I was playing around with bool variables. I am aware that boolean is used to represent a true (any other number besides 0) or false (the number 0). I realized that the variables create random integer numbers. I was wondering if one can one use boolean variables to generate random numbers? Can someone please elaborate on this behaviour caused by the keyword boolean?
My code is as follows:
#include<stdio.h>
#include<stdbool.h>
int main()
{
bool a,b,c,d,e,f,g,h,i,j,k;
printf("%d\n",a);
printf("%d\n",b);
printf("%d\n",c);
printf("%d\n",d);
printf("%d\n",e);
printf("%d\n",f);
printf("%d\n",g);
printf("%d\n",h);
printf("%d\n",i);
printf("%d\n",j);
printf("%d\n",k);
}
This is not behaviour specific to a bool. This is caused because you are using an uninitialized variable. This is not only not random, it is not safe. This is undefined behaviour, you should avoid it at all costs.
Boolean is not randomly defined. These variables take false as default value. In compiler. The default value is consistent rather than random 0 or 1 from time to time.
Related
I am getting an output of 24 which is the factorial for 4, but I should be getting the output for 5 factorial which is 120
#include <stdio.h>
int factorial(int number){
if(number==1){
return number;
}
return number*factorial(--number);
}
int main(){
int a=factorial(5);
printf("%d",a);
}
Your program suffers from undefined behavior.
In the first call to factorial(5), where you have
return number * factorial(--number);
you imagine that this is going to compute
5 * factorial(4);
But that's not guaranteed!
What if the compiler looks at it in a different order?
What it if works on the right-hand side first?
What if it first does the equivalent of:
temporary_result = factorial(--number);
and then does the multiplication:
return number * temporary_result;
If the compiler does it in that order, then temporary_result will be factorial(4), and it'll return 4 times that, which won't be 5!. Basically, if the compiler does it in that order -- and it might! -- then number gets decremented "too soon".
You might not have imagined that the compiler could do things this way.
You might have imagined that the expression would always be "parsed left to right".
But those imaginations are not correct.
(See also this answer for more discussion on order of evaluation.)
I said that the expression causes "undefined behavior", and this expression is a classic example. What makes this expression undefined is that there's a little too much going on inside it.
The problem with the expression
return number * factorial(--number);
is that the variable number is having its value used within it, and that same variable number is also being modified within it. And this pattern is, basically, poison.
Let's label the two spots where number appears, so that we can talk about them very clearly:
return number * factorial(--number);
/* A */ /* B */
At spot A we take the value of the variable number.
At spot B we modify the value of the variable number.
But the question is, at spot A, do we get the "old" or the "new" value of number?
Do we get it before or after spot B has modified it?
And the answer, as I already said, is: we don't know. There is no rule in C to tell us.
Again, you might have thought there was a rule about left-to-right evaluation, but there isn't. Because there's no rule that says how an expression like this should be parsed, a compiler can do anything it wants. It can parse it the "right" way, or the "wrong" way, or it can do something even more bizarre and unexpected. (And, really, there's no "right" or "wrong" way to parse an undefined expression like this in the first place.)
The solution to this problem is: Don't do that!
Don't write expressions where one variable (like number) is both used and modified.
In this case, as you've already discovered, there's a simple fix:
return number * factorial(number - 1);
Now, we're not actually trying to modify the value of the variable number (as the expression --number did), we're just subtracting 1 from it before passing the smaller value off to the recursive call.
So now, we're not breaking the rule, we're not using and modifying number in the same expression.
We're just using its value twice, and that's fine.
For more (much more!) on the subject of undefined behavior in expressions like these, see Why are these constructs using pre and post-increment undefined behavior?
