I'm taking CS courses, and wrote a simple program to find an average of the given (input) amount of inputs.
#include <stdio.h>
#include <cs50.h>
int main(void)
{
int n = get_int("Please enter the number of scores: ");
int scores = [n];
for(int i=0; i<n; i++)
{
int scores[i] = get_int("Please, enter the score: ");
}
printf("Average: %f\n", average(n, scores));
}
float average(int length, int array[])
{
int sum = 0;
for(int i = 0; i < length; i++)
{
sum += sum + array[i];
}
return sum / (float)length;
}
When compiling, "expected expression" error occurs on line 10 (i.e. {int scores = [n];} —
any help or suggestions would be appreciated!!
This is invalid syntax:
int scores = [n];
If you want to define scores as an array of size n you want:
int scores[n];
Also, this doesn't do what you might expect:
int scores[i] = get_int("Please, enter the score: ");
This creates an array called scores of size i, masking the array defined previously, and attempts to initialize it with a single value instead of a list of values. If you want to assign to the existing array you want:
scores[i] = get_int("Please, enter the score: ");
This a dot_product function of 2 vectors of the same length.
I don't understand how to build the array because how the machine will know which input goes to which input (for example i want a={1,2,3} but the input of 123 will come a[0]= 123)...
How do I make end of array[index] input and how do I make end of the whole array.
#include <stdio.h>
#include <stdlib.h>
#define MAXINPUT 100
int dot_product(int v[], int u[], int n)
{
int result = 0;
int i;
for (i=0; i < n; i++)
result += v[i]*u[i];
return result;
}
int main(){
int v1[MAXINPUT];
int v2[MAXINPUT];
int count = 0
int i,print;
printf(" first vector:");
for(i=0;i<MAXINPUT;i++){
scanf("%d", &v1[i]);
count +=1;
}
printf(" second vector:");
for(i=0;i<MAXINPUT;i++)
scanf("%d", &v2[i]);
print = dot_product(v1, v2, count);
printf("v1*v2:%d",print);
return 0;
}
The first problem I observe here is with
count +=1;
where count is an uninitialized automatic local variable, which makes it's initial value indeterminate. Attempt to use that value invokes undefined behavior.
You should be initializing count to 0.
That said, here, you're depending on the user to input the second array with exact same dimension of that of the first one. In case that does not happen, your program will blow up, as you did not initialize the arrays, again.
I tried to solve the Small Factorial problem on SPOJ and got 'Wrong answer'. I wrote the following code :
#include <stdio.h>
int main() {
int t,i, num;
unsigned long long int fact=1;
scanf ("%d", &t);
while (t-- >0) {
fact=1;
scanf ("%d", &num);
for (i=num; i>0; i--) {
fact*=i;
}
printf ("%llu\n",fact);
}
return 0;
}
This code is not finding factorial for large inputs like 100. What are the changes required?
In C and C++ you have the maximum allowed range from -2^63+1 to +2^63-1 which is for long long data type. As you can see this range is still too small to store >20 digits.
You have to use a char array to store single digit per array index.
one way to do this in C++:
#include<stdio.h>
#include<iostream>
int main()
{
int t;
char a[200]; //array will have the capacity to store 200 digits.
int n,i,j,temp,m,x;
scanf("%d",&t); //enter total elements
while(t--)
{
scanf("%d",&n); // enter no. one at a time
a[0]=1; //initializes array with only 1 digit, the digit 1.
m=1; // initializes digit counter
temp = 0; //Initializes carry variable to 0.
for(i=1;i<=n;i++)
{
for(j=0;j<m;j++)
{
x = a[j]*i+temp; //x contains the digit by digit product
a[j]=x%10; //Contains the digit to store at position j
temp = x/10; //Contains the carry value that will be stored on later indeces
}
while(temp>0) //while loop that will store the carry value on array.
{
a[m]=temp%10;
temp = temp/10;
m++; // increments digit counter
}
}
for(i=m-1;i>=0;i--)
printf("%d",a[i]);
printf("\n");
}
return 0;
}
This is an activity given by my instructor.
Create a program that accepts numeric input from the user. If the user enters an even number, store it to an array for even numbers. If the user enters an odd number, store it to another array for odd numbers. Input terminates if the user entered 10 numbers already. Display the size of each array and their elements.
Example:
Input: 5, 6, 12, 10, 0, 3, 4, 100, -1, 7
Even numbers (6): 6 12 10 0 4 100
Odd numbers (4): 5 3 -1 7
and this is the code I've come up with.
#include <stdio.h>
int sort(int);
int main(){
int input, count;
for(count=0;count!=10;count++){
printf("Enter 10 digits: ");
scanf("%d", &input);
sort(input);
}
printf("%d", input);
return 0;
}
int sort(int inp){
int odd[10];
int even[10];
if(inp%2==0){
odd[]=inp;
}
else
even[]=inp;
return 0;
}
Please help me on how to store the numbers into two separate arrays. Any tips will be greatly appreciated.
Check the below code. It's self-explaining.
#include <stdio.h>
#include <stdlib.h>
#define NUM 10
int main()
{
int input, i;
int oddcounter = 0, evencounter =0;
int oddarr[NUM];
int evenarr[NUM];
printf("Enter 10 integers\n");
for (i = 0; i < NUM; i++)
{
if ( scanf("%d", &input) == 1 )
{
if ((input % 2) == 0)
{
evenarr[evencounter++] = input;
}
else
{
oddarr[oddcounter++] = input;
}
}
}
printf("Number of elem in oddarray : %d, evenarray : %d\n\n", oddcounter, evencounter);
printf("Odd elements are :");
for (i = 0; i < oddcounter ; i++) printf("%d\t", oddarr[i]);
printf("\n");
printf("Even elements are :");
for (i = 0; i < evencounter; i++) printf("%d\t", evenarr[i]);
printf("\n");
return 0;
}
In addition to Sourav's comment which indicates that you shouldn't have the int[] arrays be local to sort, this syntax isn't correct for assigning to arrays in C:
odd[]=inp;
On my compiler, it generates the following error:
24:9: error: expected expression
odd[]=inp;
To store to odd, you need to indicate the index at which you'd like to store, for example:
odd[1]=inp;
which also means you'll need to keep track the latest index you wrote to for each array!
You need to tell the compiler which index of the array you are storing your data to. In sort:
if(inp%2==0){
odd[]=inp;
}
else
even[]=inp;
return 0;
}
Should look something like:
if(inp%2==0){
odd[endofoddindex]=inp;
}
else
even[endofevenindex]=inp;
return 0;
}
That said, you won't get much use out of the arrays being local variables, since they are deallocated on each call. Your best bet is to declare the arrays in main and pass them in.
Your even and odd arrays are both local. This means that they exist as long as the function exists. So you won't be retrieve the data you have stored(You also don't store it correctly).
So you need both the arrays in main and also two other variables for using as the index of both the array(i and j in the below program). The modified program is given below:
#include <stdio.h>
int sort(int);
int main(){
int input, count,i=0,j=0; //i and j to be used as array indices
int odd[10];
int even[10]; //arrays in main
for(count=0;count!=10;count++){
printf("Enter 10 digits: ");
scanf("%d", &input);
if(sort(input)) //if odd number was found
odd[i++]=input;
else //even number found
even[j++]=input;
}
printf("%d", input);
//print even and odd arrays here
return 0;
}
int sort(int inp){
if(inp%2==0)
return 0; //if number is even,return 0
return 1; //else return 1
}
You need to either have your arrays as globals or pass them into your sort function. Where they are they currently they get recreated every time the sort function is called and are inaccessible to the rest of your program.
You will also need to keep track of the max number of ints in each array and the current number.
Your test in sort would be something like this:
if( inp % 2 == 0)
{
//TODO check that currentEvenCount < maxEvenCount
even[ currentEvenCount ] = inp;
currentEvenCount++
}
else
{
//TODO check that currentOddCount < maxOddCount
odd[ currentOddCount ] = inp;
currentOddCount++;
}
To declare your arrays as globals just move the declaration outside of any function above anywhere they are referenced
int even[10];
int odd[10];
int main() ...
To pass them as parameters to sort function you could declare sort like this:
sort( int inp, int even[], int maxEvenCount, int* currentEvenCount, int odd[]. int maxOddCount, int* currentOddCount)
{
...
if( inp % 2 == 0)
{
//TODO check that currentEvenCount < maxEvenCount
even[ *currentEvenCount ] = inp;
(*currentEvenCount)++
}
}
The * in front of currentEventCount is dereferencing the pointer and getting/setting the actual value pointed to.
You would then call sort like so:
int main()
{
int evenArray[10];
int oddArray[10];
int currentEvenCount = 0;
int currentOddCount = 0;
...
sort( input, evenArray, 10, ¤tEvenCount, oddArray, 10, ¤tOddCount);
}
There is no any sense to define the arrays as local variables of function sort because each time the function is called the arrays are created anew.
The program could look the following way
#include <stdio.h>
#define N 10
enum Type { Even, Odd };
enum Type sort( int x )
{
return x % 2 == 0 ? Even : Odd;
}
int main( void )
{
int odd[N];
int even[N];
int odd_count = 0;
int even_count = 0;
int i;
printf( "Enter %d numbers: ", N );
for( i = 0; i < N; i++ )
{
int num;
scanf( "%d", &num );
switch ( sort( num ) )
{
case Even:
even[even_count++] = num;
break;
case Odd:
odd[odd_count++] = num;
break;
}
}
printf( "Even numbers (%d):", even_count );
for ( i = 0; i < even_count; i++ ) printf( " %d", even[i] );
printf( "\n" );
printf( "Odd numbers (%d):", odd_count );
for ( i = 0; i < odd_count; i++ ) printf( " %d", odd[i] );
printf( "\n" );
return 0;
}
If to enter
5 6 12 10 0 3 4 100 -1 7
then the output will be
Even numbers (6): 6 12 10 0 4 100
Odd numbers (4): 5 3 -1 7
Simply copy, paste and investigate the program.:)
Hope this program will solve you issue. Here is the working code.
