How to format output correctly? - c

Question

I cannot seem to get my output to work correctly. I want to make it so that the last remainder in parenthesis is lined up perfectly in a column following the first line's spacing. I have most of the output right except for this last part and it has to be exactly the way my professor wants it.
The best I've been able to come up with is EDITED:
Marcus Lorenzana
314156 = 19634 * 16 + 12 (C)
19634 = 1227 * 16 + 2 (2)
1227 = 76 * 16 + 11 (B)
76 = 4 * 16 + 12 (C)
4 = 0 * 16 + 4 (4)
0x4CB2C
This is the output I want:
Marcus Lorenzana
314156 = 19634 * 16 + 12 (C)
19634 = 1227 * 16 + 2 (2)
1227 = 76 * 16 + 11 (B)
76 = 4 * 16 + 12 (C)
4 = 0 * 16 + 4 (4)
0x4CB2C
But as you can see it the output is not exactly correct.
Here is my program EDITED:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BUFFER 50
static int base = 16;
int main(int argc, char * argv[]) {
printf("Marcus Lorenzana\n");
if (argc == 2) {
char hexstr[] = "0123456789ABCDEF";
int i = 0;
long long oldresult;
int remainder;
char remainders[BUFFER];
char w_num[BUFFER];
long long value = atoll(argv[1]);
//Get width of original number for formatting purposes
int vwidth = strlen(argv[1]);
char oldwidth[BUFFER];
//Convert the decimal to hexadecimal
while(value != 0) {
oldresult=value;
remainder = value%base;
value = value/base;
//Store the remainder in an array for later use
remainders[i]=hexstr[remainder];
char line[BUFFER];
//Get length of line for formatting purposes
int w = sprintf(line,"%*lld = %-*lld * %2d + %2d", \
vwidth,oldresult,vwidth,value,base,remainder);
printf("%s (%c)\n", line,hexstr[remainder]);
i++;
}
//Print the the hexadecimal number
int x = i;
printf("0x");
while(x > 0) {
printf("%c",remainders[--x]);
}
printf("\n");
} else {
printf("Error: Wrong arguments\n");
return 1;
}
return 0;
}

Answer 1

Modify based on your code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BUFFER 50
static int base = 16;
int main(int argc, char * argv[]) {
printf("Marcus Lorenzana\n");
if (argc == 2) {
char hexstr[] = "0123456789ABCDEF";
int i = 0;
long long oldresult;
int remainder;
char remainders[BUFFER];
char w_num[BUFFER];
long long value = atoll(argv[1]);
//Get width of original number for formatting purposes
int vwidth = strlen(argv[1]);
char oldwidth[BUFFER];
int wMax = 0;
//Convert the decimal to hexadecimal
while(value != 0) {
oldresult=value;
remainder = value%base;
value = value/base;
//Store the remainder in an array for later use
remainders[i]=hexstr[remainder];
char line[BUFFER];
//Get length of line for formatting purposes
int w = sprintf(line,"%*lld = %-lld * %2d + %-2d", \
vwidth,oldresult,value,base,remainder);
wMax = w > wMax ? w:wMax;
printf("%s %*s(%c)\n", line,wMax-w,"",hexstr[remainder]);
i++;
}
//Print the the hexadecimal number
int x = i;
printf("0x");
while(x > 0) {
printf("%c",remainders[--x]);
}
printf("\n");
} else {
printf("Error: Wrong arguments\n");
return 1;
}
return 0;
}

Answer 2

Your idea to print the right-hand side to a temporary string is good, but for the output you want, you should just print the stuff to thr right of the equals sign to that string. Also, because you don't want the operands lined up, strip all formatting information, i.e. the widths, from the string.
snprintf(line, BUFFER, "%lld * %d + %d", value, base, remainder);
printf("%*lld = %*s (%c)\n", vwidth, oldresult,
-(vwidth + 10), line, hexstr[remainder]);
The width of the rhs is calculated based on the initial length of the number, vwidth plus two times two for the numbers plus two times three for the operands with surrounding spaces. The width has to be negative, because you want the RHS to be left-aligned and passed with spaces to the right.
If you let printf do the padding, there's no need to store the string length from the sprintf call, w.