C programming with arrays - c

Question

This was the question
int num,square;
int array1[100];
int array2[100];
for(int i=0;i<100;i++){
printf("please enter the number %d:",i+1);
scanf("%d",&num);
if(num==-1){
break;
}
else
array1[i]=num;
}
for(int j=0;j<100;j++){
square=array1[j]*array1[j];
array2[j]=square;
}
for(int m=0;m<100;m++){
printf("number %d is %d\n",m,array2[m]);
}
this was the program i made but then the user might input only few int so i want to run the loop which prints the second array only the amount of tym the user inputted how can i do that?

Answer 1

Declare the first loop variable (i) outside the for.
This will remember the point at which the first loop exited.
Then use that for ending second loop instead of a hard-coded value i.e. 100.
int num,square;
int array1[100];
int array2[100];
int i=0;
for( i=0; i < 100; i++ )
{
printf("please enter the number %d:",i+1);
scanf("%d",&num);
if(num==-1)
{
break;
}
else
array1[i]=num;
}
for( int j=0; j < i; j++ )
{
square = array1[j] * array1[j];
array2[j] = square;
}
for(int m=0; m < i; m++)
{
printf("number %d is %d\n", m, array2[m]);
}

Answer 2

Your problem is when the user enters -1 and breaks the input loop. That will leave parts of array1 uninitialized. You are not allowed to access the uninitialized variables. Therefore you need to change the next two for-loops like:
for(int j=0;j<i;j++){
// ^ notice: i instead of 100
square=array1[j]*array1[j];
array2[j]=square;
}
for(int m=0;m<i;m++){
// ^ notice: i instead of 100
printf("number %d is %d\n",m,array2[m]);
}
This requires that the first loop is written like:
int i;
for(i=0;i<100;i++){
so that i is still valid when reaching the next loops.
In this way you avoid accessing uninitialized parts of array1 and you only print the same amount of values as the user did input (before -1).
Beside that... doing:
scanf("%d",&num);
is bad. You shall always check the value returned be scanf to make sure that it scanned the expected number of values, i.e.
if (scanf("%d",&num) != 1)
{
// Illegal input
....
Add error handling here
....
}

Answer 3

Just keep the number of entered numbers in a variable.
To store a product of two objects of the type int it is better to use type long long int.
Also it is not a good idea to use magic numbers as 100 throughout the code.
The program can look the following way
#include <stdio.h>
#define N 100
int main( void )
{
int array1[N];
long long int array2[N];
size_t n = 0;
for ( ; n < N; n++ )
{
int num;
printf("please enter the number %zu: ", n + 1);
if (scanf("%d", &num) != 1 || num == -1) break;
array1[n] = num;
}
for ( size_t i = 0; i < n; i++ )
{
long long int square = (long long int )array1[i] * array1[i];
array2[i] = square;
}
for ( size_t i = 0; i < n; i++ )
{
printf( "number %zu is %lld\n", i, array2[i]);
}
return 0;
}

Answer 4

I have updated your code.
int num,square;
int array1[100];
int array2[100];
int i=0;
for(i=0;i<100;i++){
printf("please enter the number %d:",i+1);
scanf("%d",&num);
if(num==-1){
break;
}
else
array1[i]=num;
}
/* This will only calculate square upto the numbers you have entered
I have changed the condition to <i
*/
for(int j=0;j < i;j++){
square=array1[j]*array1[j];
array2[j]=square;
}
/* This will only print upto the numbers you have entered
I have changed the condition to <i
*/
for(int m=0;m<i;m++){
printf("number %d is %d\n",m,array2[m]);
}