如果条件为真，则在 do while 循环中发布增量（最佳方式？）

Question

It is quite simple, indeed.

int main()
{
    int n,i,j;
    n = 20;
    i = 0;
    char ch[8];

    do
    {
        ch[i] = (n%2) + '0';
        n /= 2;
        // SIMPLE WAY
        if(n != 0)
            i++;
    }
    while(n != 0);

    for(j=0; j<=i; j++)
    {
        printf("%c",ch[i-j]);
    }

    return 0;
}

But i don't like this WAY

I tried the below WAY but code is bad

int main()
{
    int n,i,j;
    n = 20;
    i = 0;
    char ch[8];

    do
    {
        ch[i] = (n%2) + '0';
        n /= 2;
    }
    while(n != 0 && i++); // THIS

    for(j=0; j<=i; j++)
    {
        printf("%c",ch[i-j]);
    }

    return 0;
}

How to get the value incremented only when the loop is true with BEST WAY? OR just correct 2nd way while(n != 0 && i++)

Answer 1

How to get the value incremented only when the loop is true with BEST WAY?
OR just correct 2nd way while(n != 0 && i++)

I recommend neither approach.

There is no need for a special test to see if code should increment i .

do {
  ch[i] = (n%2) + '0';
  n /= 2;
  i++;
} while(n != 0);
// Just decrement after the loop
i--;

for(j=0; j<=i; j++) {
    printf("%c",ch[i-j]);
}

Alternatively do not decrement at all

do {
  ch[i] = (n%2) + '0';
  n /= 2;
  i++;
} while(n != 0);

for(j=0; j<i; j++) {
  printf("%c",ch[i-1-j]);
}

Or use a do ... while for printing also.

do {
  ch[i++] = (n%2) + '0';
  n /= 2;
} while(n);

do {
  printf("%c",ch[--i]);
} while (i);

---

Notes:

char ch[8]; is too small for an int outside the range [-255 ... 255] . For 32-bit int , use char ch[32]; or wider. In general, use char ch[sizeof(int) * CHAR_BIT];

ch[i] = (n%2) + '0'; will certainly incur unexpected results when n < 0 . Consider unsigned types instead.

Answer 2

Well, you can always do:

do {
   // code
} while(n != 0 && ++i);

This way, i is incremented right after the first part of && evaluates to True, and && is applied to n != 0 and the updated version of i .

Answer 3

In C++, you might do:

std::cout << std::bitset<8>(n); // 00010100 for n = 20

if leading 0 are ok.

Answer 4

It can be like this:

int n = 20;
int i = 7;
char ch[] = "00000000";
while (n) // if n already = 0 then do nothing
{
    ch[i--] = (n % 2) + '0'; // assign then increment
    n /= 2;                // shift to right same as n >>= 1;
}
printf ("%08s", ch);

Answer 5

do
{ .......... }while(n && i++ ? 1 : n);

this stuff about && actions called short-circuit evaluation