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iOS Objective-C Basics (Retired) Functional Programming in C Control Flow - For Loops

Joe Voll
Joe Voll
304 Points

What does "sum += many[i]" do?

It's an increment but what exactly is it supposed to do. And why is sum+= to many[i]?

3 Answers

Hi there,

That takes the value of sum and adds the value of many[i] to it, and assigns that back into the value of sum.

I assume that iterates through the array many[] using an index of i so that you end up with sum having the value of all the elements of many[] added up.

Hope that helps!

Steve.

So, sum++ adds one to sum and stores that value in sum. Using sum += value increments sum by the value of value..

sum += many[i]

In this case, many is an array. You can tell because it has the square brackets []. The variable i is a way to access one of the indexes of that array.

So, sum += many[i] is saying, take what is currently in sum, and add whatever is inside many at index i.

It could also be written as sum = sum + many[i].

I is often used as a looping variable for a for loop. So, with this sum += code snippet inside a loop and the index starting at zero, and running until the size of the array, you would be able to sum up all the integers in the array.

Richard Then
Richard Then
10,690 Points

Hi,

I had to watch the video several times, and go through many of the Q&As to finally understand it. I broke it down the loops into steps according to my understanding, hopefully this helps.

int many [] = {2, 4, 8};

int sum = 0;

for (int i=0; i (< 3); i++)

{sum += many [i];

printf("sum %d\n", sum); }

         //Remember, *when:  
         0 refers to 2,  
         1 refers to 4,   
         2 refers to 8

——————————————————————————————————

Loop 1:

1a. i=0

1b. i (0) < 3

         // 0 is less than 3, loop continues.

1c. sum += many[i]

         // translation, everything in parenthesis. 

         many [i], pulls input from step 1a. 

          = sum (0) += many[i (*when: 0)]  = new_sum_A = 2  

1d. printf("sum %d\n", sum)

          // “sum (new_sum_A)”  = Output A: sum 2

1e. i++ = (i=0) + 1 = 1

         // i=1

——————————————————————————————————

Loop 2:

2a. i=1

2b. i (1) < 3

         // 1 is less than 3, loop continues.

2c. sum += many[i]

         // translation, everything in parenthesis. 

         many [i], pulls input from step 2a.

         = new_sum_A (2) += many[i (*when: 1)] = new_sum_B = 6

2d. printf("sum %d\n", sum)

         // “sum (new_sum_B)” = Output B: sum 6

2e. i++ = (i=1) + 1 = 2

         // i=2

——————————————————————————————————

Loop 3:

3a. i=2

3b. i (2) < 3

         // 2 is less than 3, loop continues.

3c. sum += many[i]

         // translation, everything in parenthesis. 

         many [i], pulls input from step 3a.

         = new_sum_B (6) += many[i (*when: 2)] = new_sum_C (14) 

3d. printf("sum %d\n", sum)

         // “sum (new_sum_C)” = Output C: sum 14

3e. i++ = (i=2) + 1 = 3

         // i=3

——————————————————————————————————

(non) Loop 4:

4a: i=3

4b. i (3) < 3

         // 3 is not less than 3, loop stops.

——————————————————————————————————

sum 2 sum 6 sum 14