How to find the factorial of a number;
function factorial(n) {
if(n == 0 || n == 1 ) {
return 1;
}else {
return n * factorial(n-1);
}
//return newnum;
}
console.log(factorial(3))
#include<stdio.h>
int fact(int);
int main()
{
int a, b;
printf("enter a number : ");
scanf("%d",&a);
b=fact(a);
printf("\n%d",b);
}
int fact(int y )
{
int d=1;
for(int i = 1;i<=y;i++)
d*=i;
d= d>0 ? d : 0;
}
If I remove the last statement , O/P is a+1.
I have checked this with other functions and the function returns correct values if I use if statement or conditional operator.
I want to know why this happens.
Thank You.
6.9.1 Function definitions
...
12 If the } that terminates a function is reached, and the value of the function call is used by
the caller, the behavior is undefined.
Simply put, the behavior you're seeing is purely accidental. There's no good reason why you should get that particular result or any other.
Undefined means that the code is erroneous, but neither the compiler nor the runtime environment are required to handle it in any particular way. What's likely happening in your case is that the register used to return a value from a function is also being used to store the value of d, but that doesn't have to be true. If you change up the code, you may get a different result.
I'm fairly new to coding and am currently learning C. In my C programming class, my instructor gave us the assignment of writing a program that uses a function which inputs five integers and prints the largest. The program is fairly simple even for me, but I'm facing some problems and was hoping to get some advice.
#include <stdio.h>
int largest(int x);
int main(void) {
int integer1;
largest(integer1);
return 0;
}
int largest(int x) {
int i;
for (i = 0; i < 5; i++) {
printf("Enter an integer: ");
scanf_s("%d", &x);
}
return x;
}
This is the code that I have written. The main problem that I am having is that in my main method, the IDE tells me to initialize the value of integer1. However, I'm not really sure how to do that because I'm supposed to input the value within the largest() method via the scanf_s function. How may I solve this?
The problem is here, the warning message is to warn you about the potential pitfall of using the value of an uninitialized automatic local variable. You made the call like
largest(integer1);
but you ignore the return value, so the integer1 remains uninitialized.
Remember, in view of largest(), x is a local copy of the actual argument passed to that function, any changes made to x won't be reflecting to the caller.
That said, your code is nowhere near your requirement, sorry to say. A brief idea to get there would be
Create a function.
Create a variable (say, result) and initialize with minimum possible integer value, INT_MIN
Loop over 5 times, take user input, compare to the result value, if entered value found greater, store that into result, continue otherwise.
return result.
I know that normally help for assignments shouldn't be given but I have to say that you might need to rethink what you want to do.
You are inputting an integer to the function named largest. But why are you only inputting a single integer to a function that should return the largest value. You can't do much with a single number in that case.
You should instead be inputting say an array of 5 values(as said in your assignment) to the function and let it return the largest.
The order would then be:
Read 5 values and save to an array
Call the function largest with the array as input
Let the function do it's work and return the largest value
Do what ever you want with the largest value
But if you only want to remove the warning simply type
int integer1 = 0;
I wrote the below code to find the sum of all digits in C, and when I compiled and ran it, it was successful.
But, only later I realized that i had not entered any value for the variable 'n' in the for loop's condition.
I'm confused on how this program works, even when there is no value assigned to the condition variable.
I would like to be clarified of the same.
#include<stdio.h>
void main()
{
int no,a,b,n,sum=0;
printf("Enter the number to be added");
scanf("%d",&no);
for(int i=0;i<n;i++)
{
a=no%10;
b=no/10;
sum=a+sum;
no=b;
}
printf("The sum is %d",sum);
}
I'm confused on how this program works
Well, "works" is a very poor observation / decision here. This is undefined behavior.
You're attempting to use the value of an automatic local variable n while it is indeterminate. This invokes the UB.
To quote the C11 standard, chapter ยง6.7.9
If an object that has automatic storage duration is not initialized explicitly, its value is
indeterminate. [...]
So, in your case, n meets the criteria described above, and hence the content is indeterminate.
Now, after that, in case you try to use a variable while it holds indeterminate value and either
does not have the address taken
can have trap representation
the usage will lead to undefined behavior. That is exactly the case here.
