C: 2d char array issue - c

Question

i've a problem with an exercise. The purpose is to print a specific picture after an int input (It is assumed that the input is > 3).
Example:
https://i.stack.imgur.com/KRl0R.png
This is my code:
#include <stdio.h>
int main(void){
int i, j, n, simbolo;
printf("Inserire un numero: ");
scanf("%d", &n);
char mat[n + 1][n + 2];
simbolo = n+2;
//initialization with blank space and setting 3 * diagonal
for(i = 0; i < n + 1; i ++){
for(j = 0; j < n + 2; j++){
mat[i][j] = ' ';
mat[n - 2][0] = '*';
mat[n - 1][1] = '*';
mat[n][2] = '*';
}
}
//Add * diagonal of n length
for(i = 0; i < n + 1; i++){
mat[i][simbolo] = '*';
for(int x = i, y = 0; y < n + 2; y++){ //Print current line
printf("%c", mat[x][y]);
}
printf("\n");
simbolo--;
}
return 0;
}
The output isn't correct, it's added an extra '*' in mat[1][0]:
https://i.stack.imgur.com/4Z5Fl.png
Thanks in advance for you help

Answer 1

According to image, you want your output to have n rows, not n+1. This works correctly:
#include <stdio.h>
int main(void){
int i, j, n, simbolo;
printf("Inserire un numero: ");
scanf("%d", &n);
char mat[n][n + 2];
simbolo = n+1;
//initialization with blank space and setting 3 * diagonal
for(i = 0; i < n; i ++){
for(j = 0; j < n + 2; j++){
mat[i][j] = ' ';
}
}
mat[n - 3][0] = '*';
mat[n - 2][1] = '*';
mat[n - 1][2] = '*';
//Add * diagonal of n length
for(i = 0; i < n; i++){
if(i<(n-1))
mat[i][simbolo] = '*';
printf("%.*s\n", n+2, mat[i]);
simbolo--;
}
return 0;
}

Answer 2

Some other people have already pointed out the issue, so I thought I would just post a cleaned up version for your reference:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
printf("n: ");
int n = 0;
if (scanf("%d", &n) != 1) {
fprintf(stderr, "ERROR: couldn't read 'n'\n");
return EXIT_FAILURE;
}
char mat[n][n + 2];
// Initialize with spaces.
int i;
int j;
for (i = 0; i < n; i++)
for (j = 0; j < n + 2; j++)
mat[i][j] = ' ';
// Draw the first 2 * on the left.
mat[n - 3][0] = '*';
mat[n - 2][1] = '*';
// Draw the diagonal on the right.
j = n + 1;
for (i = 0; j >= 2; i++, j--)
mat[i][j] = '*';
// Print out the result.
for (i = 0; i < n; i++) {
for (j = 0; j < n + 2; j++)
printf("%c", mat[i][j]);
printf("\n");
}
return EXIT_SUCCESS;
}

Answer 3

I just changed
mat[i][simbolo] = '*';
for
mat[i+1][simbolo-1] = '*';
I also changed the spaces for '-' to make the output clearer.
The problem is the one mentioned in the comments: your indices were off.
Remember, in C, indexing is 0 based, so the last element of any array has index n-1.
Usually, in C when you go further the pointer just goes back to the beginning.
#include <stdio.h>
int main(void){
int i, j, n, simbolo;
printf("Inserire un numero: ");
scanf("%d", &n);
char mat[n + 1][n + 2];
simbolo = n+2;
//printf("Simbolo: %d",simbolo);
//initialization with blank space and setting 3 * diagonal
for(i = 0; i < n + 1; i ++){
for(j = 0; j < n + 2; j++){
mat[i][j] = '-';
mat[n - 2][0] = '*';
mat[n - 1][1] = '*';
mat[n][2] = '*';
}
}
//Add * diagonal of n length
for(i = 0; i < n+1 ; i++){
mat[i+1][simbolo-1] = '*';
for(int y = 0; y < n + 2; y++){ //Print current line
//printf(",x< %d, y: %d",i,y);
printf("%c", mat[i][y]);
}
printf("\n");
//printf("Simbolo: %d",simbolo);
simbolo--;
}
return 0;
}