Prerequisite: Pre-increment and post-increment in C/C++
In C++, pre-increment (or pre-decrement) can be used as l-value, but post-increment (or post-decrement) can not be used as l-value.
For example, following program prints a = 20 (++a is used as l-value)
l-value is simply nothing but the memory location, which has an address.
CPP
#include <cstdio>
int main()
{
int a = 10;
++a = 20;
printf ( "a = %d" , a);
printf ( "\n" );
--a = 10;
printf ( "a = %d" , a);
return 0;
}
|
Output:
a = 20
a = 10
Time Complexity: O(1)
The above program works whereas the following program fails in compilation with error “non-lvalue in assignment” (a++ is used as l-value)
CPP
#include <cstdio>
int main()
{
int a = 10;
a++ = 20;
printf ( "a = %d" , a);
return 0;
}
|
Error:
prog.cpp: In function 'int main()':
prog.cpp:6:5: error: lvalue required as left operand of assignment
a++ = 20; // error
^
How ++a is Different From a++ as lvalue?
It is because ++a returns an lvalue, which is basically a reference to the variable to which we can further assign — just like an ordinary variable. It could also be assigned to a reference as follows:
int &ref = ++a; // valid
int &ref = a++; // invalid
Whereas if you recall how a++ works, it doesn’t immediately increment the value it holds. For clarity, you can think of it as getting incremented in the next statement. So what basically happens is that, a++ returns an rvalue, which is basically just a value like the value of an expression that is not stored. You can think of a++ = 20; as follows after being processed:
int a = 10;
// On compilation, a++ is replaced by the value of a which is an rvalue:
10 = 20; // Invalid
// Value of a is incremented
a = a + 1;
That should help to understand why a++ = 20; won’t work. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.