so I'm fairly new to C and programming in general and also new to the website. Looking for guidance/help on figuring it all out, including help on how to properly use and ask questions on here.
So what I wanted to figure out is how to keep track of user defined numbers entered and also the value of the numbers added together. I was playing around with it for awhile looking for solutions, i was thinking there is a problem with my ways of keeping a count and not so much, there are no errors or warnings but the compiler stops after asking for an integer.
Any help or advice is much appreciated.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int num, sumodd, sumeven, countodd, counteven;
char name;
printf("Please enter your name. \n");
scanf("%c", &name);
printf("Please enter an integer. \n");
do {
scanf("%d", &num);
if (num%2 == 0)
sumeven = sumeven + num;
counteven = counteven + num++;
if (num%2 == 1)
sumodd = sumodd + num;
countodd = countodd + num++;
} while (num > 0);
printf("%c, the numbers you entered are broken down as follows: \n", name);
printf("You entered %d even numbers with a value totaling at %d", counteven, sumeven);
printf("You entered %d odd numbers with a value totaling at %d", countodd, sumodd);
return 0;
}
As pointed out in comments you were doing some mistakes like:
Not initializing sumodd, sumeven, countodd, counteven to zero
Not using { } in if else blocks of odd even checking
Using a character for a name variable
adding num++ to count variables instead of 1, and thereby changing its value.
I modified your code slightly :
int num, sumodd = 0, sumeven = 0, countodd = 0, counteven = 0; // initialize to zero
char name[200];
printf("Please enter your name. \n");
scanf("%s", name);
printf("Please enter an integer. \n");
do {
scanf("%d", &num);
if (num%2 == 0) {
sumeven = sumeven + num;
counteven = counteven + 1;
}
if (num%2 == 1) {
sumodd = sumodd + num;
countodd = countodd + 1;
}
} while (num > 0);
printf("%s, the numbers you entered are broken down as follows: \n", name);
printf("You entered %d even numbers with a value totaling at %d\n", counteven, sumeven);
printf("You entered %d odd numbers with a value totaling at %d\n", countodd, sumodd);
When i ran it i was able to see desired output:
~/Documents/src : $ ./a.out
Please enter your name.
Rohan
Please enter an integer.
1
2
3
0
Rohan, the numbers you entered are broken down as follows:
You entered 2 even numbers with a value totaling at 2
You entered 2 odd numbers with a value totaling at 4
Related
I wrote this loop to add numbers, and the break to get out of the loop if the number entered is less than zero, and in last print the calculated numbers without adding the negative number. but the problem is even I wrote the break statement before the addition when I enter 15 and 15 and -2 the output is 28 rather than 30
I found out how to fix that, what I want to know is why
and thank you.
#include <stdio.h>
void main()
{
int j = 1, num = 0, rslt = 0;
while (1) {
if (num < 0) break;
printf("enter a number : ");
scanf("%d", &num);
rslt = rslt + num;
}
printf("the resluts are %d\n", rslt);
}
Currently, you are effectively testing the input of the previous iteration, after already adding it to your result. Instead, check the number immediately after the user enters it, before you perform any calculations.
#include <stdio.h>
int main(void)
{
int num = 0, rslt = 0;
while (1) {
printf("enter a number : ");
scanf("%d", &num);
if (num < 0)
break;
rslt += num;
}
printf("the results are %d\n", rslt);
}
You might also want to check that scanf returns the number of successful conversions you were expecting (in this case one), to handle the event where the user enters invalid input.
if (1 != scanf("%d", &num))
break;
I'm a newbie!
I'm supposed to get 2 integers from the user, and print the result(sum of all numbers between those two integers).
I also need to make sure that the user typed the right number.
The second number should be bigger than the first one.
And if the condition isn't fulfilled, I have to print "The second number should be bigger than the first one." and get the numbers from the user again until the user types right numbers that meet the condition.
So if I programmed it right, an example of the program would be like this.
Type the first number(integer) : 10
Type the second number(integer) : 1
The second number should be bigger than the first one.
Type the first number(integer) : 1
Type the second number(integer) : 10
Result : 55
End
I think that I have to make two loops, but I can't seem to figure out how.
My English is limited, to help your understanding of this quiz, I'll add my flowchart below.
I tried many different ways I can think of, but nothing seems to work.
This is the code that I ended up with now.
But this doesn't work either.
#include <stdio.h>
void main(void)
{
int a = 0;
int b = 0;
int total_sum = 0;
printf("Type the first number : \n");
scanf("%d", &a);
printf("Type the second number : \n");
scanf("%d", &b);
while (a > b) {
printf("The second number should be bigger than the first one.\n");
printf("Type the first number : \n");
scanf("%d", &a);
printf("Type the second number : \n");
scanf("%d", &b);
}
while (a <= b) {
total_sum += a;
a++;
}
printf("Result : \n", total_sum);
}
Instead of using loop to sum the numbers, we can use mathematical formula.
