Pre-increment (or pre-decrement) in 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)

// CPP program to illustrate
// Pre-increment (or pre-decrement)
#include <cstdio>
  
int main()
{
    int a = 10;
  
    ++a = 20; // works
  
    printf("a = %d", a);
    getchar();
    return 0;
}

a = 20

The above program works whereas the following program fails in compilation with error “non-lvalue in assignment” (a++ is used as l-value)

// CPP program to illustrate
// Post-increment (or post-decrement)
#include <cstdio>
  
int main()
{
    int a = 10;
    a++ = 20; // error
    printf("a = %d", a);
    getchar();
    return 0;
}

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 brevity, 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 which 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.



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Improved By : SangeethSudheer

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