUNDEEP
c
user1304516
user1304516

Reputation: 13

Using an array to decode numbers to letters

For my task I have to print numbers to the screen and decode them into their specific letters. I'm using only the letters a-l in this code just to keep it simple so I can understand it.

The problem I'm having is that when I, for example, put in the number 0 which corresponds to the first entry to the array which is a, it will take out a and print b-l.

How do I make it so if I put in the number 0, the code will print only a to the screen?

#include <stdio.h>

int main()
{
char code[] = "abcdefghijkl";
int i, j, k;
printf("how many letters does your code contain?: ");
scanf("%d", &j);
for(i=0; i<j; ++i){
    printf("enter a number between 0 and 11\n");
    scanf("%d", &k);
    printf("%s\n", &code[k]);
}
}

Upvotes: 0

Views: 611

Answers (3)

Nagaraju Badaeni
Nagaraju Badaeni

Reputation: 900

printf("%c\n", code[k]); instead of printf("%s\n", &code[k]);

Upvotes: 0

Naveen
Naveen

Reputation: 73443

%s format specifier is used for printing strings, you need to use %c specifier which prints a character to the screen.

Upvotes: 0

nos
nos

Reputation: 229088

You print only the character at that location, so change 

printf("%s\n", &code[k]);

to 

printf("%c\n", code[k]);

You should also check that the value you read into k is >= 0 && < 11 , otherwise you'll access the array outside its bounds.

Upvotes: 4

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