odd , even
Sat, 2012-06-30 14:31
i knew while typing the code that it wont work , and it did'nt , so here's the code
#include <stdio.h>
int main()
{ int i;
int x;
float y;
printf("enter a number");
scanf("%d",i);
if(i/2 == x)
printf("%d is even",i);
else if (i/2 == y)
printf("it is odd");
else printf("its not a number");
return 0;
}
i think i know the problems but still i'd like anyone to explain in detain all that went wrong
thanks in advance
Sun, 2012-07-01 08:41
#2
Thanks , i thought that too . I dont know about mod , i'll learn it later .
any other way to do the program without mod?
Fri, 2012-07-06 16:41
#3
Are you just trying to write a program that will tell the user if their number is even or odd? If so, you don't need x, or y or anything... Just the user's number. Here is working code, let me know what you think:
#include <stdio.h>
int main() {
int i;
char* result = NULL;
printf("Please enter a number: ");
scanf("%d", &i);
if(i % 2 == 0)
result = "even";
else
result = "odd";
printf("%d is an %s number.\n", i, result);
return 0;
}
Fri, 2012-07-06 18:17
#4
ok thanks . Thats much simpler .
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I think that your problem is that you've declared the type for int x and float y, but not the value.
You seem to be saying that x is an integer and y is a float. Compare i to x to see if it is an integer.
However, C thinks you are saying that whatever is in x is an integer, but you never said what value integer it should be. So C picks a random number - 45678, and if I make i = 4, it is not the same as 45678.
I think your program will say that i is not x and not y, unless you win the lottery!
The way I would do the program would be to use the % which is the mod command.
4 % 2 == 0 is true because 4/2 has no remainder. 5 % 2 == 1 means that there is a remainder of 1 and so 5 is odd. Anything else is not a number. (Although I am not sure what C will make of characters in this case.)
Hope I'm right!