Question

Write a program that reads 8 floats into an array and then prints out the second, fourth, sixth and eighth members of the array, and the sum of the first, third, fifth and seventh members of the array.

Answer 1

What have you got so far?

Answer 2

Also, what language are you using :-D

you can paste your code in a site like http://ideone.com

Answer 3

[If he shows us what he got so far, I was assuming we could deduce the language from that :-)]

Answer 4

OK. Its not my assignment. Its my friend Alex's assignment. Due to electricity shortfall he has lost his PC (In PC he had saved his assignment) and after 3 hours he has to submit his assignment.

He asked me that if there anyone who could volunteerily make at least 2 programs for him.
So I said yes may be experts on openstudy could help you.

It is Turbo C

Honestly I know C language. but till functions. not array..

Kindly people tell me what should I do now? :(

Answer 5

unlucky guy :(

To begin, you can declare a float array of size 8.

Then, you can use a for loop, using the scanf() function to read floating point values one-by-one into the array.

For the last two (first printing even elements, and then printing odd elements), you can use two more for loops to read values from the array.

Answer 6

If I try program here. then could you help me step by step @agdgdgdgwngo ? I will appreciate your kindness for me.

Answer 7

sure

Answer 8

Thanks

Answer 9

#include<stdio.h>
#include<conio.h>
void main(void)
{
float my [] = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8}







Is this declaration is right?

Answer 10

it would work, but since you are reading values from stdin you don't really have to fill the array with {1,2,3,4,5,6,7,8}.

you can use an empty array instead, like:

float my[8];

Answer 11

ok you mean like this.

#include<stdio.h>
#include<conio.h>
void main(void)
{
float my [8];

Answer 12

right!

alright that's part 1 of the problem. now you need to read floats from input.

Answer 13

#include<stdio.h>
#include<conio.h>
void main(void)
{
clrscr();
float my[8]
for(int i=0;i<8;i++)
scanf("%f",my[i]);
getch();
}


Is it right? I have taken input.

Answer 14

it's almost right. you just need to watch out at scanf. :)

the second argument to scanf must be a pointer. instead you are using a float.

you must do some pointer arithmetic to get the i'th address of your float array. You can do

scanf("%f", my + i);
or you can do

scanf("%f", &(my[i]));

Answer 15

#include<stdio.h>
#include<conio.h>
void main(void)
{
clrscr();
float my[8];
for(int i=0;i<8;i++)
scanf("%f",&my[i]);
getch();
}


Is it right ? Now I would have ith address.

Answer 16

it was 2nd step. But I am confused at 3rd step, what you said about even and odd :(

Answer 17

he abstracted from "second, fourth, sixth and eighth" and "first, third, fifth and seventh "

Answer 18

First, the 2nd, 4th, 6th, and 8th items of the array need to be printed out.

You need to loop the array from i = 1 (the second item) to i = 7 (the last item), incrementing i by 2.

something like

for (i = 1; i < 8; i += 2)
{
/* print out the ith element of the array */
}

Answer 19

@nczempin yes I know about even and odd. but how to do this in array.. kindly give some hint.

Answer 20

just one additional for loop would suffice. You can do the printing and the summing within the 1 loop

Answer 21

OK now I try as agdgdgdgwngo says.

Answer 22

for (int i =0; i<8; i++){ //sorry, Java programmer
if (i%2==0){ // i is even, so the position within the array is odd
//do your printing
} else{
//do your summing
}
}

Answer 23

#include<stdio.h>
#include<conio.h>
void main(void)
{
clrscr();
float my[8];
for(int i=0;i<8;i++)
{
scanf("%f",&my[i]);
}
for(i=1;i<8;i+=2)
{
printf("%f",my[i])
}
getch();
}



Fine?

Answer 24

right; that prints out the even-indexed elements.

now print out the odd ones (first, third, fifth, seventh)

Answer 25

@agdgdgdgwngo we are using here two for loops. and nczempin used one. So What you think Which one is easy.

I am asking because you are expert people in languages..

Answer 26

they both work; whatever you find easier to understand

Answer 27

my bad :-P I got it wrong

do it nczempin's way; you are supposed to sum the odd ones :-P

Answer 28

The % operator may not be on everyone's radar ;-)

Answer 29

Its OK agdgdgdgwngo.. But be here. I need both of you experts. Now I am trying nczempin's way.

Answer 30

well, just change "now print the odd ones" to "now sum the odd ones"

Answer 31

lol. standing by, at your service :-)

Answer 32

void main() is not standard.

