guys help i'm having a problem in getting all the product of odd nums, whenever i enter 1, 3, 2, 2, the product of all odd nums is 0, it's not executing 1*3 =3... can you please point it out why, Thanks in advance...
#include <stdio.h>
#include <conio.h>
int main()
{
int arr[2][2], r, c, sum=0, prod, sumD=0, count=0, count2=0, sumAveEven;
for (r=0; r < 2; r++)
{
for (c=0; c < 2; c++)
{
printf("Enter Numbers: ");
scanf("%d", &arr[r][c]);
if(r==c)
{
sumD = sumD + arr[r][c]; //sum of diagonal pattern
}
if(arr[r][c]%2==0)
{
sum = sum + arr[r][c]; //sum ofeven nums
count= count +1;
sumAveEven = sum / count;
}
else //(arr[r][c]%2 !=0)
{
prod = prod * arr[r][c]; //prod of odd nums
}
//printf("%d ", arr[r][c]);
}
// printf("%d\n", sum);
}
printf("The Sum of All Even Numbers is: %d \n", sum);
printf("The AVERAGE of All Even Numbers is: %d \n", sumAveEven);
printf("The product of All Odd Numbers is: %d \n", prod);
printf("The Sum of Elements in pattern Diagonal is: %d \n", sumD);
//printf("Counter: %d \n", count);
getch();
return 0;
}
prod is not initialized. You need to initialize it to 1
This may not be the only problem but prod is never initialised so
prod = prod * arr[r][c];
will always give unpredictable results. As noted in other answers, you should initialise it to 1 when it is declared.
int ... sum=0, prod=1, ...
You're not initializing prod, which means it could be anything. It was PROBABLY zero in your run, and since anything times zero is zero, that's what you were getting.
In actuality, prod = prod * arr[r][c]; is returning something unknown.
It looks like you probably want to initialize prod to 1 so you have a starting point when calculating your product.
You're not initializing prod. initialize it to 1
int arr[2][2], r, c, sum=0, prod=1, sumD=0, count=0, count2=0, sumAveEven;
You didn't initialize prod. Most of the time it will be set to 0...
Related
i have a homework but i cant get the answer
I need to write a program in C...
Here is what is needed: You need to enter "n" natural number as input , and from all the natural numbers smaller than "n" , its needed to print the number which has the highest sum of devisors.
For exp: INPUT 10 , OUTPUT 8
Can anyone help me somehow?
I would really appreciate it !
i tried writing a program for finding the devisor of a number but i cant get far from here
#include <stdio.h>
int main() {
int x, i;
printf("\nInput an integer: ");
scanf("%d", &x);
printf("All the divisor of %d are: ", x);
for(i = 1; i < x; i++) {
if((x%i) == 0){
printf("\n%d", i);
}
}
}
I have implemented using function which will takes input number from user and then return the sum of divisor. hope this is one you looking for
/* function to return of sum of divisor
** input: x: integer number from user input
** return sum: sum of divisor of x
*/
int sum_of_divisor(int x)
{
int sum = 0;
for(int i = 1; i < x; i++)
{
if((x%i) == 0)
{
printf("%d\n", i);
sum = sum+i;
}
}
return sum;
}
int main() {
int x, i;
printf("\nInput an integer: ");
scanf("%d", &x);
printf("All the divisor of %d are: ", x);
printf("the sum of divisor is %d ", sum_of_divisor(x));
return 0;
}
Output:
Input an integer: 10
All the divisor of 10 are: 1
2
5
the sum of divisor is 8
After checking if i is a divisor of x, you should then store that value in another variable, for example m.
Repeat until a new divisor i is higher than that number. Add this new value to m.
I am learning C on my own with a book and I cannot for the life of me figure out how to solve this exercise. I'm obviously looking at it in the wrong way or something. Here is an explanation below.
Listed below are some functions and the main function at the bottom. This program is compiled to generate a certain number of random numbers and determine the min and the max of the random numbers. If you copy and paste this code, you will see how it works. Anyways, an exercise asks me to go to the function "prn_random_numbers()" and change the for loop from "for (i = 1; i < k; ++i)" to for (i = 2; i <= k; ++i). This causes the first line format to print incorrectly. The exercise is to further modify the program in the body of the for loop to get the output to be formatted correctly.
To sum it up, the "prn_random_numbers()" function is written to print out 5 random numbers before moving to the next line. Hence the" i % 5" if statement. Now, for some reason, when you make the slight adjustment to the for loop, as the exercise asks above, it causes the first line to only print 4 numbers before moving to the next line. I have tried a number of things, including trying to force it to print the 5th number, but it only duplicated one of the random numbers. I even tried "i % 4" to see if it would print 4 numbers for each row, but it only prints 3 numbers for the first row instead of 4! So it always prints one less number on the first line than it is supposed to. I have n clue why it is doing that and the book does not give an exercise. Do you have any idea?
