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Why doesn't pop_back() have a return value? I have Googled regarding this and found out that it makes it more efficient. Is this the only reason for making it so in the standard?
869
+1 this is the correct answer. It tells us the underlying principle which led to this design of pop_back with no return value.
pop_back() is used to remove/pop the element from the back or the last of the list container. When we use pop_back then it remove/pops the last element and the element before the last element becomes the last element and the size of the list container is reduced by 1.
vector::pop_back()() pop_back() function is used to pop or remove elements from a vector from the back. The value is removed from the vector from the end, and the container size is decreased by 1.
No. pop_back() will not shrink the capacity of vector.
Efficiency has little (or nothing, really) to do with it.
This design is the outcome of an important paper by Tom Cargill, published in the 90s, that raised quite a few eyebrows back then. IIRC, in it Cargill showed that it is impossible to design an exception safe stack pop function.
78
I think there is something related to the fact that copying an instance of the last object could throw an exception. When doing so, you're losing your object, since pop_back() did remove it from your container. Better with a few lines of code:
std::vector<AnyClass> holds = {...} ; try { const AnyClass result = holds.pop_back(); // The copy Ctor throw here! } catch (...) { // Last value lost here. } If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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