Understanding Why gcc Expands int16_t to int32_t on Unary Minus Operator скачать в хорошем качестве

Understanding Why gcc Expands int16_t to int32_t on Unary Minus Operator

Discover the reasoning behind `gcc` expanding `int16_t` to `int32_t` when applying the unary minus operator, along with the principles of integral promotion in C++. --- This video is based on the question https://stackoverflow.com/q/74478648/ asked by the user 'Byron Hawkins' ( https://stackoverflow.com/u/351410/ ) and on the answer https://stackoverflow.com/a/74480313/ provided by the user 'jcmvbkbc' ( https://stackoverflow.com/u/1943346/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions. Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Why does gcc expand int16_t to int32_t on unary minus operator? Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l... The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license. If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com. --- Understanding Why gcc Expands int16_t to int32_t on Unary Minus Operator If you've ever worked with C++ programming, particularly with the gcc compiler, you might have come across an interesting behavior: when applying the unary minus operator to a variable of type int16_t, gcc produces an int32_t result. This might be puzzling, particularly if your expectations are based on the size of the data type you are working with. In this guide, we will break down this phenomenon, explore the underlying logic, and help clarify how type promotion works in C++. The Problem in a Nutshell Consider the following example code: [[See Video to Reveal this Text or Code Snippet]] Expected vs. Actual Output You might expect the output to reflect that both i and -i are of type int16_t, hence it should display: [[See Video to Reveal this Text or Code Snippet]] However, the actual output you get is: [[See Video to Reveal this Text or Code Snippet]] So why does this happen? Let’s delve into the solution. The Logic Behind the Promotion The key to understanding this behavior lies in the C++ standards regarding arithmetic types and integral promotion. Let's break it down into some key points: Integral Promotion According to the C++ standard (specifically C++11 5.3.1:7), when using the unary minus operator: The operand (the variable being negated) must have an arithmetic or unscoped enumeration type. The result will be the negation of its operand. This means that if your operand is an integral type (like int16_t), it undergoes a process called integral promotion. What Is Integral Promotion? Integral promotion is a process defined in the C++ standard whereby certain smaller integer types are automatically converted to a larger type to facilitate operations. The relevant section (C++11 4.5:1) states: A prvalue of an integer type, other than bool, char16_t, char32_t, or wchar_t, whose integer conversion rank is less than that of int, is converted to int. If int cannot represent all the values of the source type, it can be converted to unsigned int. In simpler terms, since int16_t has a rank lower than int, it gets promoted to int before any operation is applied. Result of the Unary Minus Operation When you apply the unary minus operator on i, the following occurs: The int16_t i is promoted to int (which is typically int32_t on most platforms). The unary minus operator is applied to this promoted type. Thus, the result of -i is of type int, leading to the observed output of sizeof(-i): 4. Conclusion Understanding type promotions, especially in the context of unary operators, can clarify many unexpected behaviors that arise when we compile our code with gcc. The important takeaway is that the C++ language standard specifies that smaller integer types are promoted to int, which is why our unary minus operator does not produce the expected size of int16_t in the output. This behavior ensures that arithmetic operations are safe from overflow for smaller types, aligning with C++’s emphasis on performance and type safety. If you ever find yourself puzzled by similar issues, remember that integral promotion is often at play! Now that you're informed about this intrinsic behavior in C++, you can write your code with clearer expectations. Happy coding!