Bear with me if you think this is a stupid question. I am just learning on my own and I want to make sure I have a good foundation and understand everything as I learn it, before moving forward. I appreciate any help or advice!
prn_random_numbers(k) /* print k random numbers */
int k;
{
int i, r, smallest, biggest;
r = smallest = biggest = rand();
printf("\n%12d", r);
for (i = 1; i < k; ++i)
{
if (i % 5 == 0)
printf("\n");
r = rand();
smallest = min(r, smallest);
biggest = max(r, biggest);
printf("%12d", r);
}
printf("\n\n%d random numbers printed.\n", k);
printf("Minimum:%12d\nMaximum:%12d\n", smallest, biggest);
}
int main()
{
int n;
printf("Some random numbers are to be printed.\n");
printf("How many would you like to see? ");
scanf("%d", &n);
while (n < 1)
{
printf("ERROR! Please enter a positive integer.\n");
printf("How many would you like to see? ");
scanf("%d", &n);
}
prn_random_numbers(n);
return (EXIT_SUCCESS);
}
the following proposed code:
properly initializes the random number generator
cleanly compiles
properly checks for and handles errors
performs the desired functionality
avoids having to list instructions twice
follows the axiom: Only one statement per line and (at most) one variable declaration per statement.
does not use undefined functions like: max() and min()
and now the proposed code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void prn_random_numbers(int k)
{
int count = 1;
int r;
int smallest;
int biggest;
r = smallest = biggest = rand();
printf("\n%12d", r);
for ( int i = 2; i <= k; i++, count++)
{
if (count % 5 == 0)
{
count = 0;
printf("\n");
}
r = rand();
smallest = (r < smallest)? r : smallest;
biggest = (r > biggest)? r : biggest;
printf("%12d", r);
}
printf("\n\n%d random numbers printed.\n", k);
printf("Minimum:%12d\nMaximum:%12d\n", smallest, biggest);
}
int main( void )
{
int n;
srand( (unsigned)time( NULL ) );
do
{
printf("Please enter a positive integer, greater than 0.\n");
printf("How many would you like to see? ");
if( scanf("%d", &n) != 1 )
{
fprintf( stderr, "scanf for number of random numbers failed\n" );
exit( EXIT_FAILURE );
}
} while( n < 1 );
prn_random_numbers(n);
// in modern C, if the returned value from `main()` is 0 then no `return 0;` statement needed
}
a typical run, no input problems is:
Please enter a positive integer, greater than 0.
How many would you like to see? 20
98697066 2110217332 1247184349 421403769 1643589269
1440322693 985220171 1915371488 1920726601 1637143133
2070012356 541419813 1708523311 1237437366 1058236022
926434075 1422865093 2113527574 626328197 1618571881
20 random numbers printed.
Minimum: 98697066
Maximum: 2113527574
Try to use a debugger to solve your problem, it's easy to use and really helpfull :)
SOLUTION:
Your i variable don't count the number of numbers because it is initialize at 1 (in the for statement), so you need to declare a new variable to count properly.
If you have still a problem:
void prn_random_numbers(int k)
{
int count = 1;
int i, r, smallest, biggest;
r = smallest = biggest = rand();
printf("\n%12d", r);
for (i = 2; i <= k; i++, count++) {
if (count % 5 == 0) {
count = 0;
printf("\n");
}
r = rand();
smallest = min(r, smallest);
biggest = max(r, biggest);
printf("%12d", r);
}
printf("\n\n%d random numbers printed.\n", k);
printf("Minimum:%12d\nMaximum:%12d\n", smallest, biggest);
}
#include <stdio.h>
#include <stdlib.h>
int main() {
int i;
int mult;
int n;
int ans;
ans = mult * i;
printf("Please enter a multiple you want to explore.");
scanf("%d", &mult);
printf("Please enter the number which you would want to multiply this number till.");
scanf("%d", &n);
for(i = 0; i<n; i++) {
printf("%d x %d = %d \n", &mult, &i , &ans);
}
return 0;
}
Hi guys, this is a simple code which is supposed to help the user to list the times table for n times. However, i am receiving undefined behaviour and I am quite stumped as to what is wrong with my implementation of my "for" loop.
I am receiving this as my output.
6356744 x 6356748 = 6356736
for n times in my consoles.
I want to ask
Is anything wrong with the logic of my code? (i assume i do have a problem with my code so please do enlighten me)
Would it be better(or even possible) to use pointers to point to the memory addresses of the mentioned variables when i have to change the value of the variables constantly? If yes, how do i go around doing it?
Thanks!
In printf you must provide integers. You are now giving the addresses of integers. So change
printf("%d x %d = %d \n", &mult, &i , &ans);
to
printf("%d x %d = %d \n", mult, i, ans);
and to make the table, replace ans with just mult*i, so:
printf("%d x %d = %d \n", mult, i, mult*i);
You should also check the return value of scanf to check if it has succeeded reading your input:
do {
printf("Please enter a multiple you want to explore.");
} while (scanf("%d", &mult)!=1);
do {
printf("Please enter the number which you would want to multiply this number till.");
} while (scanf("%d", &n)!=1);
The things you see are the values of the variables memory location.
