C中的十进制到八进制

Question

I have just begun teaching myself C out of KN King's C Programming: A Modern Approach (2ndEdn).

I'm enjoying it, but am hoping to post the odd question here for advice if appropriate because unfortunately I don't have a tutor and some bits raise more questions then they answer!

I'm doing a question on taking an integer entered and displaying it in octal. It says there is an easy way to do it, but that comes later in the book. I have come up with the following:

// Convert a number to octal

int n, n2, n3, n4, n5, n6;

printf("Enter a number between 0 and 32767: ");

scanf("%d", &n);

n6 = n % 8;
n5 = (n / 8) % 8;
n4 = ((n / 8) / 8) % 8;
n3 = (((n / 8) / 8) / 8) % 8;
n2 = ((((n / 8) / 8) / 8) / 8) % 8;

printf("%d%d%d%d%d", n2, n3, n4, n5, n6);

It works OK, but I'm not good at math and was wondering if there is a more efficient way of doing this or have I done it the only way possible...

If anyone else has the book it's Q4 p.71.

Thanks for your time. Andrew

PS I did look in the search engine but couldn't find anything that was doing it this 'slower' way!

Answer 1

Everyone is right in saying that there's a built-in way to do that with printf . But what about doing it yourself?

The first thing that came to mind is that one octal digit is exactly three bits. Therefore you can do the conversion this way:

• Loop while n != 0
• Isolate the leftmost 3 bits of n into d and print d
• Shift n 3 bits to the left

The code is trivial, but I 'm not providing it so you can do it yourself (you will need to be familiar with the bitwise and shift operators in order to do it).

Answer 2

The easy way is probably to use printf() 's %o format specifier :

scanf("%d", &n);
printf("%o", n);

Answer 3

Others have posted the real, production code answer, and now I see from your comments that you haven't done loops yet. Perhaps your book is trying to teach you about recursion:

void print_oct(int n)
{
    if (n != 0) {
        print_oct(n / 8);
        printf("%d", n % 8);
    }
}

This works for n > 0.

Answer 4

/* Converts a positive base_10 into base_b */
int DecimalToBase(int n, int b)
{
    int rslt=0, digitPos=1;
    while (n)
    {
        rslt += (n%b)*digitPos;
        n /= b;
        digitPos *= 10;
    }
    return rslt;
}

Answer 5

With loops you can roll up your five very similar lines like this:

for (int d = 8 * 8 * 8 * 8; d > 0; d /= 8)
    printf("%d", n / d % 8);
printf("\n");

d will start at 8 * 8 * 8 * 8 , which is the divisor you use for n2 and then step through 8 * 8 * 8 , 8 * 8 , 8 and finally 1 , which is the divisor for n6 , printing each digit along the way.

A good compiler will actually optimize this by unrolling it back into five lines, so you'll get almost the same thing you started with. The advantage of writing it as a loop is that you can't make a mistake in just one of the lines.

The compiler will also take care of replacing divisions by 8 with shifts by 3 bits. Both give the same result in binary, but the latter is faster.

Answer 6

Since only basics are introduced you don't want (at least at this point) to use functions, loops, bitwise operators, %o format specifier and all that stuff. Here is my basic solution:

int n, d1, d2, d3, d4, d5, o;

printf("Enter a number between 0 and 32767: ");
scanf("%d", &n);

d5 = n % 8;
n /= 8;
d4 = n % 8;
n /= 8;
d3 = n % 8;
n /= 8;
d2 = n % 8;
n /= 8;
d1 = n % 8;

o = 10000 * d1 + 1000 * d2 + 100 * d3 + 10 * d4 + d5;

printf("In octal, your number is: %.5d\n", o);

Note that since n is not needed in output, you can modify (divide) it for every step (thus saving divides, which are computationally and relatively expensive). You are safe up to 32767 (in octal: 77777 ), as 32768 (8*8*8*8*8 = 8^5 = (2^3)^5 = 2^15) is the first number, that requires six digits in octal: 100000 .

This o variable is not really needed, morever it will not work when int is signed 16-bit (on some ancient system), so from this point it's better to just print separate digits.

Answer 7

Existing answers aren't clean enough for my liking. Here's mine:

#include <stdio.h>

#define OCTALBASE    8
#define OCTALSIZE    8

int main(int argc, char **argv) {
  int indecimal = 1337;
  char output[OCTALSIZE + 1];
  output[OCTALSIZE] = '\0';

  int outindex = OCTALSIZE;
  int outdigit = 0;
  int outvalue = indecimal;
  while (--outindex >= 0) {
    outdigit = outvalue % OCTALBASE;
    if (outvalue > 0 || outdigit > 0)
      { output[outindex] = '0' + outdigit; }
    else { output[outindex] = ' '; }
    outvalue /= OCTALBASE;
  }

  fprintf(stdout, "{ DEC: %8d, OCT: %s }\n", indecimal, output);
  fflush(stdout);

  return 0;
}

Result:

{ DEC:     1337, OCT:     2471 }

Answer 8

Convert Decimal to Octal in C Language

#include<stdio.h>
#include<conio.h>
void main()
{
    A:
    long int n,n1,m=1,rem,ans=0;
    clrscr();
    printf("\nEnter Your Decimal No :: ");
    scanf("%ld",&n);

    n1=n;
    while(n>0)
    {
        rem=n%8;
        ans=(rem*m)+ans;
        n=n/8;
        m=m*10;
    }

    printf("\nYour Decimal No is :: %ld",n1);
    printf("\nConvert into Octal No is :: %ld",ans);

    printf("\n\nPress 0 to Continue...");
    if(getch()=='0')
        goto A;
    printf("\n\n\n\tThank You");
    getch();
}

Answer 9

Use %o format specifier inside printf

printf("Enter a number between 0 and 32767: ");
scanf("%d", &n);
printf("%o", n);