Storing input into a character array

[Bert]

The more uniform the data structure,
the simpler the program usually is.

What's a data structure and how (have I created)/(am I using) one?

 

[Ben Bacarisse]

What's a data structure and how (have I created)/(am I using) one?

Loosely speaking, a data structure is a collection of data will some
pattern of access. You have made three arrays: one for the top row,
one 2D arrays for the middles rows, and one for the bottom row. Now,
as it happens, you don't need any of that storage but, if you did, I
am saying that you program is likely to be simpler if you just stored
all the rows together in one 2D array. It was an incidental remark.
I think you'll get the best boost from going another way altogether.

 

[Bert]

I will post my solution soon

Have you done it yet? I'm just practising past problems for my own
benefit.

 

[Ben Bacarisse]

Have you done it yet?

I'd done it when I wrote before. The delay was down to worrying about
posting solutions to homework-type problems.

I'm just practising past problems for my own
benefit.

Then this may be of little use. It is decidedly "compact" because I
wanted to ensure that, if this was homework, my solution would not be
useful as an answer.

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

void zigzag_decode(int nrows, const char *input)
{
int length = strlen(input);
char output[length + 1];
for (int i = 0, row = 0; i < length; row++) {
output[row] = input[i++];
int skip[] = {nrows - 1 - row, row};
for (int j = 0, p = row; (p += 2 * skip[j]) < length; j = !j)
if (skip[j])
output[p] = input[i++];
}
output[length] = 0;
puts(output);
}

int main(int argc, char **argv)
{
if (argc == 3) {
int nr = atoi(argv[1]);
if (nr > 1)
zigzag_decode(nr, argv[2]);
else fprintf(stderr, "Number of rows must be > 1.\n");
}
else fprintf(stderr, "Try: zigzag no-of-rows input-sequence\n");
return 0;
}