Find the maximum, minimum, and average values in the array - c

Question

In my code, the program will not allowed the negative number entered, the program will stop reading, then calculate the maximum value, minimum value and average value.
That is my code
#include <stdio.h>
int main(void) {
int age[10] = {0}; // initalized an array
printf("Please enter ages: \n"); // allow user to enter numbers
for (int i = 0 ;i < 10; i++) {
scanf("%d",&age[i]);
if (age[i] < 0) { // if it is negative number, it is should stop reading
break;
}
else if (age[i] >= 0) {
continue;
}
}
int maximum = age[0];
int minimum = age[0];
float average = 0.0;
int length = sizeof(age) / sizeof(age[0]);
for (int j = 0; j < length; j++) {
if (maximum < age[j]) {
maximum = age[j];
}
else if (minimum > age[j]) {
minimum = age[j];
}
average += age[j];
}
average = average / length;
printf("%d\n", maximum);
printf("%d\n", minimum);
printf("%.1f\n", average);
return 0;
}
Please enter ages: 5 -1
expected result: max:5;min:5,average:5;
actual result: max:5;min:-1,average: 0.4;
That was a question that I met, the code should not accept any negative value.
Thank you all.
but if I add age[i] = 0; then break;
The average value will equal to 0.5.

Answer 1

You don't need an array.
You don't need both a loop variable and a length.
It's more appropriate to use ? : for updating minimum/maximum.
You don't need two loops
You need to check the int return value of scanf(), which indicates the number of items successfully scanned, so it should be 1. I'll leave that for you/OP to add (hint: replace for-loop by while-loop to avoid having to add a separate length variable again).
int main(void)
{
printf("Please enter ages: \n");
int minimum = INT_MAX;
int maximum = 0;
int sum = 0;
int count = 0;
for (count = 0; count < 10; count++)
{
int age;
scanf("%d", &age);
if (age < 0)
{
break;
}
sum += age;
minimum = (age < minimum) ? age : minimum;
maximum = (age > maximum) ? age : maximum;
}
if (count > 0)
{
printf("Min: %d\n", minimum);
printf("Max: %d\n", maximum);
printf("Avg: %.1f\n", (float)sum / count);
}
else
{
printf("You didn't enter (valid) age(s).\n");
}
return 0;
}

Answer 2

Your approach is overly complicated and wrong.
You want this:
...
int length = 0; // declare length here and initialize to 0
for (int i = 0; i < sizeof(age) / sizeof(age[0]); i++) {
scanf("%d", &age[i]);
if (age[i] < 0) // if it is negative number, it is should stop reading
break;
length++; // one more valid number
}
// now length contains the number of numbers entered
// the rest of your code seems correct
You also might need to handle the special case where no numbers are entered, e.g: the only thing entered is -1. It doesn'make sense to calculate the average or the largest/smallest number when there are no numbers.

Answer 3

A possible solution could be:
(corrections are written in the commented code)
#include <stdio.h>
int main(void){
int arraySize = 10;
int age[arraySize]; //initialize not required
//the number of existing values inside the array (effective length)
int length = 0;
printf("Please enter ages: \n"); // allow user to enter numbers
for(int i=0; i<arraySize; i++){
scanf("%d",&age[i]);
// if it is negative number, it is should stop reading
if(age[i]<0){ break; }
//the else-if is not required
//but, if the compiler goes here,
//it means that the value is acceptable, so
length++;
}
int maximum = age[0];
int minimum = age[0];
float average = 0.0;
for(int j=0; j<length; j++){
if(maximum<age[j]){ maximum = age[j]; }
else if(minimum>age[j]) { minimum = age[j]; }
average += age[j];
}
average = average / length;
printf("%d\n", maximum);
printf("%d\n", minimum);
printf("%.1f\n", average);
return 0;
}

Answer 4

OP's primary problem is the 2nd loop iterates 10 times and not i times (the number of times a non-negative was entered.
For fun, let us try a non-floating point solution as it really is an integer problem.
An array to store values is not needed.
#include <limits.h>
#include <stdio.h>
int main(void) {
// Keep track of 4 things
int min = INT_MAX; // Set min to the max int value.
int max = INT_MIN;
long long sum = 0; // Use wide type to cope with sum of extreme ages.
int count = 0;
#define INPUT_N 10
printf("Please enter ages: \n");
for (count = 0; count < INPUT_N; count++) {
int age;
if (scanf("%d", &age) != 1) {
fprintf(stderr, "Missing numeric input.");
return EXIT_FAILURE;
}
if (age < 0) {
break;
}
if (age < min) min = age;
if (age > max) max = age;
sum += age;
}
if (count == 0) {
fprintf(stderr, "No input.");
return EXIT_FAILURE;
}
printf("Maximum: %d\n", max);
printf("Minimum: %d\n", min);
// Could use FP and
// printf("Average: %.1f\n", 1.0 *sum / count);
// But for fun, how about a non-FP approach?
#define SCALE 10
#define SCALE_LOG 1
sum *= SCALE; // Scale by 10 since we want 1 decimal place.
// Perform a rounded divide by `count`
long long average_scaled = (sum + count/2) / count;
// Print the whole and fraction parts
printf("Average: %lld.%.*lld\n",
average_scaled / SCALE, SCALE_LOG, average_scaled % SCALE);
return 0;
}

Answer 5

First of all, you must record how many positive numbers you enter. Then the value of length will be correct.
Second, for the second for loop, j must be smaller than the number of positive ages. Therefore, you won't add negative age[j] to average.
You can simply modify the second for loop.
#include <stdio.h>
int main(void) {
int age[10] = {0}; // initalized an array
printf("Please enter ages: \n"); // allow user to enter numbers
int length = 0;
for (int i = 0 ;i < 10; i++) {
scanf("%d",&age[i]);
if (age[i] < 0) { // if it is negative number, it is should stop reading
break;
}
else if (age[i] >= 0) {
length++;
continue;
}
}
int maximum = age[0];
int minimum = age[0];
float average = 0.0;
for (int j = 0; j < length; j++) {
if (maximum < age[j]) {
maximum = age[j];
}
else if (minimum > age[j]) {
minimum = age[j];
}
if ( age[j] > 0.0 )
{
average += age[j];
}
}
average = average / length;
printf("%d\n", maximum);
printf("%d\n", minimum);
printf("%.1f\n", average);
return 0;
}