Integer Separation

Question

I know that when dividing integers by default, it works to discard the fractional part. For instance,

int i, n, calls = 0;
n = 1;
n /= 3;
printf("N = %i\n", n);
for (i = 1; i > 0; i /= 3) {
    calls++;
}
printf("Calls = %i\n", calls);

Above code:

N = 0
Calls = 1

Could you explain this behavior?

Answer 1

1 is divided by 3 = .3333 (repeated, of course), mathematically. You may think that the computer truncates .3333 as it performs integer arithmetic ( 0remainder 1).

The cycle foris executed because i = 1and 1 > 0. After completing the body of the loop, you divide iby three and ibecome 0, which is not more than 0.

Answer 2

while, .

i = 1;
while ( i > 0 )
{
    calls++;        
    i /= 3; //This becomes .3333, which truncates to zero
}

Answer 3

.

for , i > 0 1 > 0, , .

Answer 4

? : 1/3 = 0.33333..., , 0.

, for - :

i=1;
while(i>0)
{
    calls++;
    i/=3;
}

, i 1; while , i, 1, 0. calls 0 1, 1. i 3, 0 ( ). while , i 0, . calls 1, .

Answer 5

n - int, , double float

Answer 6

Since in integer arithmetic the answer to 1 divided by 3 is 0 with the remainder 1. If you separate two integers, you get integer arithmetic. If you need floating point arithmetic, you need at least one of the operands to be a floating point value.

Answer 7

Everything is very simple.

int i, n, calls = 0; // Set calls to 0
n = 1;               // n is now 1
n /= 3;              // n /= 3 = 1/3 = 0
printf("N = %i\n", n);
for (i = 1; i > 0; i /= 3) { // 1/3 = 0
    calls++;                 // runs once
}                            
printf("Calls = %i\n", calls);

Hope this helps.