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/* Can you create a program to check whether a number is odd or eve? > Use a ternary operator to check if the number is odd or even. > If number is odd, print "The number is Odd" > If number is even, print "The number is Even" */ #include int main() { int number; printf("Enter a number: "); scanf("%d", &number); // by using Ternary Operator (number % 2==0) ? printf("The number is Even") : printf("The number is Odd"); }
Programming Task: #include int main() { int number; printf("Enter a number: "); scanf("%d", &number); (number%2 == 0) ? printf("%d is an even number.", number) : printf("%d is an odd number.", number); return 0; }
#include int main() { int number; printf("Please enter the number:"); scanf("%d", &number); (number%2 == 0) ? printf("Number is an even number"): printf("Number is an odd number"); return 0; }
You did an excellent job explaining the switch statement. I was and am still amazed at how easy you made this so understandable. Please continue teaching us. You have an absolute gift in how you explain these concepts. Thank you sooo much!
//Can you create a program to check whether a number is odd or even? //***Use a ternary operator to check if the number is odd or even //***If the number is odd print the number is odd //***If the number is even print the number is even #include int main() { int number; printf("Enter a number: "); scanf("%d", &number); (number % 2==0) ? printf("The number is Even.") : printf("The number is Odd."); return 0; }
i did the same thing and it worked but later on i made the ternary as int result=(number % 2==0) ? printf("The number is Even.") : printf("The number is Odd."); but when i print the resut itdoes say in the output whether the number is odd or even but it always adds 13 or 14 i dont understand why there is the "14 or 13 "
@@yasminehey3854 The program then checks whether 'num' is even or odd using the modulus operator %, and uses the ternary operator to print a message indicating whether the number is even or odd. However, the use of 'printf' in the ternary operator is not recommended because it returns the number of characters printed, not the value of the condition. As a result, the value of result will be 23 or 19, depending on the message printed. A better approach would be to use a separate variable to store the message, and then print that variable and the value of result separately. Here's an updated version of the program: #include int main() { int num = 78; char* message = (num % 2 == 0) ? "Your number is EVEN" : "Your number is ODD"; printf("%s ", message); int result = (num % 2 == 0) ? 0 : 1; printf("%d ", result); return 0; } This program first declares a char* variable named message, and assigns it a string value based on whether num is even or odd. It then uses printf to print message and a newline character. Finally, it declares an int variable named result, and assigns it a value of 0 if num is even or 1 if num is odd. It then uses printf to print result and a newline character.
for the quiz, it opts C coz in the if-else statement the if part is evaluated to be true and hence gives 5 so the same principle applies Ternary:- (CONDITION) ? Statement1(Executed when True) : Statement2(Executed when False)
One thing I can't understand bro. Why we used %d is even and %d is odd. If I used without %d it isn't working. But they didn't use %d in age calculator. Plzz explain bro
Seeing the video in 2024, It's been a while since I've stopped learning C and did not know the language evolved like that. At least, I did not know, back in 2016, these features were available in C. Thank you
@@musyabnahid8269 Because as per the Required Program variable "result" must be assigned with 5 if 5 is greater than 3 and if false it must be assigned the value of 3.
The correct answer is A, since the remainder of the modulus 5%2 is 1 making it not equal to 0. However, there is an exclamation mark before the parenthesis which means NEGATION or NOT, making it like this "1 is not equal to 0" or !1==0 in C language) which would be TRUE. That's why the answer is A. Inside if. If there is no negation or the symbol exclamation mark, the answer would be false which would be letter B. Inside else.
