How to count Integer array elements like char based strlen() in C? - c

Question

I would like to count the number of elements in an integer array(sized, like: array[1000]) after input without counting manually while inputting(using scanf() which is = number of arguments passed to scanf).Though int arrays are not null terminated as well as scanf() cannot be used like getchar() or gets() and no availale function like strlen() for int arrays,is it possible to write a C program to prompt the user to input as many numbers as he wishes and the program will count them(total arguments passed to scanf()) and print the maximum using arrays or pointers?

Answer 1

Without having a termination value, you will have to count the inputs as they are made. You could do this by defining a struct to hold the array. Your program does not know how many integers you will enter, and this code allocates more memory when the array is full, keeping track of the array size and elements used.
#include <stdio.h>
#include <stdlib.h>
#define ARRAY_STEP 10 // increment in array size
typedef struct {
int maxlen; // capacity of array
int length; // elements used
int *array; // the array pointer
} istruct;
int main(void) {
istruct intarr = {0}; // empty array
int i;
printf ("Enter some integers:\n");
while (scanf("%d", &i) == 1) {
if (intarr.length >= intarr.maxlen) {
intarr.maxlen += ARRAY_STEP; // increase array size
intarr.array = realloc(intarr.array, intarr.maxlen * sizeof(int));
if (intarr.array == NULL) {
printf("Memory allocation error\n");
exit (1);
}
}
intarr.array[intarr.length++] = i;
}
printf ("You entered:\n");
for (i=0; i<intarr.length; i++)
printf ("%-10d", intarr.array[i]);
printf ("\n");
if (intarr.array)
free(intarr.array);
return 0;
}

Answer 2

strlen() iterates on the character array until the C string termination character (ASCII \0) is encountered. So it is COUNTING the elements. If you want to do that for an int array I guess you could also have a reserved 'terminator' value for your array, make room for it (allocate one more int in your array for the terminator) and implement your own int_array_len() similar to strlen. However, in your case counting the elements while inputting them seems like a much better way to go.

Answer 3

Smart vertion:
#include <stdio.h>
int main()
{
double a[100000],mx=0;
int i,j,c=0;
printf("Enter as many numbers as you wish . Press Ctrl+D or any character to stop inputting :\n");
/*for(i=0;i<100000;i++)
a[i]=0;*/
for(i=0;i<100000;i++)
{
if((scanf("%lf",&a[i]))==1)
c++;
//else break;
}
for(j=0;j<c;j++)
{
if(a[j]>mx) mx=a[j];
}
printf("You have inserted %d values and the maximum is:%g",c,mx);
return 0;
}

Answer 4

It's pretty simple.
just declare of your desired "initial size". And keep checking if input is a valid integer or not.
if((scanf("%d",&a[i]))==1) this would be true if and only if the input is a valid integer. hence the moment the input is not an int, it exits the loop. You can count the number of input values within the same loop and also the max values.
Here is the code:
#include <stdio.h>
#include<limits.h>
int main()
{
int a[100],max = INT_MIN;
int i,j,count=0;
printf("Start Entering the integers... Give any non-integer input to stop:\n");
for(i=0;i<100;i++)
{
if((scanf("%d",&a[i]))==1) {
count++;
if(a[i]>max) {
max = a[i];
}
}
else
break;
}
printf("number of input values: %d;\nThe maximum input value: %d",count,max);
return 0;
}