Program (in C) is to ask user to input integers one at a time (0 is quit indication) and find the number and total of the even inputs/odd inputs using while loop, if/else structure, and user defined function. I can't get the user defined function to print the statements needed.
(Code so far below)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
char name[30];
printf("Enter name: ");
scanf("%s", name);
int num=1, even_count=0, even_sum=0, odd_count=0, odd_sum=0;
while (num != 0)
{
printf("Enter an integer (0 to stop): ");
scanf("%d", &num);
}
if ((num % 2) == 0) {
even_count++;
}
else {
odd_count++;
}
printf("%s,your inputs are broken down as follows: \n", name);
return even_count, even_sum, odd_count, odd_sum;
}
int output_function(int even_ct, int e_sum, int odd_ct, int o_sum)
{
int count1, sum1, count2, sum2 = main();
printf("You entered %d even numbers with a total value of %d.\n", count1, sum1);
printf("You entered %d odd numbers with a total value of %d.\n", count2, sum2);
return 0;
}
You never call output_function.
Replace
return even_count, even_sum, odd_count, odd_sum;
by
output_function(even_count, eve_sum, odd_count, odd_sum);
and remove
int count1, sum1, count2, sum2 = main();
The last line to remove makes absolutely no sense.
and finally replace your printfs by this:
printf("You entered %d even numbers with a total value of %d.\n", even_ct, e_sum);
printf("You entered %d odd numbers with a total value of %d.\n", odd_ct, o_sum);
I think, first you should declare the output_function before the main:
int output_function(int even_ct, int e_sum, int odd_ct, int o_sum);
int main()
{...}
Second, the return value of the main must be a single integer value.
Third, the output_function return always a zero.
And don't forget to call the output_function.
SK
You have some nice thoughts.
You thought:
output_function is defined after main so it can call main
A function can return several variables which in order:
return even_count, even_sum, odd_count, odd_sum;
And you can get the value of them by calling the function:
int count1, sum1, count2, sum2 = main();
But C does not work like that:
main function is executed the very first, earlier than any other functions.
You cannot return several variables like that in C (in some other high level languages like Python, it is ok).
return even_count, even_sum, odd_count, odd_sum; equals to return odd_sum;
Similar above point, int count1, sum1, count2, sum2 = somethings; just bring you sum2 = somethings
Please clean these mess first and you might get the program work well.
Related
Hi I keep trying to figuure this out but my input keeps getting ignored, thanks in advance
#include <stdio.h>
#include <stdlib.h>
int main(){
float a, b, a0, b0,i;
char ans;
printf("Fibonacci search method\n\nEnter the function:\n");
printf("\nEnter the intervals over which the Fibonacci method must be applied:\n");
for (i = 1; i <= 1; i++) {
printf("a0 = ", i);
scanf("%f", & a);
printf("bo = ", i);
scanf("%f", & b);
}
printf("Narrow down on either a maximiser or a minimiser (max/min): \n", ans);
scanf(" %c", &ans);
while(ans == 'max'){
printf("maximum selected");
}
printf("minimum selected");
return 0;
}
First of all, you're comparing a single char to a whole string, so you need to modify your ans variable declaration to make it a string, like:
char ans[4]
Keep in mind, this will have a maximum size of 3. If you need to store a bigger string, you'll need to modify this.
Then, after doing this, using a while to do that comparison isn't correct. It's better to implement an if-else. And, inside that, the comparison you're doing is wrong. You need to compare strings, not chars, so you need to use strcmp() function, like:
strcmp(ans,"max") == 0
If this function returns a 0, it means both strings are equal.
Another thing to comment is that you will need to modify your scanf to scan a string, not a char, the new one will be scanf("%3s", &ans);.
And let me tell you one more thing. The for you're using has no sense. You're using a for with parameters i = 1; i <= 1; i++. That means i will start the buckle fulfilling the conditions to break it, so it will only be executed once. In other words, the code inside that for will be executed just once, no matter if it's inside or outside the for.
Anyway, and to sum up, here's your new code:
int main(){
float a, b, a0, b0,i;
char ans[4];
printf("Fibonacci search method\n\nEnter the function:\n");
printf("\nEnter the intervals over which the Fibonacci method must be applied:\n");
for (i = 1; i <= 1; i++) {
printf("a0 = ", i);
scanf("%f", & a);
printf("bo = ", i);
scanf("%f", & b);
}
printf("Narrow down on either a maximiser or a minimiser (max/min): \n", ans);
scanf("%3s", &ans);
if(strcmp(ans,"max") == 0)
printf("maximum selected");
else
printf("minimum selected");
return 0;
}
I tried going beyond just guessing random numbers. The conditions were these:
use input() numbers used from 1 to100 and if inserted numbers that are out of this range, to show a line to re-enter a number
use output() to show the output(but show the last line```You got it right on your Nth try!" on the main())
make the inserted number keep showing on the next line.
