Print elements of array between other output in C

Question

I am trying to print results of my sudoku solving program into terminal like this:

  +-------+-------+-------+
  | 1 2 3 | 4 5 6 | 7 8 9 |
  | 1 2 3 | 4 5 6 | 7 8 9 |
  | 1 2 3 | 4 5 6 | 7 8 9 |
  +-------+-------+-------+
  | 1 2 3 | 4 5 6 | 7 8 9 |
  | 1 2 3 | 4 5 6 | 7 8 9 |
  | 1 2 3 | 4 5 6 | 7 8 9 |
  +-------+-------+-------+
  | 1 2 3 | 4 5 6 | 7 8 9 |
  | 1 2 3 | 4 5 6 | 7 8 9 |
  | 1 2 3 | 4 5 6 | 7 8 9 |
  +-------+-------+-------+

I store my solution in one dimensional array and I cannot find a way to print it. This is what I came up with so far:

printf("| %c %c %c | %c %c %c | %c %c %c |\n", test[0],test[1],test[2],test[3],test[4],test[5],test[6],test[7],test[8]);

I cannot use any cycles because I need to draw the "walls" around the numbers. Is there any better way to do it? And why does

char test[] = {'1','2','3','4','5','6','7','8','9'};
int i = 0;
printf("| %c %c %c | %c %c %c | %c %c %c |\n",   test[i++],test[i++],test[i++],test[i++],test[i++],test[i++],test[i++],test[i++],test[i++]);

returns | 9 8 7 | 6 5 4 | 3 2 1 | Thanks.

is it a multidimensional array? which part you feel it's difficult for you? — Jason Hu, Aug 29 '15 at 14:59
i can tell you are using `gcc` by reading the evaluation order. why do you say you can't use a loop? think it twice. that's not true. — Jason Hu, Aug 29 '15 at 15:01
It is one dimensional array. I don't know how can I use loops since I print three numbers and then the | character. And there are the +-----+-----+ lines. I can print the picture above with hard coded numbers but I don't know how to put numbers from my array there — Milda, Aug 29 '15 at 15:07
replace this kind of line: `printf("| %c %c %c | %c %c %c | %c %c %c |\n", test[i++],test[i++],test[i++],test[i++],test[i++],test[i++],test[i++],test[i++],test[i++]);` with `printf("| %c %c %c | %c %c %c | %c %c %c |\n", test[i],test[i+1],test[i+2],test[i+3],test[i+4],test[i+5],test[i+6],test[i+7],test[i+8]); i+=9;` — user3629249, Aug 30 '15 at 15:14

melpomene · Accepted Answer · 2015-08-29T15:20:33.250

5

Here's one way to decompose this into (nested) loops:

void print_data_row(const char *p) {
    printf("|");
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            printf(" %c", p[i*3 + j]);
        }
        printf(" |");
    }
    printf("\n");
}

void print_separator(void) {
    printf("+-------+-------+-------+\n");
}

...
print_separator();
for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
        print_data_row(&board[(i*3 + j) * 9]);
    }
    print_separator();
}

edited Aug 29 '15 at 15:20

answered Aug 29 '15 at 15:17

melpomene

84,125
8
85
148

Thanks, this is exactly what I was looking for. I tried a similar approach but couldn't get the incrementation right and when I did, it seemed too complicated so I just tried the non-loop version. Does that mean that there isn't any simpler way to do this in C? Something like printf("%c %c %c", *array of elements to print*); ? – Milda Aug 29 '15 at 15:24
@Milda You can't automatically "expand" an array to a list of function arguments. You can of course pass each element separately: `printf("%c %c %c", array[i], array[i + 1], array[i + 2]);` – melpomene Aug 29 '15 at 15:29
That was basically why I asked this question. I fixed my nested-loops version (forgot to add j in one loop) and it works perfectly now. Thanks – Milda Aug 29 '15 at 15:39

score 1 · Answer 2 · edited May 23 '17 at 11:52

1

Q. Is there any better way to do it?
A. Try using loops to achieve the same.

For your second question, the answer is that the order of evaluation of function parameters is not certain in C. Hence, you never know, which of those nine test[i++] gets evaluated first. Hence, the unknown result (undefined behavior actually). More here.

edited May 23 '17 at 11:52

Community

1
1

answered Aug 29 '15 at 14:59

CinCout

9,486
12
49
67

4

It's not just an unknown result. Modifying the same variable twice (without a sequence point in between) has undefined behavior. – melpomene Aug 29 '15 at 15:07

Jens · Answer 3 · 2015-08-30T15:39:24.203

No loops? Ok. I'd be inclined to say this is even a superior solution compared to a single or nested loop. It can be seen to do what it is supposed to do.

char *t = test;
printf("+-------+-------+-------+\n");
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("+-------+-------+-------+\n");
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("+-------+-------+-------+\n");
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
t += 9;
printf("| %c %c %c | %c %c %c | %c %c %c |\n", t[0],t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8]);
printf("+-------+-------+-------+\n");

Thanks, this is something I ended up with.I was just looking for something like printf("%c %c %c", *array of elements to print*) so I wouldn't have to use nested loops since it seemed too complicated (and therefore wrong) to me. — Milda, Aug 29 '15 at 15:30

score 1 · Answer 4 · answered Aug 29 '15 at 15:36

Try this:

for(int i = 0; i < 13; i++){ // 13 because 9 numbers + 4 separators
  for(int j = 0; i < 13; i++){
    if((i % 4 == 0) && (j % 4 == 0)){
      printf("+"); // separator crossing
    } else if (i % 4 == 0){
      printf("-"); // vertical separator
    } else if (j % 4 == 0){
      printf("|"); // horizontal separator
    } else {
      printf("%d",sudoku[row(i)][row(j)]); // sudoku is the 2d result array
    }
  printf("\n"); // end of line
}

int row(j){ // row (or column) without counting separators
  int nOfSeparator = (i - (i % 4))/4;
  return (13 - nOfSeparator);
}

Print elements of array between other output in C

4 Answers4