find a number whether it is even or odd without using any
control structures and relational operators?
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Answer / gganesh
#include<stdio.h>
#include<conio.h>
void main()
{
int n;
clrscr();
printf("Enter the number : ");
scanf("%d",&n);
n%2?printf("Odd"):printf("Even");
getch();
}
note: here i used conditional operator, not a relational
operators and control structures
| Is This Answer Correct ? | 3 Yes | 1 No |
Answer / vignesh1988i
THE LAST two answers posted by two folks are correct but the
declarations have been made wrong...... we cant make use of
1D array here , if so only 'e' or 'o' only will get
printed.... but that is not our aim... so correct
declaration is using a 2D array.....
char a[][6]={{"even"},{"odd"}};
and also it is not the must to make use of array of pointers
concept...........
thank u
| Is This Answer Correct ? | 3 Yes | 2 No |
Answer / sanju uthaiah
#include<stdio.h>
int main()
{
char result[2]={"Even","Odd"};
int n=40;
printf("%d is %s",n,result[n%2]);
return 0;
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / vishal jain
#include<stdio.h>
#include<conio.h>
int main()
{
int num;
printf("Enter any Number : \n");
scanf("%d",&num);
char *s[2]={"Even","Odd"};
printf("%s",s[num&1]);
return 0;
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / vishal jain
#include<stdio.h>
#include<conio.h>
int main()
{
int num,i;
printf("Enter any Number : \n");
scanf("%d",&num);
i=num&1;
if(i==1)
{
printf("ODD");
}
else
{
printf("EVEN");
}
return 0;
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / mohd parvez 09311349697
#include<stdio.h>
#include<conio.h>
void main()
{
int num;
printf("Enter a number:");
scanf("%d",&num);
num%2&&printf("Number is ODD")||printf("Number is EVEN");
getch();
}
| Is This Answer Correct ? | 1 Yes | 2 No |
#include<stdio.h>
#include<conio.h>
main()
{
int n;
char *p[]={"Even","odd"};
clrscr();
printf("Enter the number");
scanf("%d",&n);
n=n%2;
printf("The value is %s",p[n]);
getch();
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / om
void odd_even_check(int z)
{
(z&1)?printf("\nodd\n"):printf("\neven\n");
}
| Is This Answer Correct ? | 1 Yes | 5 No |
Answer / asad
int oddeven(int n)
{
if(n&1)
return 1; //odd
else
return 0; //even
}
| Is This Answer Correct ? | 2 Yes | 7 No |
Answer / satyanarayana
#include<stdio.h>
void main()
{
int p;
printf("number:");
scanf("%d",&p);
while(p%2)
{
printf("odd");
}
printf("even");
}
| Is This Answer Correct ? | 2 Yes | 10 No |
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