Basically, the program should be made to show like this :
insert a number : 70
bigger than 0 smaller than 70.
insert a number : 35
bigger than 35 smaller than 70.
insert a number : 55
bigger than 55 smaller than 70.
insert a number : 60
bigger than 55 smaller than 60.
insert a number : 57
You got it right on your 5th try!
I've been working on this already for 6 hours now...(since I'm a beginner)... and thankfully I've been able to manage to get the basic structure so that the program would at least be able to show whether the number is bigger than the inserted number of smaller than the inserted number.
The problem is, I am unable to get the numbers to be keep showing on the line. For example, I can't the inserted number 70 keep showing on smaller than 70.
Also, I am unable to find out how to get the number of how many tries have been made. I first tried to put it in the input() as count = 0 ... count++; but failed in the output. Then I tried to put in in the output(), but the output wouldn't return the count so I failed again.
I hope to get advice on this problem.
The following is the code that I wrote that has no errors, but problems in that it doesn't match the conditions of the final outcome.
(By the way, I'm currently using Visual Studio 2017 which is why there is a line of #pragma warning (disable : 4996), and myflush instead of fflush.)
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#pragma warning (disable : 4996)
int input();
int random(int);
void myflush();
void output(int, int);
int main()
{
int num;
int i;
int ran;
srand((unsigned int)time(NULL));
i = 0;
while (i < 1) {
ran = 1 + random(101);
++i;
}
num = input();
output(ran, num);
printf("You got it right on your th try!");a
return 0;
}
int input()
{
int num;
printf("insert a number : ");
scanf("%d", &num);
while (num < 1 || num > 100 || getchar() != '\n') {
myflush();
printf("insert a number : ");
scanf("%d", &num);
}
return num;
}
int random(int n)
{
int res;
res = rand() % n;
return res;
}
void myflush()
{
while (getchar() != '\n') {
;
}
return;
}
void output(int ran, int num) {
while (1) {
if (num != ran){
if (num < ran) {
printf("bigger than %d \n", num); //
}
else if (num > ran) {
printf("smaller than %d.\n", num);
}
printf("insert a number : ");
scanf("%d", &num);
}
else {
break;
}
}
return;
}
There are many problem and possible simplifications in this code.
use fgets to read a line then scanf the line content. This avoids the need of myflush which doesn’t work properly.
the function random is not needed since picking a random number is a simple expression.
if the range of the random number is [1,100], you should use 1+rand()%100.
there is no real need for the function output since it’s the core of the main program. The input function is however good to keep to encapsulate input.
you should test the return value of scanf because the input may not always contain a number.
Here is a simplified code that provides the desired output.
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#pragma warning (disable : 4996)
int input() {
char line[100];
int num, nVal;
printf("insert a number : ");
fgets(line, sizeof line, stdin);
nVal = sscanf(line, "%d", &num);
while (nVal != 1 || num < 1 || num > 100) {
printf("insert a number : ");
fgets(line, sizeof line, stdin);
nVal = sscanf(line, "%d", &num);
}
return num;
}
int main()
{
int cnt = 0, lowerLimit = 0, upperLimit = 101;
srand((unsigned int)time(NULL));
// pick a random number in the range [1,100]
int ran = 1 + rand()%100;
while(1) {
cnt++;
int num = input();
if (num == ran)
break;
if (num > lowerLimit && num < upperLimit) {
if (num < ran)
lowerLimit = num;
else
upperLimit = num;
}
printf("bigger than %d and smaller than %d\n", lowerLimit, upperLimit);
}
printf("You got it right on your %dth try!\n", cnt);
return 0;
}
I am unable to find out how to get the number of how many tries have been made.
Change the output function from void to int so it can return a value for count, and note comments for other changes:
int output(int ran, int num) {//changed from void to int
int count = 0;//create a variable to track tries
while (1) {
if (num != ran){
count++;//increment tries here and...
if (num < ran) {
printf("bigger than %d \n", num); //
}
else if (num > ran) {
printf("smaller than %d.\n", num);
}
printf("insert a number : ");
scanf("%d", &num);
}
else {
count++;//... here
break;
}
}
return count;//return value for accumulated tries
}
Then in main:
//declare count
int count = 0;
...
count = output(ran, num);
printf("You got it right on your %dth try!", count);
With these modifications, your code ran as you described above.
