errors in fizzbuzz in C using String approach?

Question

So I was trying to solve the FizzBuzz challenge in various languages using the string method to improve the code. I got stuck in C because things work differently here. This is my code, I'm getting errors, can anyone explain them to me and help to get the correct code.

#include<stdio.h>
#include<string.h>
int main()
{
    int i,n;
    char output;
    printf("Enter Range: ");
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        if(i%3==0)
            strcat(output,"Fizz");
        if(i%5==0)
            strcat(output,"Buzz");
        if(output=="\0")
            strcat(output,i);

        printf("\ni");
    }
    printf("\nEnd.\n");
    return 0;
}

Thanks.

Answer 1

@Ashwini Singh

There are some mistakes in your code ,

1) You declared the output variable as char datatype & concatenate string in Fizz/Buzz in char . so how can be value of string(which is a array of character) placed in character output .

2) You are concatenate the integer value i with character output eg strcat(output,i) . we need to first typecast the integer value i in to char/string datatype & then concatenate with output .

The condition of the FizzBuzz programs are,

1) If number is multiple of 3 then add Fizz in the resultant string

2) If number is multiple of 5 then add Buzz in the resultant string

3) If number is neither multiple of 3 nor multiple of 5 then add number in the resultant string

Code :

#include<stdio.h>
#include<string.h>
int main()
{
int i,n;
char output[100]=" ";
printf("Enter Range: ");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
    char s[] = {'0' + i, '\0'};
    if(i%3==0)
        strcat(output,"Fizz ");
    else if(i%5==0)
        strcat(output,"Buzz ");
    else
        strcat(strcat(output,s)," ");
}
puts(output);
return 0;
}

Output :

Enter Range: 10                                                                  
 1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz