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How it's executed if(i&1)
5 odpowiedzi
+ 14
for(int i = 0; i < 10; i++) {
if(i&1) cout << i;
}
Brute force output:
"Cycle #1"
0000 = i = 0
0001 &
______
0000 = 0
no output.
///////
"Cycle #2"
0001 = i = 1
0001 &
______
0001 = 1
output: 1
//////
"Cycle #3"
0010 = i = 2
0001 &
______
0000 = 0
output still 1
///////
"Cycle #4"
0011 = i = 3
0001 &
______
0001 = 1
new output: 11
//////
"Cycle #5"
0100 = i = 4
0001 &
______
0000 = 0
output still 11
///////
"Cycle #6"
0101 = i = 5
0001 &
______
0001 = 1
new output: 111
///////
"Cycle #7"
0110 = i = 6
0001 &
______
0000 = 0
output still 111
////////
"Cycle #8"
0111 = i = 7
0001 &
______
0001 = 1
new output: 1111
///////
"Cycle #9"
1000 = i = 8
0001 &
______
0000 = 0
output still 1111
///////
"Cycle #10"
1001 = i = 9
0001 &
______
0001 = 1
new output: 11111
As you see, when i is an odd number (1, 3, 5, ...) you have output.
+ 2
& here is bitwise AND.
A bitwise AND takes two binary representation and performs the logical ANDoperation on each pair of the corresponding bits, by multiplying them. Thus, if both bits in the compared position are 1, the bit in the resulting binary representation is 1 (1 × 1 = 1); otherwise, the result is 0 (1 × 0 = 0 and 0 × 0 = 0).
+ 1
Let i be 5 or any odd number
then 5&1 = 101&1 [ 5 in binary is 101 ]
or
101
×1
———
101
Last bit is 1 so it returns true
Let i be 6 or any even number
then 6&1 = 110&1 [ 6 in binary is 110 ]
or
110
×1
———
110
Last bit is 0 so it returns false
+ 1
what's the meaning of & operator
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Слишком глубоко для ответа на кратковременый вопрос: ,,Какой выход?''