Pre-increment (or pre-decrement) in C++

• Difficulty Level : Medium
• Last Updated : 12 Jul, 2018

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 assignmenta++ = 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; // validint &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 incrementeda = a + 1;