So my program asks for name and numbers, then tallies up the number of even entries and odd entries, then giving a total of all the even entries and a total for all the odd.
It all works except the calculations, it tells me I have all odd entries and only values them as 1 when adding them up.
I feel like it has something to do with my variables and referencing in my calc function
#include <stdio.h>
int number = 1;
int evencount, oddcount;
int eventotal, oddtotal;
int main () {
char name[256];
printf("Enter your name \n");
scanf("%s", name);
printf("Enter numbers within 1-100 \n");
printf("Enter 0 to quit\n");
calc(number);
printf("%s,the numbers you have entered are broken down as follows: \n", name);
printf("%d odd entries \n", oddcount);
printf("%d even entries\n", evencount);
printf("You entered even numbers with a total value of %d \n", eventotal );
printf("You entered odd numbers with a total value of %d \n", oddtotal);
return 0;
}
int calc (int input) {
while (number != 0) {
scanf("%d", &number);
if (input%2 == 1) {
oddcount++;
oddtotal += input;
}
else {
evencount++;
eventotal += input;
}
};
}
scanf("%d", &number);
if (input%2 == 1) {
{
oddcount++;
oddtotal += input;
}
I think you probably want number % 2 rather than input % 2,
oddtotal += number rather than oddtotal += input, etc.
In your calc function, you're mixing up input, which is a parameter, and number, which is a global.
Using a global here for your input variable is bad form, and in fact you don't need to pass anything into this function.
Also, better to use do..while instead of while since the loop must run at least once:
You're also using the function before it's declared, so you should have a function prototype. And since calc doesn't need to return anything, set its return type to void.
#include <stdio.h>
int evencount, oddcount;
int eventotal, oddtotal;
void calc();
int main()
{
...
calc();
...
}
void calc()
{
int number;
do {
scanf("%d", &number);
if (number%2 == 1) {
oddcount++;
oddtotal += number;
} else {
evencount++;
eventotal += number;
}
} while (number != 0);
}
Related
This program will determine the count, minimum, maximum, sum and average of all valid numbers entered. It will stop asking for new values when zero is entered and don't accept negative values. After asking for 1st number it just stops working. I will be very happy if anyone can give me some tips how to write it more efficiently and some for just coding.
#include <stdio.h>
#include <stdlib.h>
void display(void);
void menu(void);
int main(void)
{
display();
menu();
system("pause");
return 0;
}
void display(void)
{
printf("program will determine the count, minimum, maximum, sum and average of all valid numbers entered.\n");
return;
}
void menu(void)
{
int count;
float min = 0;
float max = 0;
float sum;
float avg;
float user = 0;
int index;
printf("Please enter the number ==> ");
do
{
// I think I scew up here plz check it.
scanf("%f", user);
if (user > 0)
{
if (user > max)
{
max = user;
}
if (user < min)
{
min = user;
}
}
else
{
printf("Please enter a positive number");
}
sum += user;
count = index;
index++;
} while (user != 0);
avg = sum / count;
printf("The minimum number ==> %f", min);
printf("The maximum number ==> %f", max);
printf("The number of enteries ==> %i", count);
printf("The sum of all values ==> %f", sum);
printf("The average of all values ==> %f", avg);
}
scanf("%f", user); should be scanf("%f", &user);
You'd not want to increase these if input - user is not correct:
sum += user;
count = index;
index++;
Moreover, your count is not being computed correctly. To be specific, you're calculating 1 less than actual count. So put this block inside if statement like this
if (user > 0)
{
if (user > max)
{
max = user;
}
if (user < min)
{
min = user;
}
sum += user;
count = index + 1;
index++;
}
else
{
printf("Please enter a positive number");
}
Be cautious when using %i. %d and %i are same when used in printf but not in scanf.
printf("The number of enteries ==> %d", count); //Notice %d
This line is wrong:
scanf("%f", user);
You are not storing the input into user. You can fix it like this:
scanf("%f", &user);
Here's your error:
scanf("%f", user);
The %f format specifier to scanf expects the address of a float, i.e. a float *. All function parameters in C are pass by value, meaning that changing a parameter isn't reflected in the calling function. By passing a pointer, the function can dereference the pointer to write to the location it points to.