Sum of first N integers= N*(N+1)/2
#include <stdio.h>
int main(void)
{
int a = 0;
int b = 0;
int sum;
//Run infinite loop untill a>b
while(1)
{
printf("Type the first number : ");
scanf("%d", &a);
printf("Type the second number : ");
scanf("%d", &b);
if(a>b)
{
printf("The second number should be bigger than the first one.\n");
}
else
{
break;
}
}
//Reduce comlexity of looping
sum=((b*(b+1))-(a*(a-1)))/2;
printf("Result : %d " , sum);
return 0;
}
After corrections your code should run. The community has pointed out many mistakes in your code. Here's an amalgamated solution:
#include <stdio.h>
int main(void)
{
int a = 0;
int b = 0;
int correctInput=0;
int total_sum = 0;
do
{
printf("Type the first number : \n");
scanf("%d", &a);
printf("Type the second number : \n");
scanf("%d", &b);
if(a<b)
correctInput=1;
else
printf("The second number should be bigger than the first one.\n");
}
while (correctInput ==0) ;
while (a <= b) {
total_sum += a;
a++;
}
printf("Result : %d \n" , total_sum);
return 0;
}
Factorials are used frequently in probability problems. The factorial of a positive integer n (written n! and pronounced "n factorial") is equal to the product of the positive integers from 1 to n: n! = 1 x 2 x 3 x x n Write a program that takes as input an integer n and computes n!.
#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!)
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 6 years ago.
Improve this question
I'm not sure what i'm doing wrong but the for loop is not initializing
The code just goes immediately to displaying the printfs. That have no values in them since the for loop didn't activate
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#define PAUSE system("Pause")
main() {
// INITALIZE VARIABLES
int number = 0;
int i = 0;
int odd = 0;
int even = 0;
int totalNum = 0;
int tempNum = 0;
int count;
printf("Enter a number between 2 and 25\n");
scanf("%i", &number);
do{
if (number < 2 || number > 25)
printf("That was an invalid number please try again\n");
scanf("%i", &number);
} while (number < 2 || number > 25);
printf("Enter how many numbers you want to input\n");
scanf("%i", &count);
for (i = 1; i == count; ++i){
printf("input numbers\n");
scanf("%i", &tempNum);
if (tempNum % 2 == 0)
even++;
else
odd++;
totalNum = totalNum + tempNum;
} // END FOR LOOP
// DISPLAY OUTPUT
printf("You entered %i numbers\n", count);
printf("The sum of the %i numbers is %i\n", count, totalNum);
printf("The average of the %i numbers is %i\n", count, totalNum / count);
printf("You entered %i odd numbers and %i even numbers\n", odd, even);
PAUSE;
} // END MAIN
Your loop will only execute at best once, when count == 1 as you initialize i to 1.
If you enter a 1 for count,
printf("Enter how many numbers you want to input\n");
scanf("%i", &count);
the loop will run exactly once, until i increments to 2
You probably want:
for (i = 1; i <= count; ++i){
do{
if (number < 2 || number > 25)
printf("That was an invalid number please try again\n");
scanf("%i", &number);
} while (number < 2 || number > 25);
it should be...
do{
if (number < 2 || number > 25){
printf("That was an invalid number please try again\n");
scanf("%i", &number);
}
} while (number < 2 || number > 25);
else it asks always another number
i = 1, so i == count; gives false therefore the loop is ignored.
A for loop in C works like this:
for ( variable initialization; condition; variable update ) {
/* Do something... */
}
The loop will execute for as long as condition is true. So, when you do:
for (i = 1; i == count; ++i)
The loop will execute for as long as i == count is true. So, unless count holds 1 when this line is executed, the loop will never run.
As others pointed out, you probably want this:
for (i = 1; i <= count; ++i)
So your loop will run for all values of i, until it reaches count.
As a side note i should point out that the usual way to write for loops in C is something like this:
for (i = 0; i < count; i++)
We start with i = 0 because C arrays are zero-based, so the Nth element of an array has index n-1
You were so close. In addition to fixing your loop test clause for (i = 1; i <= count; i++), I would suggest using " %d" for your format specifier. Your do loop need only be a while loop to avoid printing your invalid number message every time.
Additionally, While not an error, the standard coding style for C avoids caMelCase variables in favor of all lower-case. See e.g. NASA - C Style Guide, 1994.
With those changes, (and changing your odd/even check to a simple &) you could write your code as follows.
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
// #define PAUSE system("Pause")
int main (void)
{
int number, i, odd, even, totalnum, tempnum, count;
number = i = odd = even = totalnum = tempnum = count = 0;
printf ("enter a number between 2 and 25: ");
scanf (" %d", &number);
while (number < 2 || number > 25) {
printf ("invalid number, again (between 2 and 25): ");
scanf (" %d", &number);
}
printf ("numbers to input: ");
scanf (" %d", &count);
for (i = 1; i <= count; i++) {
printf ("input number %2d: ", i);
scanf (" %d", &tempnum);
if ((tempnum & 1) == 0)
even++;
else
odd++;
totalnum = totalnum + tempnum;
}
printf ("You entered %d numbers\n", count);
printf ("The sum of the %d numbers is %d\n",
count, totalnum);
printf ("The average of the %d numbers is %d\n",
count, totalnum / count);
printf ("You entered %d odd numbers and %d even numbers\n",
odd, even);
// PAUSE;
return 0;
}
note: main is type int (e.g. int main (int argc, char **argv) or simply int main (void) to indicate no arguments taken). Since it is type int it will return a value to the shell. While historic implementations may have allowed void main that is no longer the case for portable code.
Example Use/Output
$ /bin/forskipped
enter a number between 2 and 25: 4
numbers to input: 4
input number 1: 1
input number 2: 2
input number 3: 3
input number 4: 4
You entered 4 numbers
The sum of the 4 numbers is 10
The average of the 4 numbers is 2
You entered 2 odd numbers and 2 even numbers
Look it over and let me know if you have any questions.
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.