(However, th doesn't work so well though for the 1st, 2nd or 3rd tries)
If you want the program to always display the highest entered number that is lower than the random number ("bigger than") and the lowest entered number that is higher then the random number ("smaller than"), then your program must remember these two numbers so it can update and print them as necessary.
In the function main, you could declare the following two ints:
int bigger_than, smaller_than;
These variables must go into the function main, because these numbers must be remembered for the entire duration of the program. The function main is the only function which runs for the entire program, all other functions only run for a short time. An alternative would be to declare these two variables as global. However, that is considered bad programming style.
These variables will of course have to be updated when the user enters a new number.
These two ints would have to be passed to the function output every time it is called, increasing the number of parameters of this function from 2 to 4.
If you want a counter to count the number of numbers entered, you will also have to remember this value in the function main (or as a global variable) and pass it to the function output. This will increase the number of parameters for the function to 5.
If you don't want to pass so many parameters to output, you could merge the contents of the functions output and input into the function main.
However, either way, you will have to move most of the "smaller than" and "bigger than" logic from the function output into the function main, because that logic is required for changing the new "bigger_than" and "smaller_than" int variables which belong to the function main. The function output should only contain the actual printing logic.
Although it is technically possible to change these two variables that belong to the function main from inside the function output, I don't recommend it, because that would get messy. It would require you to pass several pointers to the function output, which would allow that function to change the variables that belong to the function main.
I have now written my own solution and I found that it is much easier to write by merging the function output into main. I also merged all the other functions into main, but that wasn't as important as merging the function output.
Here is my code:
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#pragma warning (disable : 4996)
int main()
{
const char *ordinals[4] = { "st", "nd", "rd", "th" };
int num_tries = 0;
int bigger_than = 0, smaller_than = 101;
int input_num;
int random_num;
srand( (unsigned int)time( NULL ) );
random_num = 1 + rand() % 101;
for (;;) //infinite loop, equivalent to while(1)
{
printf( "Bigger than: %d, Smaller than: %d\n", bigger_than, smaller_than );
printf( "enter a number: " );
scanf( "%d", &input_num );
printf( "You entered: %d\n", input_num );
num_tries++;
if ( input_num == random_num ) break;
if ( input_num < random_num )
{
if ( bigger_than < input_num )
{
bigger_than = input_num;
}
}
else
{
if ( smaller_than > input_num )
{
smaller_than = input_num;
}
}
}
printf( "You got it right on your %d%s try!", num_tries, ordinals[num_tries<3?num_tries:3] );
return 0;
}
Also, I made sure that the program would print "1st", "2nd" and "3rd", whereas all the other solutions simply print "1th", "2th", "3th". I used the c++ conditional operator for this.
Hey guys so I need to make a program which asks the user to enter a number as a argument and then let them know if it is a prime number or 0 otherwise. So the code I have so far is as follows but I am a little confused on how to make it run through all the possible values of the and make sure that it isn't a non-prime number. Right now what happens is that the program opens, I enter a value and nothing happens. Note: I have math in the header as I am unsure if it is needed or not at this stage.
EDIT: SO I MADE THE CHANGES SUGGESTED AND ALSO ADDED A FOR LOOP HOWEVER WHEN I GO TO COMPILE MY PROGRAM I GET AN WARNING SOMETHING ALONG THE LINES OF 'CONTROL MAY REACH END OF NON-VOID FUNCTION'. HOWEVER THE PROGRAM DOES COMPILE WHEN I GO TO ENTER A NUMBER AND HIT ENTER IRRELEVANT OT WHETHER OR NOT IT IS A PRIME NUMBER I GET AN ERROR BACK SAYING 'FLOATING POINT EXCEPTION: 8'.
