On c is there something where i could return element by element to calling funcrion and then come back to the original function

I don't understand your question, something like a foreach? maybe a macro?

like yield on python

i got a geeks for geeks link for that but cant send here 😅😁

Yeah I know python's yield, but i don't think C has something like that

On c is there something where i could return element by element to calling funcrion and then come back to the original function

Afaik you don't have anything like that! You have to return it as a whole array... Or you can write a subroutine that you will call at the end of function and then call the parent function in that function

Help me

#include<stdio.h> int sum(int arr[], int n) { int sum = 0; // initialize sum // Iterate through all elements // and add them to sum for (int i = 0; i < n; i++) sum += arr[i]; return sum; } void main() { int i=0, j=0; int a[20]; char temp; do { scanf("%d%c", &a[i], &temp); i++; } while(temp != '\n'); int b=sum(a,13); printf("%d",b); } i am reading input from user and trying to find their sum but there is some problem while passing the array the values whoe sum is finding is not what i pass if the input i am giving is 1 2 3 4 sum is some random value

the problem is i can read and print n spaced integers like this but cant use the value to find them can anyone else suggest another way to read n spaced integer

https://stackoverflow.com/questions/3764014/how-do-you-read-scanf-until-eof-in-c

the problem is i can read and print n spaced integers like this but cant use the value to find them can anyone else suggest another way to read n spaced integer

int b=sum(a,13); use i instead of 13

i=0 while(scanf("%d", &words) != EOF) { arr[i]=words; i++; } so does this work

i=0 while(scanf("%d", &words) != EOF) { arr[i]=words; i++; } so does this work

you mean using gdb right.

Dry run?

Do everything with a pen and a paper exactly the way written in code

I'm afraid I'm not grasping the equation. What am I missing here? Anything I should google for this?

And how does this convert a string to an integer

So the equation basically takes an ascii char, and gets the ascii number for it.

Kindof like this, right @bond00_7 ? int number; for (int i = 0; i < strlen(str); i++) { switch (i) { case "1": number = 1; number *= number; break; ... } return number; }

int number = 0; for (int i = 0; i < strlen(str); i++) { switch (i) { case '1': number +=1; break; ... } number *= 10; } return number; } I bolded what you missed

Woops, yeah thanks. Is it bad to leave variables unitialized like at the top??

int number = 0; for (int i = 0; i < strlen(str); i++) { switch (i) { case '1': number +=1; break; ... } number *= 10; } return number; } I bolded what you missed

int number = 0; for (int i = 0; i < strlen(str); i++) { switch (i) { case '1': number +=1; break; ... } number *= 10; } return number; } I bolded what you missed

Swich case really is not suited to this

Do you have anything else?!