What Does the Unary Plus Operator Actually Do in C?

Deeply explore the mysterious unary plus operator in C language

The unary plus operator ( ) in C is often confusing, with seemingly contradictory behavior. But in fact, it does play an important role.

Reveal its capabilities

Contrary to popular belief, the unary plus operator does not simply "do nothing". It performs subtle but important transformations on its operands, so-called "usual arithmetic transformations". These conversions produce a new value that may have a different width and sign than the original operand.

For example, consider an unsigned integer whose width is less than an integer. Applying the unary plus operator to such a value converts it to a wider-width signed integer. It is worth noting that this conversion affects program execution, as shown in the following C code:

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

void foo(int x) {
  std::cout << "x is an int" << std::endl;
}

int main() {
  unsigned short us = 65535;
  foo(us);             // calls foo(unsigned short)
  foo(+us);            // calls foo(int)  
  return 0;
}</code>

Running this code will output "x is an int". This is because the unary plus operator converts the original unsigned short operand to a signed integer, causing the function to be called with an int parameter.

Impacts and Precautions

While the unary plus operator may seem harmless, it’s important to be aware of its impact. Using the unary plus operator simply as a comment to represent positive integers may lead to unexpected behavior because it triggers type and sign conversions. Therefore, it is best to avoid using the unary plus operator as a substitute for proper documentation or variable naming conventions.

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