A Puzzle on C/C++ R-Value Expressions?

Here we will see one puzzle about R-value expressions in C. Suppose there is a program which is given below, we have to tell what will be the output and why?

Syntax

~expression;     // R-value: computes but doesn't store
variable = ~expression;  // L-value: stores the result

Example 1: R-value Expression

In this example, the complement operation is performed but not assigned to any variable −

#include <stdio.h>

int main() {
    int x = 0xab;
    ~x;  /* R-value: computed but not stored */
    printf("%x", x);
    return 0;
}

ab

So, there is no change. But why? The reason is ~x is converting x into its complemented form, but that value is not assigned to any variable. The expression is an R-value expression. Until an L-value assignment is used, the computed result will not be stored into the variable.

Example 2: L-value Assignment

If we assign the complement result back to the variable using L-value assignment, it will look like this −

#include <stdio.h>

int main() {
    int x = 0xab;
    x = ~x;  /* L-value: result is stored back to x */
    printf("%x", x);
    return 0;
}

ffffff54

Key Points

• R-value expressions compute a value but don't modify any variable unless assigned.
• L-value assignments store the computed result into a variable.
• The bitwise NOT operator ~ flips all bits of the integer.
• In a 32-bit system, ~0xab becomes 0xffffff54.

Conclusion

This puzzle demonstrates the difference between R-value expressions and L-value assignments in C. An expression like ~x computes the result but doesn't store it unless explicitly assigned to a variable.