refrence in c++
#include<stdio.h>
#include<conio.h>
int main(void)
{
clrscr();
int x=100;
int&y=x;
int&a=x;
printf("%d%d%d",x,y,a);
x++;
y++;
a++;
printf("%d%d%d",x,y,a);
while(!kbhit());
return 0;
}
output
100 100 100
103 103 103
why it is so?
it works on both increment and decrement operators
#include<conio.h>
int main(void)
{
clrscr();
int x=100;
int&y=x;
int&a=x;
printf("%d%d%d",x,y,a);
x++;
y++;
a++;
printf("%d%d%d",x,y,a);
while(!kbhit());
return 0;
}
output
100 100 100
103 103 103
why it is so?
it works on both increment and decrement operators
y and a are different names for the variable x. There is only one variable; y and a are other names you can use to mean the exact same variable.
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thanx for the reply sir ...........
i am not completly satisfy with your ans......a/c to u
x++ //adjust the value of x then there should be increment in the value of x by 1
y++ //adjust the value of y, which is just another name for x so adjust the value of y by 2
in the same way a++ is adjust the value of y which just another name for x so increment in the value of y will be 3
but here scenerao is diffrent.......
sir please clear the doubt
i am not completly satisfy with your ans......a/c to u
x++ //adjust the value of x then there should be increment in the value of x by 1
y++ //adjust the value of y, which is just another name for x so adjust the value of y by 2
in the same way a++ is adjust the value of y which just another name for x so increment in the value of y will be 3
but here scenerao is diffrent.......
sir please clear the doubt
Read the code as this...
int main(void)
{
clrscr();
int x=100;
int x=x;
int x=x;
printf("%d%d%d",x,x,x);
x++;
x++;
x++;
printf("%d%d%d",x,x,x);
while(!kbhit());
return 0;
}
int main(void)
{
clrscr();
int x=100;
int x=x;
int x=x;
printf("%d%d%d",x,x,x);
x++;
x++;
x++;
printf("%d%d%d",x,x,x);
while(!kbhit());
return 0;
}
closed account (1vRz3TCk)
JAI SINGH,
Effectively you have this:
Three names for one value.
Incrementing any of the names will increment the one value.
Reading any of the names will read the one value.
Effectively you have this:
+-----+
X = ----> | 100 |
| +-----+
Y = -+
|
A = -+ |
Three names for one value.
Incrementing any of the names will increment the one value.
Reading any of the names will read the one value.
| but here scenerao is diffrent...... |
No, it isn't different. What I said is exactly what happened.
| y++ //adjust the value of y, which is just another name for x so adjust the value of y by 2 |
What are you talking about? That's nonsense. y++ means increase the value of y by one. Not by two. The value of y was ALREADY increased when we did x++, because y is just another name for x. y and x mean the exact same object in the exact same piece of memory.
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