PROGRAMMING TASK: #include int main() { int num; printf("Enter number here: "); scanf("%d", &num); (num%2==0)?printf("The number is even."):printf("The number is odd."); return 0; } PROGRAMIZ QUIZ ANSWER: Option(B) 5>3?5:3;
#include int main (){ int num ; printf("Enter a number :"); scanf("%d", &num); if ( num % 2 == 0){ printf("The number is even."); } else if (num % 2 != 0){ printf("The number is odd."); }
Mam, aapke liye chand taare tod ke laa sakta hoo.......coding challenge to keya chiz hai ? Ye lo answer.... #include #include int main() { int num; printf("Enter the age "); scanf("%d",&num); (num%2==0)?printf("The number is even"):printf("The number is odd"); return 0; } Aapka 2nd question ka answer hai option B.
program is #include int main() { int num; printf("enter the number: "); scanf("%d", &num); (num % 2 == 0) ? printf("number is even") : printf("number is odd");
Hello madam. Mind is displaying "number is even" when I haven't insert anything. int number; (number%2==0)? printf("number is even"): printf("number is odd");
why do I get an error when I try to do the operation with a "double" data type. I expected the integer to be converted to double since the double is greater than the int in the ranking
#include int main() { double num1=1; double num2=3,result; result=1+3; printf("enter the number: "); scanf("%d", &result); (result>=4)? printf("the number is even"):printf("the numer is odd"); return 0; }
I think the answer is B. Because the condition in the if...Else statement is if (5 > 3) instead of if the result = (5 > 3); for in that case the right answer would be C. Can anyone confirm?
programming task answer void main(){ int number; printf("enter number:"); scanf("%d", &number); (number %2==0)? printf("the number is even"): printf("the number is odd"); }
// Online C compiler to run C program online #include int main() { int number; printf("enter your number "); scanf("%d",&number); (number %2==0)? printf("num is even") : printf("nub is odd");
// Online C compiler to run C program online #include int main() { // Write C code here int number = 11; int result = (number % 2 == 0) ? printf("it is even") : printf("it is odd"); return 0; }
// Online C compiler to run C program online #include int main() { int num =25; (num%2)? printf("The numberis odd") : printf("The number is even"); return 0;
the answers people are giving for the programming task that is given is kind of out of the question because she said that we should assign a value to the variable number but no one is doing that,i am trying to follow the instruction that she gave but its damn hard
I hope this will be the correct answer, I have tried it on online compiler provided by Programiz. /*This is a program to check whether a number is EVEN or ODD by enquiring an input from a user.*/ #include int main(){ int number;
printf("Enter any number! "); scanf("%d", &number);
(number%2==0) ? printf("The number %d is Even.",number) : printf("The number %d is Odd.",number);
int main() { // Write C code here int num; printf("Enter number: "); scanf("%d", &num); if (num>0){ printf("The number %d is a positive number.", num); } else if (num
//division method #include int main() { int num; printf("Enter the number:"); scanf("%d",&num); (num/2==0)? printf("It is an even number") : printf("It is an odd number"); return 0; }
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/*
Can you create a program to check whether a number is odd or eve?
> Use a ternary operator to check if the number is odd or even.
> If number is odd, print "The number is Odd"
> If number is even, print "The number is Even"
*/
#include
int main() {
int number;
printf("Enter a number: ");
scanf("%d", &number);
// by using Ternary Operator
(number % 2==0) ? printf("The number is Even") : printf("The number is Odd");
}
thanks
did you try doing this operation with the "double" data type?
@@eazyj_c
Thank You!
Thanks
Programming Task:
#include
int main() {
int number;
printf("Enter a number: ");
scanf("%d", &number);
(number%2 == 0) ? printf("%d is an even number.", number) : printf("%d is an odd number.", number);
return 0;
}
wtf
#include
int main() {
int number;
printf("Please enter the number:");
scanf("%d", &number);
(number%2 == 0) ?
printf("Number is an even number"):
printf("Number is an odd number");
return 0;
}
You did an excellent job explaining the switch statement. I was and am still amazed at how easy you made this so understandable. Please continue teaching us. You have an absolute gift in how you explain these concepts. Thank you sooo much!
hey buddy! How much part of C have you completed to date?
