Function call in for loop declaration [closed] - c++

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Could anybody explain me how this code is actually working?
when I run it, It gives me 52 as an output.
#include <iostream>
using namespace std;
int Fun(void){
static int x=7;
return x--;
}
int main(void)
{
for(Fun();Fun();Fun())
cout<<Fun();
return 0;
}

Answer 1

Every "regular" for loop has 3 parts :
initialization
condition
iteration
On entering x is 7, initialization will return 7, decrement and go on with the condition where x is now 6. Upon condition it will return 6 (which is truthy), decrement and go on with the body where x is now 5. Upon printing, it will return 5, decrement and go on with the iteration where x is now 4.
Upon iteration it will return 4, decrement and go on with the condition where x is now 3. Upon condition it will return 3 (which is truthy), decrement and go on with the body where x is now 2. Upon printing it will return 2, decrement and go on with iteration where x is now 1.
Upon iteration it will return 1, decrement and go on with the condition where x is now 0. Upon condition it will return 0, which is falsy, and thus break out of the loop.
One of the important things to understand is that x is static and thus is part of the "state" of the function. It is initialized on first call and reused on each call.

Answer 2

probably you better understand with this code, since you are doing cout << Fun(); you see only the value at that time, you can see the return value at each phase with below code.
first initialization of for you get 7,
second condition check you get 6, since its non zero enters into loop.
inside loop call Fun which prints 5
then go to update part of for loop where we receive is 4
after that go to condition check it gives 3, since its non zero enter the loop
inside loop this time it will print 2
then go to update part of for which return 1
then go to condition check which gets 0, loop terminates.
#include <iostream>
using namespace std;
int Fun()
{
static int x=7;
return x--;
}
int main( )
{
int x, y, z = 0;
for( x = Fun(); y = Fun();z = Fun())
cout<<"x = " <<x<<" y = "<<y<<" z = "<< z<<" and Fun ="<<Fun()<<endl;
return 0;
}

Answer 3

After each call of the function Fun the static variable x is decremented. The function returns the value of the variable x before decrementing.
So in this for loop
for(Fun();Fun();Fun())
cout<<Fun();
Before the body of the loop gets the control the function Fun was called two times
for(Fun();Fun();Fun())
1 2
So after these calls the value of the variable x is equal to 5. This value is outputted in the statement
cout<<Fun();
and after the call of the function used in this statement the value of x becomes equal to 4.
Then again the function Fun was called two times
for(Fun();Fun();Fun())
2 1
So the value of the x becomes equal to 2. This value is outputted in the statement
cout<<Fun();
And after the call of the function in this statement the value of x is equal to 1. Then in the first call of Fun
for(Fun();Fun();Fun())
2 1
the value of x becomes equal to 0. This value is returned by the second call of the function. So the condition evaluates to boolean false.
Thus the loop outputted two numbers 5 and 2. The result value of the static variable x is -1. (0 was returned in the last call of Fun and after that the variable x was decremented)

Answer 4

To understand, you can "unroll" the loop:
int Fun()
{
static int x = 7;
return x--;
}
int main()
{
Fun(); // x is 6
if (!Fun()) return 0; // x is 5
std::cout << Fun(); // prints 5, x is 4
Fun(); // x is 3
if (!Fun()) return 0; // x is 2
std::cout << Fun(); // prints 2, x is 1
Fun(); // x is 0
if (!Fun()) return 0; // end
}
Note that Fun() returns a value which equals x+1.