How can a referece to integer constant change value in a function?
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Could someone pls explain the below question? Q1 : In line1 the second parameter is referece to integer constant. If so how is the value in 'x' changed to 4 Question:
#include
#include
using namespace std;
int mani(int (&arr)[10], const int& x)//Line1
{
for (int i = 0; i < 1; ++i)
arr[i] = arr[i] + x;
return x; //returns 4
}int main(){
int arr[10] = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
cout< -
Could someone pls explain the below question? Q1 : In line1 the second parameter is referece to integer constant. If so how is the value in 'x' changed to 4 Question:
#include
#include
using namespace std;
int mani(int (&arr)[10], const int& x)//Line1
{
for (int i = 0; i < 1; ++i)
arr[i] = arr[i] + x;
return x; //returns 4
}int main(){
int arr[10] = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
cout<Member 3974347 wrote:
If so how is the value in 'x' changed to 4
Since the variable x is a reference to the
a[0], when ever you use x, keep in mind that you are actually usinga[0]itself. And in the for loop, when the value ofi = 0, you chnages the value ofa[0]and so the x will also points to new value. Defining the x as constant only prevents directly chaging the value of x, using the x variable itself. something likex = a[2];will be shown as error. -
Could someone pls explain the below question? Q1 : In line1 the second parameter is referece to integer constant. If so how is the value in 'x' changed to 4 Question:
#include
#include
using namespace std;
int mani(int (&arr)[10], const int& x)//Line1
{
for (int i = 0; i < 1; ++i)
arr[i] = arr[i] + x;
return x; //returns 4
}int main(){
int arr[10] = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
cout<Well, "x" an indeed an "integer constant" from the our point of view: by taking a constant reference, we obligate ourselves not to change the value of "x", we can only read it. However, it does not mean that the value of x has to be constant, it's just that _we_ can't modify the respective memory area through this variable named "x". If it's not a constant actually, then anyone else, or even us, through an other alias can modify the memory area (in this example through the non-const reference to the array), the modified value of course reflecting also when accessing the value of x. So, in this example, "x" is just an alias to the first element of the "arr" array marked const (=you can't modify the value through me), but since the function takes the array by reference (thereby obtaining an alias for the original "arr" array) and changes its first element (remember, the array is a non-const reference, it can be changed), it actually changes the first element of the original, global "arr" array, but "x" is still just an alias for this, thereby its value must be 2 + 2 = 4.