What Does the Unary Plus Operator Actually Do in C/C ?

The often-overlooked Unary Plus Operator in C/C

The unary plus operator ( ) in C and C is frequently misunderstood. While seemingly trivial, it performs a non-trivial type conversion on its operand, leading to potentially significant changes in value and data type.

At first glance, applying the unary plus operator appears to do nothing. However, it implicitly performs standard mathematical conversions on its operand, potentially resulting in a value with a wider integer type.

Specifically, if the operand is an unsigned integer with a smaller bit width than the standard int type, the unary plus operator converts it to a signed integer. This grants the ability to represent negative numbers, a crucial difference.

While seemingly minor, this conversion can be important in specific situations. Avoid using the unary plus operator simply as a visual cue to indicate a positive number; its function is more nuanced.

Consider this C example:

<code class="language-c++">void foo(unsigned short x) {
    std::cout << typeid(+x).name() << std::endl; 
}

int main() {
    unsigned short us = 10;
    foo(us); // Output will likely be something like "i" (for int)
    return 0;
}</code>

This code showcases the unary plus operator's type conversion. The output shows that x is of type int, demonstrating the conversion from unsigned short to signed int.

The unary plus operator thus acts as a powerful tool for integer manipulation, ensuring well-defined behavior and avoiding unexpected results by explicitly managing type conversions in C and C code.

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