Putting a composite statement in the condition of a for loop

Question

I have a contrived example to demonstrate the request for a specific functionality - I wonder if anyone has a clever trick to do this.

The following is a problem one

Answer 1

For both readability and performance, I think the following code can be used:

for (ii = 0;; ii++)
{
    printf("%d", ii);
    (ii < 3) ? (putchar(' ')) : ({ putchar('\n'); break; });
}

Actually, the above code is similar to your last code. It is readable, and it still has one conditional evaluation in each increment.

Answer 2

[In many ways this question should be closed as it's opinion based.]

This problem crops up often. I always opt for a solution that minimises the instructions in the iterative part.

{ /*don't pollute the outer scope with ii*/
    int ii;
    for (ii = 0; ii < 3; ++ii/*I've always preferred this to ii++*/) {
        printf("%d ", ii);
    }
    printf("%d\n", ii);
}

Ternaries, if statements etc. just obfuscate things. In my opinion.

Answer 3

int numberoftimes = 4

if numberoftimes > 1
{
  for(ii=0; ii<numberoftimes - 1; ii++) 
  {
    printf("%d ", ii);
  }
}
printf("%d\n", numberoftimes - 1);

Either you test every time in the loop, or you test once first...

Answer 4

Here's a way that requires a separate initialization pass, but that has a slightly cleaner for loop. Array sepsarray holds the separators you want (in this example, commas so you can see them). The last element of sepsarray is followed by a '\0' that terminates the loop.

#include <stdio.h>

#define MAXNUM 5
  //print up to this 

void main() {
  char sepsarray[256] = {0}; //256=arbitrary max size
  int i;
  char *seps=sepsarray;

  for(i=0;i<MAXNUM;++i) sepsarray[i]=',';   //fill in the separators
  sepsarray[i]='\n';    //the last separator

  for(i=0; printf("%d%c",i,*seps++),*seps; i++) /* pretend to do nothing */ ;
}

Output:

0,1,2,3,4,5

Hardly IOCCC material, but hey... :)

Answer 5

"Brute force" and ternary condition solution have the same complexity, but the second one is less "readable".

You can do a simple method print:

 void print() {
    int i = 0;
    for(i=0; i != size - 1; ++i) {
      printf("%i ",i);
    }
    printf("%i\n", i);
 }

I think it is efficient and readable too.
In this way you reduce cyclomatic complexity of your alghotitm.

Answer 6

I have used the following type of construct for this situation. It is not more efficient (still has a condition at each iteration), but I like it because it results in a single printf:

char *sep = " ";
for(ii=0; ii<4; ii++) {
  if ( ii == 3 )
    sep = "\n";
  printf( "%d%s", ii, sep );
}