Getting a function to return an array

[abasel]

I am trying to get a function to return an array but I can not get it right; an empty array appears to be returned as nothing is displayed after "Light1"

#include <stdio.h>
int * getTimings (int n)
{
  int *lghts=malloc(3*sizeof(int));

  lghts[0] = n;
  if (n < 60)
    {
      lghts[1] = n + 10;
      lghts[2] = n + 20;
    }
  else
    {
      lghts[1] = n + 60;
      lghts[2] = n + 120;
    }
   
  return lghts;
}

int main ()
{
   int *lights=malloc(3*sizeof(int));

  lights = getTimings(90);
    
  printf("\nLight1: %d",lights[0]
    
  return 0;
}

[laserlight]

A few notes:

• You forgot to #include <stdlib.h> for malloc.
• You forgot to free what you malloc
• You used malloc in main, but that results in a memory leak because you then overwrite the pointer to the memory with the pointer returned by getTimings.
• Although your use of malloc is correct, it would be better to write:
  

  int *lghts = malloc(3 * sizeof(*lghts));

  as this way the result of sizeof would be correct even if the type of lghts was changed.
• You seem to have made a small typo error at the end of the printf call.

Fixing all these, although with a small tweak to the output to make it more likely to be easily readable, I compiled and ran:

#include <stdio.h>
#include <stdlib.h>

int * getTimings(int n)
{
    int *lghts = malloc(3 * sizeof(int));

    lghts[0] = n;
    if (n < 60)
    {
        lghts[1] = n + 10;
        lghts[2] = n + 20;
    }
    else
    {
        lghts[1] = n + 60;
        lghts[2] = n + 120;
    }

    return lghts;
}
 
int main(void)
{
    int *lights = getTimings(90);

    printf("Light1: %d\n", lights[0]);

    free(lights);

    return 0;
}

and received the output:

Light1: 90

which was expected.

Oh, and by right you should check the return value of malloc before using it, although in this case it is unlikely to matter.

[abasel]

Thanks, that all makes sense. When I however try and implement it into my complete code I still can not get it to work. I might be introducing a typo which I cannot see. Or could it be the online compiler that I am using, GDB online Debugger | Compiler - Code, Compile, Run, Debug online C, C++

#include <stdio.h>
#include <stdlib.h>

int getMinutes (float n)
{
  return trunc (n);
}

int getSeconds (float n)
{
  float integer;
  int num;
  num = round (modf (n, &integer) * 100);
  return num;

}

int * getTimings(int n)
{
    int *lghts = malloc(3 * sizeof(lghts));

    lghts[0] = n;
    if (n < 60)
    {
        lghts[1] = n + 10;
        lghts[2] = n + 20;
    }
    else
    {
        lghts[1] = n + 60;
        lghts[2] = n + 120;
    }

    return lghts;
}

int
main ()
{
  float number;
  int min, sec, totalseconds, x;
 
  printf ("Enter a real number: ");

  scanf ("%f", &number);

  min = getMinutes (number);

  sec = getSeconds (number);

  // printf() displays the formatted output
  printf ("You entered: %.2f", number);
  printf ("\nMinutes: %d", min);
  printf ("\nSeconds: %d", sec);

  totalseconds = min * 60 + sec;

  int *lights = getTimings(totalseconds);
    
  printf ("\nTotal seconds: %d", totalseconds, "\n");
  printf("\n",lights[0]);

  for (x = 0; x < 3; x += 1)
    {
      printf ("\nLight %d", x + 1, ": ", lights[x]);
    }
    
  free(lights);
    
  return 0;
}

Output is

Enter a real number: 1
You entered: 1.00
Minutes: 1
Seconds: 0
Total seconds: 60

Light 1
Light 2
Light 3

[abasel]

I can not use

printf ("\nLight %d", x + 1, ": ",lights[x]);

Should be

printf ("\nLight %d %s %d", x + 1,": ",lights[x]);

Thanks again

[whiteflags]

Yeah, the normal set of arguments is printf(FormatString, argument1, argument2, ... argumentN);

printf("Light %d: %d\n", x + 1, lights[x]);