//Can you create a program to check whether a number is odd or even?
//***Use a ternary operator to check if the number is odd or even
//***If the number is odd print the number is odd
//***If the number is even print the number is even
#include
int main() {
int number;
printf("Enter a number: ");
scanf("%d", &number);
(number % 2==0) ? printf("The number is Even.") : printf("The number is Odd.");
return 0;
}
Thank you
i did the same thing and it worked but later on i made the ternary as
int result=(number % 2==0) ? printf("The number is Even.") : printf("The number is Odd.");
but when i print the resut
itdoes say in the output whether the number is odd or even but it always adds 13 or 14
i dont understand why there is the "14 or 13 "
@@yasminehey3854 The program then checks whether 'num' is even or odd using the modulus operator %, and uses the ternary operator to print a message indicating whether the number is even or odd.
However, the use of 'printf' in the ternary operator is not recommended because it returns the number of characters printed, not the value of the condition. As a result, the value of result will be 23 or 19, depending on the message printed.
A better approach would be to use a separate variable to store the message, and then print that variable and the value of result separately. Here's an updated version of the program:
#include
int main()
{
int num = 78;
char* message = (num % 2 == 0) ? "Your number is EVEN" : "Your number is ODD";
printf("%s
", message);
int result = (num % 2 == 0) ? 0 : 1;
printf("%d
", result);
return 0;
}
This program first declares a char* variable named message, and assigns it a string value based on whether num is even or odd. It then uses printf to print message and a newline character.
Finally, it declares an int variable named result, and assigns it a value of 0 if num is even or 1 if num is odd. It then uses printf to print result and a newline character.
in this part of your code,
" (number % 2==0) ? printf("The number is Even.") : printf("The number is Odd."); "
what does the % mean kindly🙏
@@sammuchina8189 % is could be DEVISION or as per our concept REMAINDER
Your tutorials are so helpful. i learnt switch statement and im now learning tenrary operator. keep the great work up!
Really useful content I really want to thank the teachers who are making the classes
for the quiz, it opts C coz in the if-else statement the if part is evaluated to be true and hence gives 5 so the same principle applies Ternary:- (CONDITION) ? Statement1(Executed when True) : Statement2(Executed when False)
it is option c
Love it, explantion was clear and straightforward, however, I suggest you also talk about how to implement nested Ternary operators.
how are you doing after 5 months in programming?
#include
int main();
{
Int number=2;
(number % 2 == 0) ? Printf("the number is even") : printf("the number is odd");
return 0 ;
}
you like shortcuts
Thanks for providing this useful and & well content for us....
Code
#include
int main()
{
int a;
scanf("%d", &a);
(a %2 ==0) ?
printf("%d is even", a):
printf("%d is odd", a);
}
Ans for Question is C.
One thing I can't understand bro. Why we used %d is even and %d is odd. If I used without %d it isn't working. But they didn't use %d in age calculator. Plzz explain bro
why C?
Great explanation as always! Thank you for everyone to make this happen.
how are you doing after two months?
I tried this in front of my teacher and even he doesn't know about this operator. Thank U❤️❤️❤️
Thankyou for explaining it so simply.. was waiting for your video ❤️
how are you doing after two years in programming?
Mam your explanation is very good and nice, can you make a tutorial videos on Data structures and algorithms in good explanation based on c language.
the best C language tutorial👌👌👌
Seeing the video in 2024, It's been a while since I've stopped learning C and did not know the language evolved like that. At least, I did not know, back in 2016, these features were available in C. Thank you
Thanks for providing " Good quality "content
By the way answer is option C
why c
i think option B is corrrect
@@shreyasy9164 yes boss b is ans
@@shreyasy9164 in the option B, if you change 5>3 to 3>5 , result will be 0 instead the result should be 3. So C is the answer.
b can also be the answer i checked it
Thanks for explaining this code simply
Beautiful 🤩
Thank you for explaining it so well
Answer is option c
Result=5>3 ? 5 : 3;
why its not B?
