Understanding Factorialize solution
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1
function factorialize(num) {
var count = 1;
for (var i = 1; i < num; i++) {
count = count + (count * i);
}
return count;
}
factorialize(1);
Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?
javascript algorithm
edited May 9 '16 at 17:35
Alex Pan
2,73452332
asked May 9 '16 at 17:21
wquwqu
80117
add a comment |
1
function factorialize(num) {
var count = 1;
for (var i = 1; i < num; i++) {
count = count + (count * i);
}
return count;
}
factorialize(1);
Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?
javascript algorithm
edited May 9 '16 at 17:35
Alex Pan
2,73452332
asked May 9 '16 at 17:21
wquwqu
80117
2
It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen
– juvian
May 9 '16 at 17:23
1
In theforloop,var i = 1;setsito1.i < numis the condition for this loop to run. Sincenumis1, the loop does not run, and the function simply returnscount.
– blex
May 9 '16 at 17:24
add a comment |
1
1
1
function factorialize(num) {
var count = 1;
for (var i = 1; i < num; i++) {
count = count + (count * i);
}
return count;
}
factorialize(1);
Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?
javascript algorithm
edited May 9 '16 at 17:35
Alex Pan
2,73452332
asked May 9 '16 at 17:21
wquwqu
80117
function factorialize(num) {
var count = 1;
for (var i = 1; i < num; i++) {
count = count + (count * i);
}
return count;
}
factorialize(1);
Why does this return 1 when you use 1? Shouldn't it return 2 because count = 1 + (1 * 1)?
javascript algorithm
javascript algorithm
edited May 9 '16 at 17:35
Alex Pan
2,73452332
asked May 9 '16 at 17:21
wquwqu
80117
edited May 9 '16 at 17:35
Alex Pan
2,73452332
asked May 9 '16 at 17:21
wquwqu
80117
edited May 9 '16 at 17:35
Alex Pan
2,73452332
edited May 9 '16 at 17:35
Alex Pan
2,73452332
edited May 9 '16 at 17:35
Alex Pan
2,73452332
2,73452332
asked May 9 '16 at 17:21
wquwqu
80117
asked May 9 '16 at 17:21
wquwqu
80117
asked May 9 '16 at 17:21
wquwqu
80117
80117
2
It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen
– juvian
May 9 '16 at 17:23
1
In theforloop,var i = 1;setsito1.i < numis the condition for this loop to run. Sincenumis1, the loop does not run, and the function simply returnscount.
– blex
May 9 '16 at 17:24
add a comment |
2
It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen
– juvian
May 9 '16 at 17:23
1
In theforloop,var i = 1;setsito1.i < numis the condition for this loop to run. Sincenumis1, the loop does not run, and the function simply returnscount.
– blex
May 9 '16 at 17:24
2
2
It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen
– juvian
May 9 '16 at 17:23
It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen
– juvian
May 9 '16 at 17:23
1
1
In the
for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.– blex
May 9 '16 at 17:24
In the
for loop, var i = 1; sets i to 1. i < num is the condition for this loop to run. Since num is 1, the loop does not run, and the function simply returns count.– blex
May 9 '16 at 17:24
add a comment |
1 Answer
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oldest
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For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
add a comment |
0
For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
add a comment |
0
0
0
For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
For num = 1, when we enter the loop, the check i < num returned false, so we do not get to the count = count + (count * i); and go to return with count = 1.
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
edited Nov 16 '18 at 11:43
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
answered Nov 1 '18 at 11:24
Eugene MihaylinEugene Mihaylin
1,0101725
1,0101725
add a comment |
add a comment |
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2
It is not entering the loop as i=1 < 1 is false. With <= what you expect for 1 would happen
– juvian
May 9 '16 at 17:23
1
In the
forloop,var i = 1;setsito1.i < numis the condition for this loop to run. Sincenumis1, the loop does not run, and the function simply returnscount.– blex
May 9 '16 at 17:24