Why is this for loop getting ignored? [closed] - c

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I'm not sure what i'm doing wrong but the for loop is not initializing
The code just goes immediately to displaying the printfs. That have no values in them since the for loop didn't activate
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#define PAUSE system("Pause")
main() {
// INITALIZE VARIABLES
int number = 0;
int i = 0;
int odd = 0;
int even = 0;
int totalNum = 0;
int tempNum = 0;
int count;
printf("Enter a number between 2 and 25\n");
scanf("%i", &number);
do{
if (number < 2 || number > 25)
printf("That was an invalid number please try again\n");
scanf("%i", &number);
} while (number < 2 || number > 25);
printf("Enter how many numbers you want to input\n");
scanf("%i", &count);
for (i = 1; i == count; ++i){
printf("input numbers\n");
scanf("%i", &tempNum);
if (tempNum % 2 == 0)
even++;
else
odd++;
totalNum = totalNum + tempNum;
} // END FOR LOOP
// DISPLAY OUTPUT
printf("You entered %i numbers\n", count);
printf("The sum of the %i numbers is %i\n", count, totalNum);
printf("The average of the %i numbers is %i\n", count, totalNum / count);
printf("You entered %i odd numbers and %i even numbers\n", odd, even);
PAUSE;
} // END MAIN

Answer 1

Your loop will only execute at best once, when count == 1 as you initialize i to 1.
If you enter a 1 for count,
printf("Enter how many numbers you want to input\n");
scanf("%i", &count);
the loop will run exactly once, until i increments to 2
You probably want:
for (i = 1; i <= count; ++i){

Answer 2

do{
if (number < 2 || number > 25)
printf("That was an invalid number please try again\n");
scanf("%i", &number);
} while (number < 2 || number > 25);
it should be...
do{
if (number < 2 || number > 25){
printf("That was an invalid number please try again\n");
scanf("%i", &number);
}
} while (number < 2 || number > 25);
else it asks always another number

Answer 3

i = 1, so i == count; gives false therefore the loop is ignored.

Answer 4

A for loop in C works like this:
for ( variable initialization; condition; variable update ) {
/* Do something... */
}
The loop will execute for as long as condition is true. So, when you do:
for (i = 1; i == count; ++i)
The loop will execute for as long as i == count is true. So, unless count holds 1 when this line is executed, the loop will never run.
As others pointed out, you probably want this:
for (i = 1; i <= count; ++i)
So your loop will run for all values of i, until it reaches count.
As a side note i should point out that the usual way to write for loops in C is something like this:
for (i = 0; i < count; i++)
We start with i = 0 because C arrays are zero-based, so the Nth element of an array has index n-1

Answer 5

You were so close. In addition to fixing your loop test clause for (i = 1; i <= count; i++), I would suggest using " %d" for your format specifier. Your do loop need only be a while loop to avoid printing your invalid number message every time.
Additionally, While not an error, the standard coding style for C avoids caMelCase variables in favor of all lower-case. See e.g. NASA - C Style Guide, 1994.
With those changes, (and changing your odd/even check to a simple &) you could write your code as follows.
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
// #define PAUSE system("Pause")
int main (void)
{
int number, i, odd, even, totalnum, tempnum, count;
number = i = odd = even = totalnum = tempnum = count = 0;
printf ("enter a number between 2 and 25: ");
scanf (" %d", &number);
while (number < 2 || number > 25) {
printf ("invalid number, again (between 2 and 25): ");
scanf (" %d", &number);
}
printf ("numbers to input: ");
scanf (" %d", &count);
for (i = 1; i <= count; i++) {
printf ("input number %2d: ", i);
scanf (" %d", &tempnum);
if ((tempnum & 1) == 0)
even++;
else
odd++;
totalnum = totalnum + tempnum;
}
printf ("You entered %d numbers\n", count);
printf ("The sum of the %d numbers is %d\n",
count, totalnum);
printf ("The average of the %d numbers is %d\n",
count, totalnum / count);
printf ("You entered %d odd numbers and %d even numbers\n",
odd, even);
// PAUSE;
return 0;
}
note: main is type int (e.g. int main (int argc, char **argv) or simply int main (void) to indicate no arguments taken). Since it is type int it will return a value to the shell. While historic implementations may have allowed void main that is no longer the case for portable code.
Example Use/Output
$ /bin/forskipped
enter a number between 2 and 25: 4
numbers to input: 4
input number 1: 1
input number 2: 2
input number 3: 3
input number 4: 4
You entered 4 numbers
The sum of the 4 numbers is 10
The average of the 4 numbers is 2
You entered 2 odd numbers and 2 even numbers
Look it over and let me know if you have any questions.