Here's my code:
#include <stdio.h>
int main(){
int n,sqrs,start=0,h,y,x; scanf("%i",&n);
int p[n][n];
if(n%2==0) sqrs = n/2;
else sqrs = n/2 1;
h = 1;
y = start;
x = start;
for(int t=0; t<sqrs; t ){
for( ; x<n; x ) p[y][x] = h; // THESE LOOPS
for( ; y<n; y ) p[y][x] = h;
for( ; x>=start; x--) p[y][x] = h;
for( ; y>=start; y--) p[y][x] = h;
y ; x ; h ; start ; n-=2;
}
for(int i=0; i<n; i ){
for(int j=0; j<n; j ){
printf("%i",p[i][j]);
} printf("\n");
}
}
As you can see, these loops aren't nested. They would nest though if there weren't any statements in between them, but there are.
So, the first one iterates normally but all of the following ones don't. They just skip.
I did manage to get around this, but curiously only when I added both braces and newlines, i.e. when all loops (except the first one, which doesn't seem to require it) looked like this:
for( ; y<n; y ) {
p[y][x] = h;
}
This remains a mistery to me. Of course, I'm sorry if this is a duplicate. I just couldn't find anything about this specifically.
CodePudding user response：
There are multiple mistakes in your code:
- you should undo the last increment/decrement after each of the inner loops. As coded, you store
hbeyond the end of the arrayp. nshould be decremented only by1- since
nis modified, the final loops have no effect.
Here is a modified version, keeping your corny style:
#include <stdio.h>
int main(){
int n,sqrs,start=0,end,h,y,x; scanf("%i",&n);
int p[n][n];
if(n%2==0) sqrs = n/2;
else sqrs = n/2 1;
h = 1;
y = start;
x = start;
end = n;
for(int t=0; t<sqrs; t ){
for( ; x<end; x ) p[y][x] = h; x--;
for( ; y<end; y ) p[y][x] = h; y--;
for( ; x>=start; x--) p[y][x] = h; x ;
for( ; y>=start; y--) p[y][x] = h; y ;
y ; x ; h ; start ; end--;
}
for(int i=0; i<n; i ){
for(int j=0; j<n; j ){
printf("%i",p[i][j]);
} printf("\n");
}
}
This code can be simplified as:
#include <stdio.h>
int main(){
int n; scanf("%i",&n);
int p[n][n];
int h=1,start=0,stop=n-1;
for(int t=0; t<n; t =2,h ,start ,stop--){
int x=start,y=start; p[y][x] = h;
while(x<stop) p[y][x ] = h;
while(y<stop) p[y ][x] = h;
while(x>start) p[y][x--] = h;
while(y>start) p[y--][x] = h;
}
for(int i=0; i<n; i ){
for(int j=0; j<n; j ) printf("%i",p[i][j]);
printf("\n");
}
}
And here is a more classic version:
#include <stdio.h>
int main() {
int n;
if (scanf("%i", &n) != 1 || n <= 0)
return 1;
int p[n][n];
int h = 1, start = 0, stop = n - 1;
for (int t = 0; t < n; t = 2, h , start , stop--) {
int i = start, j = start;
if (start == stop)
p[i][j] = h;
for (; j < stop; j )
p[i][j] = h;
for (; i < stop; i )
p[i][j] = h;
for (; j > start; j--)
p[i][j] = h;
for (; i > start; i--)
p[i][j] = h;
}
for (int i = 0; i < n; i ) {
for (int j = 0; j < n; j )
printf("%i", p[i][j]);
printf("\n");
}
return 0;
}
CodePudding user response：
It appears that you are trying to make a pattern of concentric boxes in a matrix. Your logic is wrong. Following is an example using the Ada programming language. Translating that into C should be easy.
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
procedure Main is
type matrix is array (Positive range <>, Positive range <>) of Positive;
Side : Positive;
begin
Put ("Enter the number of elements on a side of the array: ");
Get (Side);
declare
Box : matrix (1 .. Side, 1 .. Side);
Center : Positive;
Start : Positive := 1;
Stop : Positive := Side;
Value : Positive := 1;
begin
Center := (if Side mod 2 = 0 then side / 2 else side / 2 1);
for turn in 1 .. Center loop
-- sides
for I in Start .. Stop loop
Box (Start, I) := Value;
Box (I, Start) := Value;
Box (Stop, I) := Value;
Box (I, Stop) := Value;
end loop;
-- adjust for inner box values
Start := Start 1;
Stop := Stop - 1;
Value := Value 1;
end loop;
for I in Box'Range (1) loop
for J in Box'Range (2) loop
Put (Item => Box (I, J), Width => 1);
end loop;
New_Line;
end loop;
end;
end Main;
The output of a sample execution is:
Enter the number of elements on a side of the array: 7
1111111
1222221
1233321
1234321
1233321
1222221
1111111
CodePudding user response：
It's not clear from either the OP code or the description what it is you are trying to achieve.
Based on another recent SO question, perhaps this is what you are seeking. This takes advantage of the symmetry of the (I believe) desired output. Four array elements are assigned in each iteration until the centre of the pattern is reached.
int main() {
const uint8_t n = 9;
uint8_t mat[n][n];
for( int r = 0; r < (n 1)/2; r )
for( int c = 0; c < (n 1)/2; c )
mat[r ][c ] =
mat[n-1-r][c ] =
mat[r ][n-1-c] =
mat[n-1-r][n-1-c] =
1 (c<r?c:r);
for( int i = 0; i < n; i ) {
for( int j = 0; j < n; j )
printf("%d ", mat[i][j]);
puts( "" );
}
return 0;
}
1 1 1 1 1 1 1 1 1
1 2 2 2 2 2 2 2 1
1 2 3 3 3 3 3 2 1
1 2 3 4 4 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 4 4 3 2 1
1 2 3 3 3 3 3 2 1
1 2 2 2 2 2 2 2 1
1 1 1 1 1 1 1 1 1
Perhaps the "not nested" is the criteria?
int main() {
const uint8_t n = 9, h = ((n 1)/2); // 'h'alf way
uint8_t mat[n][n];
for( int x = 0; x < h*h; x ) {
int r0 = x / h, r1 = n-1-r0; // top & bottom rows of this square
int c0 = x % h, c1 = n-1-c0; // left & right cols of this square
uint8_t v = 1 ( c0 < r0 ? c0 : r0 );
mat[r0][c0] = mat[r0][c1] =
mat[r1][c0] = mat[r1][c1] = v;
}
/* same nested output loops */
return 0;
}