Correct but men are more prone to it.
Answer:
"The value of the variable will remain the same which is already have when the sub-processor is called".
Explanation:
The above question said that:-
void fun(int a)
{
a=a+1;
}
void main()
{
int a=5;
fun(a);
}
//what will be the value of a in the main function after the fun function is excuted.
- Then the answer is: the value of a will be 5 in the main function.
- It is because when the fun function is called, then a variable that is defined in the fun function is a local variable for fun function. That scope after the fun function is null.
- The a variable inside the fun function is a different variable and the main function a variable is also a different variable.
- So when the user prints the value of a variable inside the fun function, it will give the result as 6.
- But when he prints the value of a variable inside the main function, then it will give the value as 5.
Answer:The answer is near field communication.
Explanation:
Answer:
C 1,2,3,4
Explanation:
This means I want to count from 0-4 and set it o the current loop I am currently on.
for i in range(0, 5):
print("i currently equals: ", i)
The result will be
i currently equals: 0
i currently equals: 1
i currently equals: 2
i currently equals: 3
i currently equals: 4
