Programming a reversed pyramid in c - c

Question

Im working on an optional test review problem for an introduction to C class, and I need to have a program that prints out the following based upon what ever number a user enters:
Enter a number: 5
5
44
333
2222
11111
000000
11111
2222
333
44
5
So far this is the code that I have written
#include <stdio.h>
int main(void){
int row,column,space;
int number;
printf("Enter a number: ");
scanf_s("%d",&number);
for (row = 1; row <= number + 1; row++){
for (space = number; space >=row; space--){
printf(" ");
}
for(column = 1;column <= row; column++){
printf("%d",space);
}
printf("\n");
}
for (row = 1; row <=number;row++){
for(space = 1;space <= row;space++){
printf(" ");
}
for(column = number;column >=row;column--){
printf("%d",space);
}
printf("\n");
}
return 0;
}
This is the output that I get
Enter a number: 5
0
11
222
3333
44444
555555
22222
3333
444
55
6
I've spent quite a few hours trying to figure out how to print the upper half of the half diamond using the user entered numbers but I can't seem to figure it out. Could anyone point me in the right direction?

Answer 1

Your numbers are just off a bit, correct the printf calls and you're done:
First one:
printf("%d", number - space);
Second one:
printf("%d", space - 1);
A slightly better (more readable and a bit more logical) way would be to use other variables instead:
First one:
printf("%d", number + 1 - row);
Second one:
printf("%d", row);
Also note that some basic math can help you to realize the following:
Total number of rows: 2 * number + 1
Number of spaces: 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5 => abs(number - row) (If starting your rows with 0)
Number to print: Same as "Number of spaces"
Number count: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 => 6 - number_of_spaces
This gives a much cleaner, more readable version with only one loop:
#include <stdio.h>
#include <math.h>
int main(void){
int row,column,space;
int number;
printf("Enter a number: ");
scanf_s("%d",&number);
for (row = 0; row <= number * 2; row++){
int number_of_spaces = abs(number - row);
int number_to_print = number_of_spaces;
int number_count = 6 - number_of_spaces;
for (space = 1; space <= number_of_spaces; space++){
printf(" ");
}
for(column = 1;column <= number_count; column++){
printf("%d", number_to_print);
}
printf("\n");
}
}
return 0;
}

Answer 2

To expand on schnaader's answer (which is perfectly fine and complete), you can improve your code even more, letting printf() do the spacing for you rather than doing a loop and several calls to printf():
printf("%*s", width, "");
Here width is replaced with the calculated space you'd like to fill. The precision * is a special placeholder that tells the function to take the actual precision/length from the parameter list. Since the string to print is empty, it will always fill the whole range with space characters.

Answer 3

for (row = 0; row <= number + number; row++) {
int t = row * (row <= number) + (number + number - row) * (row > number);
printf("%*c", number - t, ' ');
printf("%*d\n", t + 1, t);
}

Answer 4

There are numerous ways, I've just modified your code
#include <stdio.h>
int main(void){
int row,column,space;
int number;
printf("Enter a number: ");
scanf("%d",&number);
for (row = 1; row <= number + 1; row++){
for (space = number; space >=row; space--){
printf(" ");
}
for(column = 1;column <= row; column++){
printf("%d",number -space); //change1
}
printf("\n");
}
for (row = 1; row <=number;row++){
for(space = 0;space < row;space++){ //change 2
printf(" ");
}
for(column = number;column >=row;column--){
printf("%d",space);
}
printf("\n");
}
return 0;
}

Answer 5

the following code outputs the first half of the problem.
Notice the checking for errors in the call to scanf()
It compiles cleanly and works correctly
I leave it to you to complete the function for the second half of the output
which should be easy now that you have the first half
#include <stdio.h>
#include <stdlib.h>
int main()
{
int row;
int i; // loop counter/index
int n; // user input
printf( "\nEnter the wedge width: ");
if(1 != scanf(" %d", &n) )
{ // then, scanf failed
perror( "scanf failed" );
exit( EXIT_FAILURE );
}
// implied else, scanf successful
for( row = 0; row <= n; row++ )
{
// print n-row ' '
for( i=0; i<(n-row); i++)
{
printf(" ");
}
// print row+1 '(n-row)'
for( i=0; i<(row+1); i++)
{
printf( "%1d", (n-row));
}
printf( "\n" );
}
// reverse the calculations to print the lower half
return 0;
} // end function: main