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PHP LOOPING

hi i got an example of looping php and im not understanding it <?php function func($arg){ $result=0; for($i=0;$i<$arg;$i++){ $result=$result+$i; } return $result; } echo func(4); ?> The output is 6 and i don’t understand why Can anybody help please

1st Oct 2020, 1:30 AM
Ezra Tawachi
3 Answers
+ 6
Here whem function is calling you passing 4 func(4) So in arg u passed 4 and loop will work total 4 times Initially i =0 The we will check condition i<4 means 0<4 which is true then next line $result=result+$i here result is 0 and i is also 0 Then $i in third step of loop here i will increase and i become 1 Then we check condition i<4 and i =1 so 1<4 this is true so loop will run again $result=result + i =0+1 =1 and 1 will assign to result then again i++ here i will increase again so i=2 And check condition its true so again result=result + i ( here result is 1 and i is 2) so result =1+2=3 result will be 3 then in $i++ here i will increase so i=3 And result 3 $result=result +i = 3+3= 6 so final value is 6
1st Oct 2020, 2:13 AM
A S Raghuvanshi
A S Raghuvanshi - avatar
+ 2
You are calling function func() by passing the value 4 to it so your arg is 4 now. In the for loop you assign i=0 and it should iterate till i<arg which is i<4. In 1st iteration when i=0 result = 0+0 = 0 In 2nd iteration when i=1 result = 0+1 = 1 In 3rd iteration when i=2 result = 1+2 = 3 In 4th iteration when i=3 result = 3+3 = 6 In 5th iteration when i=4 The loop terminates because i<args fails because 4<4 is false. So you return the result value which is 6.
1st Oct 2020, 2:05 AM
Avinesh
Avinesh - avatar
+ 1
Please tag PHP instead of '/' up there ☝
1st Oct 2020, 3:06 AM
Ipang