So change the function call to pass in the address of user:
scanf("%f", &user);
So, I have to write a program to ask the user for an integer, and then that integer will determine how many more entries the user gets before adding all the numbers that were entered. So, if the first entered integer is "5", then the user can enter 5 more integers. Those 5 integers are then added together at the end and displayed. I have written a program with for loops, but for some reason, it is only adding first 4 integers and not the 5th one. Here is the code:
int main() { //declare main function
int c=0,n,i; //declare integers
int sum=0;
printf("\nEnter an integer: "); //ask user for input and create a label
scanf("%d",&n);
if (n>=0) { //use if statement
for (i=0;i<n;i++) //use for loop inside if statement to account for negative integers
{
sum+=c;
printf("Enter an integer: ");
scanf("%d",&c);
}
}
else {
printf("Wrong number. You can only enter positive integers!");
}
printf("The sum of the %d numbers entered is: %d",i,sum);
return 0;
}
Just change the position of
sum+=c;
to after the scanf it should work.
It is good to split the program. use functions. Not everything in the main function.
int getInteger(void)
{
char str[100];
int number;
while(!fgets(str, 100, stdin) || sscanf(str, "%d", &number) != 1)
{
printf("Wrong input. Try again:") ;
}
return number;
}
int main()
{
int nsamples;
long long sum = 0;
printf("Enter number of samples:");
while((nsamples = getInteger()) <= 0)
{
printf("Try again, entered number must be >= 0\n");
}
printf("Enter numbers:\n");
for(int i = 1; i <= nsamples; i++)
{
printf("Sample no %d:", i);
sum += getInteger();
}
printf("The sim is: %lld\n", sum);
}
I'm getting a consistent divide by zero error, even though each loop should be populating the variables. Code below:
#include <stdio.h>
void calculateAverage()
{
int grade, count, sum;
double average;
sum = 0;
count = 0;
grade = 0;
average = 0.0;
int coolvalue = 0;
while (coolvalue==0)
{
scanf("%d", &grade);
if (grade == -1)
{
sum, sizeof(double);
count, sizeof(double);
average = (sum / count);
printf("%lf", &average);
break;
}
else
{
if ((grade > 100) || (grade < -1))
{
printf("Error, incorrect input.\n");
break;
}
else
{
sum = +grade;
count = count + 1;
return count;
return sum;
}
}
}
coolvalue = 1;
}
int main(void)
{
while (1)
calculateAverage();
while (1) getchar();
return 0;
}
Even while using return, I'm not able to properly increment the value of sum or count.
There are multiple issues in your code.
scanf("%d", &grade); - you don't check the value returned by scanf(). It returns the number of values successfully read. If you enter a string of letters instead of a number, scanf("%d") returns 0 and it does not change the value of grade. Because of this the code will execute the rest of the loop using the previous value of grade. You should restart the loop if the value returned by scanf() is not 1:
if (scanf("%d", &grade) != 1) {
continue;
}
Assuming you enter 10 for grade this block of code executes:
sum = +grade;
count = count + 1;
return count;
return sum;
sum = +grade is the same as sum = grade. The + sign in front of grade doesn't have any effect. It is just the same as 0 + grade.
You want to add the value of grade to sum and it should be sum += grade. This is a shortcut of sum = sum + grade.
return count makes the function complete and return the value of count (which is 1 at this point) to the caller. The caller is the function main() but it doesn't use the return value in any way. Even more, your function is declared as returning void (i.e. nothing) and this renders return count incorrect (and the compiler should warn you about this).
return sum is never executed (the compiler should warn you about it being dead code) because the function completes and the execution is passed back to the caller because of the return count statement above it.
Remove both return statements. They must not stay here.
