how do I get the minimum to print properly? - c

Question

This is supposed to ask the user to input a number than if the number is positive it outputs the number and adds one to the count and outputs the value of count (just for the most recent input value). if the number input is negative it just asks for a new number and if the input is 0 it outputs the count, minimum, maximum, sum and average for the entire run of the program.
I've gotten most of it working I'm just having trouble with getting the minimum value for the 0 scenario. I also get a runtime error for int min
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int main() {
int input = 7;
int count = 0;
int sumInput = 0;
int lastInupt = 0;
int max;
int min;
double avg;
while (input != 0) {
printf("enter a number\n");
scanf("%i", &input);
if (input > 0) {
count++;
printf("\ninput = %i", input);
printf("\ncount = %i\n", count);
/*input1 = input;*/
sumInput += input;
//lastInupt = input;
}
else if (input < 0) {
printf("\ninput = %i", input);
printf("\ncount = %i\n", count);
}
if (input > lastInupt) {
if (input > max) {
max = input;
/*input1 = input;*/
}
}
if (lastInupt > input) {
if (min > input){
min = input;
}
}
/*while (min > 0)
{
if (input < min) {
min = input1;
}
}*/
}
avg = (float)sumInput / (float)count;
printf("you have entered %i positive numbers..\n", count);
printf("Sum of inputs: %i", sumInput);
printf("\nThe maximum # is: %i", max);
printf("\nThe average is: %.2lf", avg);
printf("\nThe minimum # is: %i", min);
getchar();
getchar();
return 0;
}
The code has to keep the same general format this is an assignment for my college.

Answer 1

Consider initializing both min and max and skip the lastInput variable.
int min = INT_MAX;
int max = INT_MIN;
.....
if (input > max)
max = input;
if (min > input)
min = input;
This should ensure that min and max are always initialized, and in the special case where the first and only input is 0 both min and max will get the value 0.
Because you initialize min to MAX_INT (the highest possible value for an int variable) the first iteration min > input will always evaluate to true and min will be updated. Similar for max and MIN_INT
if min and max only should be updated when input !=0 then modify the entire program to
int main() {
int input = 7;
int count = 0;
int sumInput = 0;
int max = INT_MIN;
int min = INT_MAX;
double avg;
while (input != 0) {
printf("enter a number\n");
scanf("%i", &input);
if (input > 0) {
count++;
printf("\ninput = %i", input);
printf("\ncount = %i\n", count);
sumInput += input;
if (input > max ) max = input;
if (input < min ) min = input;
}
else if (input < 0) {
printf("\ninput = %i", input);
printf("\ncount = %i\n", count);
}
}
if(count==0) { // only 0 entered.
printf("inform user she didnt enter any positive numbers");
} else {
avg = (float)sumInput / (float)count;
printf("you have entered %i positive numbers..\n", count);
printf("Sum of inputs: %i", sumInput);
printf("\nThe maximum # is: %i", max);
printf("\nThe average is: %.2lf", avg);
printf("\nThe minimum # is: %i", min);
}
getchar();
getchar();
return 0;
}

Answer 2

Since your runtime error was fixed, I will address the min == 0 issue.
When a 0 is input, it will fail the two first if statements input>0 and input < 0
But it still will try the last two if statements, and of course 0 will be less than the current min, so it will set min to 0 (incorrectly)
What you need to do is put this after the else if(input < 0) block
else{//input is 0
break;
}
So that it will skip checking if input is below min when input is 0. As you want to break from the loop at this point anyway.