@@musyabnahid8269 Because as per the Required Program variable "result" must be assigned with 5 if 5 is greater than 3 and if false it must be assigned the value of 3.
@@AsirPurchase right bro 😃
The great teacher of c programing
Thank you! I didn't think it would be that easy to learn.
ur way of teaching is exlent.....
Thank you! The explanation is so useful!!!
(option c)...........cool and clear content by progrmizzzz.....plz keep going......
why c??????
@@shreyasy9164 5>3? Is a condition. After evaluating the condition, value will be assigned to result variable as either 5 or 3.
Thank You!!
It's going super good because of you explaining💞💕
wonderful explaination, thank you!
Thank you for your happy programming for us🎉
Option C : result = 5> 3 ? 5 : 3;
-----------------------------------------------------------
#include
int main()
{
int num;
printf("Enter A Number: ");
scanf("%d", &num);
(num % 2 == 0) ? printf("Even") : printf("Odd");
return 0;
}
Thank you🙏 mam. Your teaching quality is awesome👏
❤❤❤thank you for explaining it so easily❤❤❤
This was really helpful
B as we have to print Result = 5
but in C the value of just 5 is will be as output
Thank you so much for such wonderful explanation!
how are you doing after 1 year in programming?
The correct answer is A, since the remainder of the modulus 5%2 is 1 making it not equal to 0. However, there is an exclamation mark before the parenthesis which means NEGATION or NOT, making it like this "1 is not equal to 0" or !1==0 in C language) which would be TRUE. That's why the answer is A. Inside if.
If there is no negation or the symbol exclamation mark, the answer would be false which would be letter B. Inside else.
Great video. Thank you so much
I am going to write a test about this tomorrow🙄but I finally understant🥳
GREAT EXPLANATION!
Subscribed coz you guys are brilliant and have a great presentation style
Edit: Answer is C
was waiting for you mam ... please upload the next video soon
PROGRAMMING TASK:
#include
int main() {
int num;
printf("Enter number here: ");
scanf("%d", &num);
(num%2==0)?printf("The number is even."):printf("The number is odd.");
return 0;
}
PROGRAMIZ QUIZ ANSWER: Option(B) 5>3?5:3;
Ma'am thank you so much, I really appreciate ur work, and it really help me so much❤
Thank you, padma ❤
Programming Task:
#include
#include
int main() {
int number;
printf("Enter Number: ");
scanf("%d", &number);
(number % 2==0) ? printf("EVEN #") : printf("ODD #");
return 0;
}
How is the flow chart drawn?
But is this advisable to use in an exam
#include
int main (){
int num ;
printf("Enter a number :");
scanf("%d", &num);
if ( num % 2 == 0){
printf("The number is even.");
}
else if (num % 2 != 0){
printf("The number is odd.");
}
return 0;
}
it is coded in the common style.
VERY GREAT CONTENT
excellent teacher
option B.(5>3)? 5:3;
6:20 the answer is C
Mam, aapke liye chand taare tod ke laa sakta hoo.......coding challenge to keya chiz hai ? Ye lo answer....
#include
#include
int main()
{
int num;
printf("Enter the age
");
scanf("%d",&num);
(num%2==0)?printf("The number is even"):printf("The number is odd");
return 0;
}
Aapka 2nd question ka answer hai option B.
whats conio
I have a question can I? create a program to integrate into a program like array and basic computer calculator etc. I hope you can answer thank you
program is #include
int main() {
int num;
printf("enter the number: ");
scanf("%d", &num);
(num % 2 == 0) ? printf("number is even") :
printf("number is odd");
return 0;
}
Hello madam.
Mind is displaying "number is even" when I haven't insert anything.
int number;
(number%2==0)? printf("number is even"):
printf("number is odd");
why do I get an error when I try to do the operation with a "double" data type. I expected the integer to be converted to double since the double is greater than the int in the ranking
Thanks 👍
The real MVP
its going good !!