If you enter -1 for grade, this block of code is executed:
sum, sizeof(double);
count, sizeof(double);
average = (sum / count);
printf("%lf", &average);
break;
sum, sizeof(double) is an expression that does not have any effect; it takes the value of sum then discards it then takes the value of sizeof(double) (which is a constant) and discards it too. The compiler does not even generate code for it.
the same as above for count, sizeof(double);
average = (sum / count);:
the parenthesis are useless;
because both sum and count are integers, sum / count is also an integer (the integral result of sum / count, the remainder is ignored).
you declared average as double; to get a double result you have to cast one of the values to double on the division: average = (double)sum / count;
if you enter -1 as the first value when the program starts, count is 0 when this code is executed and the division fails (division by zero).
printf("%lf", &average); - you want to print the value of average but you print its address in memory. Remove the & operator; it is required by scanf() (to know where to put the read values). It is not required by printf(); the compiler generates code that passes to printf() the values to print.
break; - it passes the execution control after the innermost switch or loop statement (do, while or for). It is correct here and makes the variable coolvalue useless. You can simply remove coolvalue and use while (1) instead.
All in all, your function should look like:
void calculateAverage()
{
int sum = 0;
int count = 0;
int grade = 0;
double average = 0.0;
while (1) {
if (scanf("%d", &grade) != 1) {
// Invalid value (not a number); ignore it
continue;
}
// A value of -1 signals the end of the input
if (grade == -1) {
if (count > 0) {
// Show the average
average = (double)sum / count;
printf("Average: %lf\n", average);
} else {
// Cannot compute the average
puts("You didn't enter any value. Cannot compute the average.\n");
}
// End function
return;
}
if ((grade < -1) || (100 < grade)) {
puts("Error, incorrect input.\n");
// Invalid input, ignore it
continue;
}
sum += grade;
count ++;
}
}
Quite a few corrections need to be made.
The while loop in the calculateAverage() function. That's an infinite loop buddy, because you are not changing the value of that coolValue variable anywhere inside, instead you make it 1 only when it exits the loops, which it never will.
So, use while(1) {...}, and inside it, check for the stopping condition, i.e, if (grade == -1) { ... } and inside it calculate and print the average and return. This will automatically break the while.
You're not checking if the input grade is actually a valid integer or not. Check the value of scanf for that, i.e, use if (scanf("%d", &grade) != 1) { ... }
The expression sum = +grade; is just another way of writing sum = 0+grade which in turn is nothing but sum = grade. Replace this with sum += grade;. This is the right way to write a shorthand for addition.
Two return statements..a very wrong idea. First of all, a function can have just one return(in an obvious way I mean, at once). Secondly, the function calculateAverage() is of return-type void. there's no way how you can return double value from it. So remove these two statements.
I have attached the code below which works. Also do go through the output which I have attached.
CODE:
#include <stdio.h>
void calculateAverage()
{
int grade, count = 0, sum = 0;
double average;
printf("\nenter the grades... enter -1 to terminate the entries\n.");
while (1) {
printf("\nEnter the grade: ");
if (scanf("%d", &grade) != 1) {
printf("\nInvalid characters entered!!!");
continue;
}
else if(((grade > 100) || (grade < -1))) {
printf("\nInvalid grade entered!!!");
continue;
}
else {
if (grade == -1) {
average = sum/count;
printf("\nAverage value of grades: %.3lf",average);
return;
}
else {
sum += grade;
count++;
}
}
}
}
int main(void)
{
calculateAverage();
return 0;
}
OUTPUT:
enter the grades... enter -1 to terminate the entries.
Enter the grade: 50
Enter the grade: 100
Enter the grade: 60
Enter the grade: -1
Average value of grades: 70.000
Perhaps it is better for the function to be of type double instead of void. Although it is not my favorite solution it is close to what you want.
