Why Does the Erase-Remove Idiom Fail to Eliminate Pairs with .first Set to 4 in this Vector?

Harnessing the Power of Erase-Remove Idiom for Vector Inhabitation

When attempting to eliminate pairs from a vector of pair elements, the erase-remove_if idiom emerges as a versatile tool. However, proper implementation is crucial to ensure desired results. In this question, we explore a modification to the provided code to achieve intended functionality.

The original code sought to remove pairs with .first set to 4 from a vector using erase-remove_if. However, the code failed to achieve this due to an incorrect specification of the erase range in the erase() method.

The correct syntax, as provided in the answer, is:

stopPoints.erase(std::remove_if(stopPoints.begin(),
                                stopPoints.end(),
                                [](const stopPointPair stopPoint)-> bool { return stopPoint.first == 4; }), 
                 stopPoints.end());

This modification extends the erase range to include the entire segment of elements that matches the predicate defined in the lambda function. This effectively removes all matching elements and preserves the desired vector composition, as illustrated by the corrected output:

- 2, Up
- 6, Up

The erase-remove idiom relies on the interplay of remove_if and erase functions. remove_if identifies and shifts non-matching elements towards the beginning of the vector, while erase removes the range of matching elements starting from the iterator returned by remove_if. Understanding this mechanism is key to effectively employing this idiom.

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