That said, for a hosted environment, the conforming signature of main() is int main(void), at least.
An uninitialized variable MUST contain some value - every address in a computer must have some combination of 1's and 0's even if those values are useless. Therefore it is given a random one when first created if you do not initialize it to something yourself. Often it isn't even "given" data so much as it "takes on" whatever value was at the address it was given to live at when created, but different environments will handle this "non-initialization" differently. You are essentially getting lucky right now under your current conditions.
For that reason this kind of code is considered to have undefined behavior because you are not guaranteed to get lucky like that in every situation. Be safe and always initialize values for variables because this can be a hard thing to catch later when code that used to work suddenly doesn't.
n is not initialized, so is worth anything present at the allocated memory location when run. So if you're lucky, after a few iterations no is 0 and the result is correct but it's just luck.
In your case, you don't need n, just stop when division yields 0 (that's probably why you forgot to initialize n)
while(no!=0)
{
a=no%10;
b=no/10;
sum+=a;
no=b;
}
Whenever we call a function returning value why it is not required to catch the value?
consider the following C code,
int main()
{
int i;
scanf("%d",&i);
printf("Value of i is: ",i);
return 0;
}
Here scanf() returns value 1, but as it is not catched in anywhere why didn't the error pops up?
What is the reason to allow such programming?
Primarily because expressions in C also yield values. For example: x = 1; yields the value 1. Sometimes you use that for multiple assignment like x = y = 1;, but more often you don't.
In early C, the void return type hadn't been invented either, so every function returned some value, whether it was generally useful or not (for example, your call to printf also returns a value).
The rules of the language don't make this an error (doing so would lose compatibility with virtually existing code) and since this is common and rarely indicates a problem, most compilers don't warning about it either. A few lint tools do, which has led a few misguided programmers to write things like (void)printf("whatever"); (i.e., casting the unused return to void to signal that it really, truly was intentional when it was ignored. This, however, rarely does any good, and frequently does quite a bit of harm, so (thankfully) it's rarely seen.
Functions that return a value have that functionality for the use and convenience of the programmer. If you aren't interested in the return value, then don't use it.
Do you need the return value in this example? NO. So you have not used that in this case. But in another situation the return value might be important. For example if you want to read as long as some integer in available in input stream, then you can do something like this:
while (scanf("%d", &i) == 1) {
// do something
}
If there is an EOF then this loop will break. Here return value is needed.
So the summary is use return value when needed, and don't use when not needed. Such programming is allowed because both scenario is possible.
A lot of library functions return values you might not think about. Imagine having to use the return value of every printf call! (printf returns the number of characters printed.)
Pretty much every native c function returns a int. scanf and printf included. It would be really annoying if you always had to "capture" it to satisfy the compiler; In a large program you would end up creating thousands of variables just for storing return values that you never look at.
The reason is that C has no other established way of handling errors than by return value. These returned values, that is to say those return to report success or failure, should almost always be checked (unless you're just doodling around or you have a proof that the function will not fail).
Now since return values are also used for other things than returning success/failure information there might be, and are, situations where you will not be interested in the value a function returns, but just the side effects of executing it. In this case forcing the programmer to inspect/bind the returned value would become quite tideous.
Simple: if you don't use the return value, you don't use it.
It is not mandated that you do.
(That said, in a case like this, you should: you have no idea at present whether scanf encountered an error.)
It is not the case that everytime a value is evaluated,& it must be stored or returned,because the value you have obtained may be used by some other functions,to evaluate different kind of things.A value may be a measure...for example consider the following simple programm,where we want to check that the number entered is even or not
'int main()
{
int a;
printf("Enter a number");
scanf("%d",&a);
if(a%2==0)
printf("even number");
else
printf("odd no");
return 0;
}'
here the variable 'a' is not necessarily to be returned,because we just want to check that the number is even or odd...no need of returning