Thank you very much!!
good quality 😊
#include
int main()
{
int number = 3;
(number % 2==0) ? printf("the number is odd") : printf("the number is even");
return 0;
}
#include
int main() {
double num1=1;
double num2=3,result;
result=1+3;
printf("enter the number: ");
scanf("%d", &result);
(result>=4)? printf("the number is even"):printf("the numer is odd");
return 0;
}
6:21 answer is C C C
I think the answer is B. Because the condition in the if...Else statement is if (5 > 3) instead of if the result = (5 > 3); for in that case the right answer would be C. Can anyone confirm?
Love it❤
option c is the correct answer
The answer is option C)
i’m learning sm
programming task answer
void main(){
int number;
printf("enter number:");
scanf("%d", &number);
(number %2==0)? printf("the number is even"): printf("the number is odd");
}
dont forget to include the library header
// Online C compiler to run C program online
#include
int main() {
int number;
printf("enter your number ");
scanf("%d",&number);
(number %2==0)? printf("num is even") : printf("nub is odd");
return 0;
}
Nice❤❤❤
The answer is option B.
// Online C compiler to run C program online
#include
int main() {
// Write C code here
int number = 11;
int result = (number % 2 == 0) ? printf("it is even") : printf("it is odd");
return 0;
}
#include
int main(){
int num=3;
(num%2==0)? printf("Number is even"):printf("Number is odd");
return 0;
}
Thanks ma'am
Option c
// Online C compiler to run C program online
#include
int main() {
int num =25;
(num%2)? printf("The numberis odd") : printf("The number is even");
return 0;
good work
C is the right answer
the answers people are giving for the programming task that is given is kind of out of the question because she said that we should assign a value to the variable number but no one is doing that,i am trying to follow the instruction that she gave but its damn hard
#include
int main () {
int x = 3;
(x%2 == 0) ?
printf("The number is even"):
printf("The number is odd");
return 0;
}
Correct ans is B
Answer for the que?
I hope this will be the correct answer, I have tried it on online compiler provided by Programiz.
/*This is a program to check whether a number is EVEN or ODD by enquiring an input from a user.*/
#include
int main(){
int number;
printf("Enter any number!
");
scanf("%d", &number);
(number%2==0) ? printf("The number %d is Even.",number) : printf("The number %d is Odd.",number);
return 0;
}
//Thank U for helping us.
#include
int main()
{
int num = 4;
(num%2==0)?
printf("This is an even number"):
printf("This is an odd number");
return 0;
}
int main() {
// Write C code here
int num;
printf("Enter number: ");
scanf("%d", &num);
if (num>0){
printf("The number %d is a positive number.", num);
}
else if (num
thanks a lot
int a;
printf("Enter the a:");
scanf("%d",&a);
(a%2==0)?printf("even number"):printf("odd number");
return 0;
programming task
int number;
printf("Enter a number: ");
scanf("%d", &number);
int resulte = (number%2 == 0) ? printf("
an even number") : printf("
an odd number");
Thank youu
Answer is C. result = 5 > 3 ? 5 : 3;
Looks like i am revising javascript
//division method
#include
int main()
{
int num;
printf("Enter the number:");
scanf("%d",&num);
(num/2==0)? printf("It is an even number") : printf("It is an odd number");
return 0;
}
I think you should write (num%2==0), because in that way we are looking at the number that is left after dividing, no the actual result.
Option C mam!!
WHY C ???
@@shreyasy9164 you solv the ternary operator ! you must ans in option c
C is the answer
It just shows error in char line
#include
int main() {
int number = 21;
(number%2==0) ? printf("Even ") :printf("Odd");
return 0;
}
int main() {
int num;
scanf("%d", &num);
( num % 2 == 0) ? printf("Number is even") : printf("Number is odd");
return 0;
C option for quiz mam