Change your lines inside for loop as below
ans = mult * i;
printf("%d x %d = %d \n", mult, i, ans);
There are some mistakes in your code .
you are using the & operator in print statement which is used to print the address of the variable.
Initiate the loop with the value '1' instead of '0' & execute the loop till 'i' less than equal to 'n'.
instead of using the ans variable outside the loop , use it inside the loop as it evaluate the multiplication result in each iteration of the loop.
#include <stdio.h>
int main()
{
int i;
int mult;
int n;
int ans;
printf("Please enter a multiple you want to explore.");
scanf("%d", &mult);
printf("Please enter the number which you would want to multiply this number till.");
scanf("%d", &n);
for(i = 1; i<=n; i++) {
ans = mult*i ;
printf("%d x %d = %d \n", mult, i , ans);
}
return 0;
}
#include <stdio.h>
#include <conio.h>
int getn(int n, int i);
int main()
{
int n, i;
getn(n, i);
getch();
return 0;
}
int getn(int n, int i)
{
int even = 0;
int odd = 1;
int avg;
printf("Enter ten integers: \n");
for (i = 1 ; i <= 10 ; i++)
{
printf("Integer %d: ", i);
scanf("%d", &n);
if ( n % 2 == 0 )
{
even = even + n;
}
else
{
odd = odd * n;
}
}
avg = even / 10;
printf("\n\nAverage of even numbers: %d", avg);
printf("\nProduct of odd numbers: %d", odd);
}
It seems the even calculations worked but when it comes to odd it gives the wrong answer. Please help
Our instructor wants us to use looping or iterations. No arrays. Please help me
First, your C code needs some correction:
at least give the prototype of getn before using it
getn is defined to return an int and doesn't return anything. Either replace int with void or return a value.
Second,
Your code computes the product of ten numbers, if this product is too big, it cannot be store as-is in an int. For example, it works well if you enter ten times number 3, the result is 59049, but if you enter ten times number 23, it will answer 1551643729 which is wrong because 23^10=41426511213649 but that can't be stored in an int. This is known as arithmetic overflow.
Your average is bad, because you sum ints, but the average is (in general) a rational number (average(2,3)=2.5 isn't it ?). So double avg = out/10.0; (means compute a floating division) and printf("Average %f\n",avg); would be better.
I have a problem, I tried to write a program to show the whole sum from 1 to 22 and after that, to do 2 while loops. The first one is supposed to perform the sum of some numbers given by the user, as an example: you type 10, 30 and 40 then as you enter a 0 the program sums the first three numbers. Unfortunetly the first while loop is not working. It goes directly to the last while loop where it is supposed to type a decimal numbers like (10.20 30.50 40.55) and after you type 0 again it sum those numbers and add and multipli every entry with 1.19. So far the last loop is working properly, unfortunately the second loop does not, if I move printf and scanf over the while it let me write but just start writing w/o stopping the number I wrote . Thank You in advance!
Here is the code :
#include <stdio.h>
int main()
{
int sum = 0;
int a;
int b;
double i;
double sum1 = 0;
for (a= 0; a <= 22; a++) {
sum = sum + a;
printf("the sum from 1 till 22 : %i\n ", sum);
}
while (b != 0) {
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
}
printf("the sum is : %i\n", sum);
while(i !=0) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
sum1 += i*1.19;
printf("%lf\n", i);
}
printf("The decimal summ is: %lf\n",sum1);
return 0;
}
You don't initialise i to any value before entering the loop with
while(i != 0)
i might very well be zero at this point, so your loop won't be entered even once. Initialising i to a non-zero value should fix this particular problem. The same holds for the variable b.
You should turn on warnings in your compiler, so it can show you problems like this one.
The first time the condition of the second while is evaluated, b has undefined value, since it wasn't initialized. The same applies to the third while.
Whether or not both loops are executed is only a question of chance.
Initialize both variables with non-zero values to ensure both whiles are entering. Or use a do-while:
do {
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
} while (b != 0);
Don't test b with while, test it after the user enters the number. Then you can use break to exit the loop.
while (1) {
printf("type a number:");
scanf("%i", &b);
if (b == 0) {
break;
}
sum += b;
printf("%i\n", b);
}
while(1) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
if (i == 0.0) {
break;
}
sum1 += i*1.19;
printf("%lf\n", i);
}
Your only issues are initialization: see edits in the code below. (it compiles and runs)
Did you get any compiler warnings for these? If not, you should change your settings so you do.
#include <stdio.h>
int main()
{
int sum = 0;
int a;
int b=-1; //initialize (any non-zero value will work)
double i;
double sum1 = 0;
for (a= 0; a <= 22; a++) {//a initialized in for(...) statement, (this is good)
sum = sum + a;
printf("the sum from 1 till 22 : %i\n ", sum);
}
while (b != 0) { //b Needs to be initialized before using (done above)
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
}
printf("the sum is : %i\n", sum);
i=-1; //initialize i to any non-zero value
while(i !=0) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
sum1 += i*1.19;
printf("%lf\n", i);
}
printf("The decimal summ is: %lf\n",sum1);
getchar();
return 0;
}