EDIT 2: THE FLOATING POINT ERROR HAS BEEN FIXED HOWEVER NOW THE PROGRAM SEEMS TO THINK THAT EVERY NUMBER IS NON - PRIME AND OUTPUTS IT THIS WAY. I CAN'T SEEM TO SEE WHY IT WOULD DO THIS. I AM ALSO STILL GETTING THE 'CONTROL MAY REACH END OF NON-VOID FUNCTION' WARNING
#include <stdio.h>
#include <math.h>
int prime(int a){
int b;
for(b=1; b<=a; b++){
if (a%b==0)
return(0);
}
if(b==a){
return(1);
}
}
int main(void){
int c, answer;
printf("Please enter the number you would like to find is prime or not= ");
scanf("%d",&c);
answer = prime(c);
if(answer==1){
printf("%d is a prime number \n",c);
}
else
printf("%d is not a prime number\n",c);
}
1. You never initialized i (it has indeterminate value - local variable).
2. You never call function is_prime.
And using a loop will be good idea .Comparing to what you have right now.
I just modified your function a little. Here is the code
#include <stdio.h>
#include <math.h>
int prime(int a)
{
int b=2,n=0;
for(b=2; b<a; b++)
{
if (a%b==0)
{
n++;
break;
}
}
return(n);
}
int main(void)
{
int c, answer;
printf("Please enter the number you would like to find is prime or not= ");
scanf("%d",&c);
answer = prime(c);
if(answer==1)
{
printf("%d is not a prime number \n",c);
}
else
{
printf("%d is a prime number\n",c);
}
return 0;
}
Explanation-
In the for loop, I am starting from 2 because, I want to see if the given number is divisible by 2 or the number higher than 2. And I have used break, because once the number is divisible, I don't want to check anymore. So, it will exit the loop.
In your main function, you had not assigned properly for the printf() statement. If answer==1, it is not a prime number. (Because this implies that a number is divisible by some other number). You had written, it is a prime number(which was wrong).
If you have any doubts, let me hear them.
I suggest you start with trial division. What is the minimal set of numbers you need to divide by to decide whether a is prime? When can you prove that, if a has a factor q, it must have a smaller factor p? (Hint: it has a prime decomposition.)
Some errors your program had in your prime finding algorithm:
You start the loop with number 1 - this will make all numbers you test to be not prime, because when you test if the modulo of a division by 1 is zero, it's true (all numbers are divisible by 1).
You go through the loop until a, which modulo will also be zero (all number are divisible by themselves).
The condition for a number to be prime is that it must be divisible by 1 and itself. That's it. So you must not test that in that loop.
On main, the error you're getting (control reaches end of non-void function) is because you declare main to return an int.
int main(void)
And to solve that, you should put a return 0; statement on the end of your main function. Bellow, a working code.
#include <stdio.h>
#include <math.h>
int prime(int a)
{
int b;
for (b = 2; b < a; b++) {
if (a % b == 0)
return (0);
}
return 1;
}
int main(void)
{
int c, answer;
printf
("Please enter the number you would like to find is prime or not= ");
scanf("%d", &c);
answer = prime(c);
if (answer == 1) {
printf("%d is a prime number \n", c);
} else {
printf("%d is not a prime number\n", c);
}
return 0;
}
On a side note, don't use the CAPSLOCK to write full sentences. Seems like you're yelling.
Mathematically the maximum divisor of a number can be as a large as the square of it, so we just need to loop until sqrt(number).
A valid function would be:
//Function that returns 1 if number is prime and 0 if it's not
int prime(number) {
int i;
for (i = 2; i < sqrt(number); i++) {
if (a % i == 0)
return (0);
}
return 1;
}
#include<stdio.h>
int main()
{
int n , a, c = 0;
printf ("enter the value of number you want to check");
scanf ("%d", &n);
//Stopping user to enter 1 as an input.
if(n==1)
{
printf("%d cannot be entered as an input",n);
}
for(a = 2;a < n; a++)
{
if(n%a==0)
{
c=1;
break;
}
}
if(c==0 && n!=1)
{
printf("%d is a prime number \n",n);
}
else
{
if(c!=0 && n!=1)
{
printf("%d is not a prime number \n",n);
}
}
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x,i;
printf("enter the number : ");
scanf("%d",&x);
for ( i=2; i<x;i++){
if ( x % i == 0){
printf("%d",x);
printf(" is not prime number ");
printf("it can be divided by : ");
printf("%d",i);
break;
}[this is best solution ][1]
}
if( i>=x) {
printf("%d",x);
printf(" is prime number");
}
}
I wrote this program, which returns the biggest integer inserted by the user. Now, I would like the program to return the 2nd biggest integer. I created a new variable (called "status"), which is supposed to increment 1 unit every time the cycle repeats. Then, after the break condition happens, I would step back 1 unit in status variable, in order to retrieve the 2nd biggest number. I would like to follow this line of thought (if it is reliable in C) and I ask you fellows what is wrong with my current implementation.