#include <stdio.h>
int sort(int[]);
int main(){
int input[10], count;
printf("Enter 10 digits: ");
for(count=0;count<10;count++){
scanf("%d", &input[count]);
}
sort(input);
return 0;
}
int sort(int inp[]){
int odd[10];
int even[10];
int oddCount=0, evenCount=0;
int i;
for(i=0; i<10;i++)
{
if(inp[i]%2==0){
even[evenCount]=inp[i];
evenCount++;
}
else
{
odd[oddCount]=inp[i];
oddCount++;
}
}
printf("ODD COUNT is %d \n", oddCount);
for(i=0; i<oddCount;i++)
{
printf("ODD VALUE %d \n", odd[i]);
}
printf("EVEN COUNT is %d \n", evenCount);
for(i=0; i<evenCount;i++)
{
printf("EVEN VALUE %d \n", even[i]);
}
return 0;
}
Your variable input is just an int, it's not an array. You need two arrays:
int even[10], odd[10];
Then you need to keep track of the number of numbers you've read so far, and for each number check which array to store it in.
I don't see a need to do sorting, so not sure why you have a function called sort().
It should just be something like:
int even[10], odd[10];
int oddindex = 0, evenindex = 0;
while(scanf(" %d", &x) == 1)
{
if(x % 2 == 0)
even[evenindex++] = x;
else
odd[oddindex++] = x;
if((evenindex + oddindex) >= 10)
break;
}
/* Loop here to print numbers. */
An answer suitable for an assignment question:
int main()
{
int i,c,o[10],e[10];int oc=0;int ec=0;int*pc;for(c=0;c<10;c++){scanf("%d",&i);pc=(i&1)?&o[oc++]:&e[ec++];*pc=i;}
// Now print out the values as requested in oc, o, ec and e.
}
I have an assignment that requires me to first set up integer arrays to store arbitrarily large numbers. By using each element of an int array, I can hold one digit per element because int variable types are 32 bits and can only reach up to about 2 billion before failing. I know that there are other libraries using BigInt but I wanted to make my own arrays instead.
I tested my code out, but it doesn't seem to be accepting values past 9 digits until it starts to fail.
int * toArray(int, int);
int main()
{
int n, *res, counter, i;
printf("Input: ");
scanf("%d", &n);
while(n>0){
n/=10;
counter++;
}
res = toArray(n, counter);
for(i=1;i<=counter;i++){
printf("%d", *(res)+i);
}
printf("\n");
}
int * toArray(int n, int counter)
{
int i;
int *numberArray = malloc(sizeof(int) * counter);
for(i=0; i< counter; i++)
{
numberArray[i] = n % 10;
}
return numberArray;
}
I want to be able to accept close to twenty digits at the very least. I know this can be done with int arrays, although char arrays (or strings) are also a possibility. But I wanted to use int arrays instead. Any help in understanding why it fails around 9 digits and suggestions for going past that would be very much appreciated. Thank you.
The problem is that you are reading an int from keyboard
scanf("%d", &n);
so therefore no matter how many digits you enter, you still will only get 9 digits.
In order to be able to enter an arbitrary number you would have to read it as a string instead, then convert it to your int array.
EDIT:
this way (for instance) would allow for 20 digits
char ch;
int digits[20];
int i = 0;
do
{
ch = fgetc(stdin);
if ( isdigit(ch) )
{
digits[i++] = ch - 48; // convert ASCII to int
}
}
while (isdigit(ch) && i < sizeof(digits)/sizeof(digits[0]));
#include<stdio.h>
#include<stdlib.h>
int *toArray(long long int n, int counter); int main() {
long long int n=0,copy_of_n=0;
int counter=0, i=0;
int *res=NULL;
printf("Input: ");
scanf("%lld", &n);//Refer note in the answer
copy_of_n=n;//see the while loop why use this.
while(n>0){
n/=10;//Good, but you are not storing it. copy_of_n does that.
counter++;//You didn't initialize this
}
res=toArray(copy_of_n, counter);
for(i=counter-1;i>=0;i--){//This ensures you get the numbers in a proper order and since index starts from 0 in the toArray function
printf("%d", res[i]);
}
printf("\n");
free(res);//You forgot this
return 0; }
int *toArray(long long int n, int counter) {
int i=0;
int *numberArray = malloc(sizeof(int) * counter);
memset(numberArray,0x00,counter*sizeof(int));//Good to do this
//Check for memory allocation
for(i=0; i< counter; i++)
{
numberArray[i] = n % 10;
n=n/10LL;//You have to remove the digits too
}
return numberArray; }
NOTE: Read in a while loop the integers in small parts because long long has limits of how many digits it can store. Another approach would be to store in a string and send split the string into exactly the max. no. of digits that can be stored in long long in the n variable and send it to the function, part by part.
long long size is implementation dependent, hence ensure you check the max. no. of digits it can accomodate.(It will ofcourse bypass your 9-digit limit as asked in question)