Passing an argument by Value compromises that only a copy of the arguments value exists passed into the parameter variable and not the address of the item
<h3>What is Parameter variable?</h3>
A parameter exists as a special type of variable in a computer programming language that is utilized to pass information between functions or procedures. The actual information passed exists called an argument. A parameter exists as a named variable passed into a function. Parameter variables exist used to import arguments into functions.
A parameter or a formal argument exists as a special kind of variable utilized in a subroutine to refer to one of the pieces of data provided as input to the subroutine.
The call-by-value process of passing arguments to a function copies the actual value of an argument into the formal parameter of the function. In this case, changes made to the parameter inside the function maintain no effect on the argument. By default, C++ utilizes call-by-value to pass arguments.
Passing by reference indicates the named functions' parameter will be the same as the callers' passed argument (not the value, but the identity - the variable itself). Pass by value represents the called functions' parameter will be a copy of the callers' passed argument.
Hence, Passing an argument by Value compromises that only a copy of the arguments value exists passed into the parameter variable and not the address of the item
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nu it would not but if they call microsoft to check it then yez the history would show up for them but dont try to make it notice able so they dont have to call
Answer:
<u>Call by reference</u>
In an function if the variables are passed as reference variables this means that the variables are pointing to the original arguments.So the changes made in the function on the reference variables will be reflected back on the original arguments.
For example:-
#include<stdio.h>
void swap(&int f,&int s)
{
int t=f;
f=s;
s =temp;
}
int main()
{
int n,m;
n=45;
m=85;
swap(n,m);
printf("%d %d",m,n);
return 0;
}
the values of m and n will get swapped.
<u>
Call by value</u>
In this program the values of m and n will not get swapped because they are passed by value.So duplicate copies of m and n will be created and manipulation will be done on them.
#include<stdio.h>
void swapv(int f,int s)
{
int t=f;
f=s;
s=temp;
}
int main()
{
int n,m;
n=45;
m=85;
swapv(n,m);
printf("%d %d",n,m);
return 0;
}
Explanation:
Age = 23.
To convert a base 10 number to hexadecimal number we have to repeatedly divide the decimal number by 16 until it becomes zero and store the remainder in the reverse direction of obtaining them.
23/16=1 remainder = 5
1/16=0 remainder = 1
Now writing the remainders in reverse direction that is 15.
My age in hexadecimal number is (15)₁₆.
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