#include <stdio.h>
int main()
{
int x, tmp=0, status=0, bigger;
printf("Insert numbers:\n");
do{
status+=1;
scanf("%d", &x);
if (tmp>=x)
bigger=tmp;
else {
bigger=x;
tmp=x;
}
}while (x!=0);
status-=1;
printf("The second biggest number is %d.\n", bigger);
return 0;
}
If you need only two biggest integers, you can just save them in each iteration, such that you have the max and max2 values:
#include <stdio.h>
int main()
{
int x, max=0, max2=0;
printf("Insert numbers:\n");
do {
scanf("%d", &x);
if (x > max) {
max2 = max; // Save the previous max, as the second largest value
max = x ; // Save the new max
}
else if (x > max2) {
max2 = x; // The input is not max, but greater than second max
}
}while (x!=0);
printf("The second biggest number is %d.\n", max2);
return 0;
}
The status in your code is just a counter. To solve your problem :
Use an array to store all the data.
Make a variable and test it if it is different than the biggest and bigger than all other entries
try this :
int main()
{
int x, status=0, bigger=0, secondbigger=0, Array[50];
while(true){
scanf("%d",&x)
if(x==0)break;
Array[status]=x;
status++;
}
for(i=0;i<status-1;i++){
if(Array[i]>bigger) bigger=Array[i];
}
for(i=0;i<status-1;i++){
if(Array[i]>secondbigger && Array[i]!=bigger) secondbigger=Array[i];
}
printf("The second biggest number is %d.\n", secondbigger);
return 0;
}
I want a program that can get two integers from user and put the sum of those inputs in a variable, after that checks that is sum more than 5 or not ? (I know I can do it with if , ... but I want to do it with while). I myself did it but it has some problems, would you mind saying what is the problem and how can I debug it ? Here is my code :
#include <stdio.h>
int main()
{
int ui1;
int ui2;
puts("Please enter two numbers:");
scanf("%2i", &ui1, &ui2);
int sum;
sum = ui1+ui2;
while(sum > 5) {
printf("Whats up !");
}
return 0;
}
This line is only scanning for 1 integer (%i with a 2 format, indicating only take 2 digits.):
scanf("%2i", &ui1, &ui2);
But it seems you expected to receive two integers.
This will leave the second argument, ui2, uninitialized.
(It should fill ui1 successfully, at least)
Try instead:
scanf("%i %i", &ui1, &ui2);
Try including the scanf statement into the loop, it will no longer be an infinite loop... (also need to dereference the integers, see EDIT)
#include <stdio.h>
int main()
{
int ui1;
int ui2;
puts("Please enter two numbers:\n");
//scanf("%2i", &ui1, &ui2);
int sum = 10;//(so that it will enter the loop at least once)
//sum = ui1+ui2;
while(sum > 4)
{
printf("enter number 1:\n");
scanf("%i", &ui1); //EDIT &
printf("enter number 2:\n");
scanf("%i", &ui2); //EDIT &
sum = ui1+ui2;
}
printf("result is: %d\n", sum);
getchar();//so you can see the result;
getchar();
return 0;
}
Actually while is a loop stmt not a conditional checker
if you want conditional checker use if...else series , switch etc
Note: in your code loop starts if (sum > 5) and never ends (infinate "Whats up !")
sum = ui1+ui2;
while(sum > 5) ///loop starts if (sum > 5) and never ends (infinate "Whats up !")
{
printf("Whats up !"); // (infinate "Whats up !")
}
if(sum > 5)
{
//greater stuff
}
else
{
//lower stuff
}
See Tutorial Here conditionals Stmts
You need to reset the "sum", because otherwise the while loop will be true FOREVER.
Second the input scanf is simply wrong.
Here the correct code
#include <stdio.h>
int main()
{
int ui1;
int ui2;
puts("Please enter two numbers:");
scanf("%d %d", &ui1, &ui2);
int sum;
sum = ui1+ui2;
while(sum > 4) { printf("Whats up !");
sum=0;}
return 0;
}
I'm not sure that i got what you want to do... but if you simply want to check the sum of the two integers using the while statement, you can put a break inside the while loop and everything will work :)
#include <stdio.h>
int main()
{
int ui1;
int ui2;
puts("Please enter two numbers:");
scanf("%2i", &ui1, &ui2);
int sum;
sum = ui1+ui2;
while(sum > 5) {
printf("Whats up !");
break;
}
return 0;
}
As others told you, using a if is the best solution