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Please look through the code given below; there might be something wrong. You are asked to state what parts are missing or wrong, explain how to fix the code. Then tell us what the output will be

** This is an essay question **

Code

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
	int a, i;
	printf("Enter a number: ");
	scanf
	
	if(a=0){
		printf("Please do not enter a negative number or zero")
	}
	
	else
	
	for(i=1;i<=a;i+){
		if(i%2!=0){
		printf("\n %d--Odd", i);
		
		printf("\n %c--Even", i);
	}
	
	return 0;
}

 

****    % = reminder of division

[Exercise] Essay (Open question) - asked in Chapter 3: Branching, Looping and Functions by (5.9k points)
ID: 30244 - Available when: Unlimited - Due to: Unlimited
| 2k views
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What do you mean by reminder of division? do we have to divide by 2?
0 0
Hi.

It's just an explanation of an operation called 'Moderator'. By using it, it keeps a remainder number. Such as
2%4 = 2        remainder = 0
2%5 =  2        remainder = 1

14 Answers

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50/100 answered by (221 points)
edited by
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50/100 answered by (244 points)
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50/100 answered by (183 points)
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Hidden content!
#include <stdio.h>

#include <stdlib.h>



int main(int argc, char *argv[]){



int a, i;



  printf("Enter a number: ");



  scanf("%d",&a);



    if(a<=0){



  printf("Please do not enter a negative number or zero");



}



 else{



 for(i=1;i<=a;i++){


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Conclusion: This code is to check the number is odd or even. If the input: 0 or negative number, it will say:Please do not enter a negative number or zero.
0/100 answered
edited by
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Hidden content!
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100/100 answered by (243 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>



#include <stdlib.h>



int main(int argc, char *argv[])



{



  int a, i;



  printf("Enter a number: ");



  scanf("%d",&a);



  



  if(a==0){



  printf("Please do not enter a negative number or zero");



}



 else{



 for(i=1;i<=a;i++){


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  return 0;



}
50/100 answered by (269 points)
0 like 0 dislike
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50/100 answered by (153 points)
edited by
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100/100 answered by (283 points)
edited by
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50/100 answered by (269 points)
0 like 0 dislike
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100/100 answered by (273 points)
0 0
Excellent job
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