#include <stdio.h>
double calculateAverage(void)
{
double average;
int sum = 0, count=0, grade;
while (1)
{
scanf("%d", &grade);
if ((grade > 100) || (grade < -1))
printf("Error, incorrect input.\n");
else if (grade != -1)
{sum = sum+ grade; count = count + 1;}
else
break;
}
if (count==0)
average=-1.0; //none valid input. Notify the main()
else
average=(double)sum/count;
return average;
}
int main(void)
{
double result;
result= calculateAverage();
if (result!=-1.0)
printf("\n average= %lf",result);
else
printf("No grades to calculate average");
getchar();
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int add_even(int);
int add_odd(int);
int main() {
int num, result_odd, result_even, even_count, odd_count;
char name;
printf("What is your name?\n");
scanf("%s", &name);
while (num != 0) {
printf("Enter a number:\n");
scanf("%d", &num);
if (num % 2 == 1) {
printf ("odd\n");
odd_count++;
} else
if (num == 0) {
printf("%s, the numbers you have entered are broken down as follows:\n",
name);
result_even = add_even(num);
printf("You entered %d even numbers with a total value of %d\n",
even_count, result_even);
result_odd = add_odd(num);
printf("You entered %d odd numbers with a total value of %d\n",
odd_count, result_odd);
} else {
printf("even\n");
even_count++;
}
}
return 0;
}
int add_even(int num) {
static int sum = 0;
if (num % 2 != 0) {
return 0;
}
sum += add_even(num);
return sum;
}
int add_odd(int num) {
static int sum = 0;
if (num % 2 == 0) {
return 0;
}
sum += add_odd(num);
return sum;
}
Can anyone give me some insight as to what I did wrong exactly?
The point of the code is to get inputs from the user until they decide to stop by inputting 0. Separating the evens from the odd. Tell them how many even/odd they put and the total of all the even/odd numbers.
I understand how to separate the evens from the odds. I think my issue is with my function.
There are multiple problems in your code:
scanf() causes undefined behavior when trying to store a string into a single character. Pass an array and specify a maximum length.
you should check the return value of scanf(): if scanf() fails to convert the input according to the specification, the values are unmodified, thus uninitialized, and undefined behavior ensues. In your case, if 2 or more words are typed at the prompt for the name, scanf("%d",...) fails because non numeric input is pending, no further characters are read from stdin and num is not set.
num is uninitialized in the first while (num != 0), causing undefined behavior.
functions add_even() and add_odd() are only called for num == 0, never summing anything.
functions add_even() and add_odd() should always return the sum and add the value of the argument num is it has the correct parity. They currently cause undefined behavior by calling themselves recursively indefinitely.
odd_count and even_count are uninitialized, so the counts would be indeterminate and reading their invokes undefined behavior.
In spite of all the sources of undefined behavior mentioned above, the reason your program keeps prompting without expecting an answer if probably that you type more than one word for the name. Only a single word is converted for %s, leaving the rest as input for numbers, which repeatedly fails in the loop. These failures go unnoticed as you do not verify the return value of scanf().
Here is a corrected version:
#include <stdio.h>
#include <stdlib.h>
int add_even(int);
int add_odd(int);
int main(void) {
int num, result_odd, result_even, even_count = 0, odd_count = 0;
char name[100];
printf("What is your name? ");
if (scanf("%99[^\n]", name) != 1)
return 1;
for (;;) {
printf("Enter a number: ");
if (scanf("%d", &num) != 1 || num == 0)
break;
if (num % 2 == 1) {
printf("odd\n");
odd_count++;
add_odd(num);
} else {
printf("even\n");
even_count++;
add_even(num);
}
printf("%s, the numbers you have entered are broken down as follows:\n", name);
result_even = add_even(0);
printf("You entered %d even numbers with a total value of %d\n",
even_count, result_even);
result_odd = add_odd(0);
printf("You entered %d odd numbers with a total value of %d\n",
odd_count, result_odd);
}
return 0;
}
int add_even(int num) {
static int sum = 0;
if (num % 2 == 0) {
sum += num;
}
return sum;
}
int add_odd(int num) {
static int sum = 0;
if (num % 2 != 0) {
sum += num;
}
return sum;
}
You declared:
char name; // One single letter, such as 'A', or 'M'
printf("What is your name?\n"); // Please enter a whole bunch of letters!
scanf("%s", &name); // Not enough space to store the response!
