is aligning with printf() via a variable value possible?

Code:

void PrintMatrix(int **matrix, int rij, int kolom)
{
	int i, j;
	
	printf("Resultaat: \n");
	
	for (i = 0; i < rij; i++)
	{
		printf("[");
		for (j = 0; j < kolom - 1; j++)
			printf("%-10i", *(*(matrix + i)+j));
		printf("%-10i", *(*(matrix + i)+j));             <-------
		printf("]\n");
	}
	
	printf("\n");
}

Is it possible to have a variable alignment instead of a constant (10 in this example)?
so something like this:

printf("%-(%i)i", numbers(*(*(matrix + i)+j)), *(*(matrix + i)+j)); <-------

[ where numbers is a function that calculates the #amount of digits an integer has.
numbers(321) = 3. ]

I don't think so.

i suppose you could do something like this

Code:

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

int main (void) {

   int number = 8;
   char output[50] = "start of string %";
   char* str2 = "%d";
   char* str3 = "d end of string";

   sprintf(&output[17], str2, number);
   sprintf(&output[18], str3);
   printf("'%s'\n\n", output);
   printf(output, 123);

   return 0;

}

so yes, it is possible

A couple of work arounds, maybe:

Code:

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

int main() {
  int i; 
  int num[4] = { 1000, 100, 10, 1 };
  char spc[4] = {""};
  printf("\n\n");
  for(i=0;i<4;i++) {
    printf("\n%d: %s%d", i, spc, num[i]);
    spc[i] = ' ';
  }
  printf("\n\n\n");
  spc[0]=spc[1]=spc[2]=spc[3]='\0';  //reset the space array
  for(i=0;i<4;i++) {
    if(num[i] > 999) 
      spc[0]='\0';
    else if(num[i] > 99)
      spc[0]=' ';
    else if(num[i] > 9)
      strcpy(spc, "  ");
    else 
      strcpy(spc, "   ");
    printf("\n%d: %s%d", i, spc, num[i]);
    spc[i]='\0';
  }

  printf("\n\n\t\t\t     press enter when ready");

  i = getchar(); ++i;
  return 0;
}

@LordPc - Welcome to the Forum!

Hey, I think we could use a righteous contributor.

Originally Posted by LordPc

so yes, it is possible

Yeah, good idea! The first arg to printf is actually just a C string, so you could pre-construct the whole thing.

Originally Posted by manual

The field width

An optional decimal digit string (with non-zero first digit) specifying a minimum field width. If the converted value has fewer characters than the field width, it will be padded with spaces on the left (or right, if the left-adjustment flag has been given). Instead of a decimal digit string one may write '*' or '*m$' (for some decimal integer m) to specify that the field width is given in the next argument, or in the m-th argument, respectively, which must be of type int. A negative field width is taken as a '-' flag followed by a positive field width. In no case does a non-existent or small field width cause truncation of a field; if the result of a conversion is wider than the field width, the field is expanded to contain the conversion result.

So for example

printf("%*d\n", 5, myvalue );

At least, we now know how many people are not printf expert!

unrelatd tip, but

Code:

*(*(matrix + i)+j)

is equivelent to

Code:

matrix[i][j]

A lot of people I see use the first one because they think you can't use subscript notation when the array is passed in as int **.


Of course, if you prefer the other way, then ignore me ^_^

Thanks for all the response, Salem's in particular.

Originally Posted by KBriggs

unrelatd tip, but

Code:

*(*(matrix + i)+j)

is equivelent to

Code:

matrix[i][j]

A lot of people I see use the first one because they think you can't use subscript notation when the array is passed in as int **.


Of course, if you prefer the other way, then ignore me ^_^

I was acctually doing a matrix project to refresh the pointer notation ^_^