What is wrong with the printf statement in this code:

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This code came from Zyooks I was to add a printf statement to get this answer: xyz xzy yxz yzx zxy zyx I'm a newbie trying to understand char and integer literals.

#include <stdio.h>
int main(void) {
   char a;
   char b;
   char c;
   scanf(" %c ", &a);
   scanf(" %c ", &b);
   scanf(" %c ", &c);

//this is the line I added to get this output    xyz  xzy   yxz  yzx   zxy  zyx  
  printf("%c %c %c %c %c %c," abc, acb, bac, bca, cab, cba);  
						       
printf("\n");
return 0;
}

What I have tried:

This is C code. I have tried several changes, but none work.

Posted 18-Sep-19 13:00pm

Member 14597159

Updated 18-Sep-19 14:03pm

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2 solutions

Solution 2

Quote:
What is wrong with the printf statement in this code

Many things, S1 code is how you should do.

C++

printf("%c %c %c %c %c %c," abc, acb, bac, bca, cab, cba);
//      ^ this is for 1 char only, but you want 3 chars at a time
//                       ^ the comma, here, should be after double quptes
//                          ^ here is a new undefined variable, not a combination of 3 variables

Posted 18-Sep-19 14:03pm

Patrice T

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honey the codewitch · Accepted Answer · 2019-09-18T13:16:00

Solution 1

you can't concat variables like that

the compiler sees abc as an undefined variable.

what you need is like.

C++

printf("%c%c%c ",a,b,c);
printf("%c%c%c ",b,c,a);
printf("%c%c%c ",c,a,b);

etc

Posted 18-Sep-19 13:16pm

honey the codewitch

Updated 18-Sep-19 13:18pm

v3

Comments

Member 14597159 19-Sep-19 14:37pm

Thanks. You solved this 84 year old Newbie's problem.