What you really want is more like
char name[31]; // Up to 30 letters, and an End-of-String marker
printf("What is your name?\n"); // Please enter a whole bunch of letters!
scanf("%s", name); // name is the location to put all those letters
// (but not more than 30!)
I'm self-studying C and I'm trying to make 2 programs for exercise:
the first one takes a number and check if it is even or odd;
This is what I came up with for the first one:
#include <stdio.h>
int main(){
int n;
printf("Enter a number that you want to check: ");
scanf("%d",&n);
if((n%2)==0)
printf("%d is even.",n);
else
printf("%d is odd.",n);
return 0;
}
the second one should take n numbers as input and count the number of even numbers, odd numbers, and zeros among the numbers that were entered. The output should be the number of even numbers, odd numbers, and zeros.
I would like to ask how to implement the loop in this case: how can I set an EOF value if every integer is acceptable (and so I cannot, say, put 0 to end)? Can you show me how to efficiently build this short code?
#include <stdio.h>
int main(void) {
int n, nEven=0, nOdd=0, nZero=0;
for (;;) {
printf("\nEnter a number that you want to check: ");
//Pressing any non-numeric character will break;
if (scanf("%d", &n) != 1) break;
if (n == 0) {
nZero++;
}
else {
if (n % 2) {
nEven++;
}
else {
nOdd++;
}
}
}
printf("There were %d even, %d odd, and %d zero values.", nEven, nOdd, nZero);
return 0;
}
Check the return value of scanf()
1, 1 field was filled (n).
0, 0 fields filled, likely somehtlig like "abc" was entered for a number.
EOF, End-of-file encountered (or rarely IO error).
#include <stdio.h>
int main(void) {
int n;
for (;;) {
printf("Enter a number that you want to check: ");
if (scanf("%d",&n) != 1) break;
if((n%2)==0)
printf("%d is even.",n);
else
printf("%d is odd.",n);
}
return 0;
}
Or read the count of numbers to subsequently read:
int main(void) {
int n;
printf("Enter the count of numbers that you want to check: ");
if (scanf("%d",&n) != 1) Handle_Error();
while (n > 0) {
n--;
printf("Enter a number that you want to check: ");
int i;
if (scanf("%d",&i) != 1) break;
if((i%2)==0) {
if (i == 0) printf("%d is zero.\n",i);
else printf("%d is even and not 0.\n",i);
}
else
printf("%d is odd.\n",i);
}
return 0;
}
hey look at this
#include<stdio.h>
#include<conio.h>
void main()
{
int nodd,neven,num,digit ;
clrscr();
printf("Count number of odd and even digits in a given integer number ");
scanf("%d",&num);
nodd = neven =0; /* count of odd and even digits */
while (num> 0)
{
digit = num % 10; /* separate LS digit from number */
if (digit % 2 == 1)
nodd++;
else neven++;
num /= 10; /* remove LS digit from num */
}
printf("Odd digits : %d Even digits: %d\n", nodd, neven);
getch();
}
You can do something like this:
#include <stdio.h>
int main(){
int n,evenN=0,oddN=0,zeros=0;
char key;
do{
clrscr();
printf("Enter a number that you want to check: ");
scanf("%d",&n);
if(n==0){
printf("%d is zero.",n);
zeros++;
}
else if((n%2)==0){
printf("%d is even.",n);
evenN++;
}
else{
printf("%d is odd.",n);
oddN++;
}
puts("Press ENTER to enter another number. ESC to exit");
do{
key = getch();
}while(key!=13 || key!=27) //13 is the ascii code fore enter key, and 27 is for escape key
}while(key!=27)
clrscr();
printf("Total even numbers: %d",evenN);
printf("Total odd numbers: %d",oddN);
printf("Total odd numbers: %d",zeros);
return 0;
}
This program ask for a number, evaluate the number and then ask to continue for another number or exit.