Answer 33

Please don't use it.

Answer 34

#include<stdio.h>
#include<conio.h>
void main(void)
{
clrscr();
float my[8];
for(int i=0;i<8;i++)
scanf("%f",my[i]);
for (int i =0; i<8; i++)
if (i%2==0)
printf("%f",my[i]);
else
HOW TWO SUM ARRAY ELEMENTS HERE?
getch();
}


Kindly look remaining code is fine?

Answer 35

FFM so how?

Answer 36

scanf("%f",my[i]); --> Last time I check if it's not a character array or pointer it needs '&'

Answer 37

@FFM sorry friend I forgot & sign. But look remaining code. except else body.

Answer 38

Google "void main() vs int main()". main() returns an integer (always).

Answer 39

for the sum, you can declare a `sum` variable before the loop and set it to zero.

float sum = 0;

and in the part of the loop where you add to the sum, you can have something like

sum += my[i];

so at the very end of the program you can print out the sum using printf

Answer 40

#include<stdio.h>
#include<conio.h>
void main(void)
{
clrscr();
float my[8]sum=0;
for(int i=0;i<8;i++)
scanf("%f",&my[i]);
for (int i =0; i<8; i++)
if (i%2==0)
printf("%f",a[i]);
else
sum+=my[i];
printf("print sum %f",m[i]);
getch();
}


Fine agdgdgdgwngo? Is it right now?

Answer 41

@FFM I have gotten what you have said about main() function. and in the last i will return 0

Answer 42

int main(){
float Sum = 0.0, F[8];
int i;

//Reading the inputs:
for( i=1; i<= 8;i++)
scanf("%f", &F[i-1]);
printf(" The Even terms are:\n");
for( i=1; i<= 8;i++){
if(i &1)
Sum += F[i-1];
else
printf("%f ",F[i-1]);
}
printf("\n Sum of Odd terms: %f\n",Sum);
return 0;
}

I haven't debugged but this should work!

Answer 43

Thanks a lot FFM but I want you to describe lines.
scanf("%f", &F[i-1]);
i-1 implies what?

Answer 44

See I have set i from 1 to n but the array is starting from 0 so just memory optimization.

Answer 45

it implies that what we call the "first" element is really the "zeroth" in C, etc. I had to make a similar adjustment in my pseudocode.

Answer 46

@Shayaan
http://ideone.com/G15BY you just need to clean up the syntax and then it should compile and run since we are doing the right steps

let's see if fffm's code works:
http://ideone.com/l7VOX

yeah it works correctly

Answer 47

and i&1 is one step above even the % operator ;-)

Answer 48

bitwise operations ftw

Answer 49

it may be a little suspicous if you're using bitwise and or modulo but have never learned it :)

Answer 50

One last thing @agdgdgdgwngo how did you compile the code?

Answer 51

so an "easier" way of doing it (checking even vs. odd) would be to divide by two and multiply back by two; if it's the same number as the original, the original was even.

So come to think of it, just doing an additional for loop, while not as elegant, is probably easier to understand for a beginner.

Answer 52

Or even easier would be
sum = my[0]+my[2]+my[4] etc.

Answer 53

for size 8 that's fine, it would just get a little tedious for 800 :-)

Answer 54

I just used http://ideone.com to build and run the code

ideone.com expects plain C thogh, so if you remove the #inlude <conio.h> and the corresponding functions (getch() and clrsr()) your code should run as well.

Answer 55

thanks all 3 of you. Even I don't know how to work with arrays but you guys are so helpful as I said to my friend.

Thanks @agdgdgdgwngo @nczempin @FoolForMath

Answer 56

Now I have one medal so whom should I give it..
hmm...
I suggest not to give it to anyone of you because I all 3 equally helped me.

Should I give it or not friends?

Answer 57

I think it is given to foolformath because you both have already one :D lol

Answer 58

yea

Answer 59

that's fine

Answer 60

But I want you people always be there for my help where I need it. OK guys?

Answer 61

It is first program that I have made with the help of you experts. Thanks once again.

Answer 62

sry, can't promise ;-)

Answer 63

I know you can't promise. But you can try atleast.

Answer 64

just one request:
Please _always_ put brackets around "if clauses". Trust me, you'll thank me eventually

Answer 65

and always follow foolformath's suggestion use int main! :-D (...well, except for a few VERY special cases)

Answer 66

@nczempin OK I will consider this for next time. You will not see again without brackets. thnx for this.

Answer 67

@agdgdgdgwngo yes. I wil follow his suggestion. Thnx