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We have the question is there a performance difference between i++ and ++i in C?
What's the answer for C++?
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++i is sometimes faster than, and is never slower than, i++. For intrinsic types like int, it doesn't matter: ++i and i++ are the same speed. For class types like iterators or the previous FAQ's Number class, ++i very well might be faster than i++ since the latter might make a copy of the this object.
According to the Google C++ Style Guide, "when the return value is ignored, the 'pre' form ( ++i ) is never less efficient than the 'post' form ( i++ ), and is often more efficient."
There is no difference in your case. --i is pre-decrement and i-- is post-decrement.
In C++ the faster is ++i; due to its object. However some compilers may optimize the post-increment operator.
[Executive Summary: Use ++i if you don't have a specific reason to use i++.]
For C++, the answer is a bit more complicated.
If i is a simple type (not an instance of a C++ class), then the answer given for C ("No there is no performance difference") holds, since the compiler is generating the code.
However, if i is an instance of a C++ class, then i++ and ++i are making calls to one of the operator++ functions. Here's a standard pair of these functions:
Foo& Foo::operator++() // called for ++i { this->data += 1; return *this; } Foo Foo::operator++(int ignored_dummy_value) // called for i++ { Foo tmp(*this); // variable "tmp" cannot be optimized away by the compiler ++(*this); return tmp; } Since the compiler isn't generating code, but just calling an operator++ function, there is no way to optimize away the tmp variable and its associated copy constructor. If the copy constructor is expensive, then this can have a significant performance impact.
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