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CSS style not inherited from class

 .subcat li {
   font-weight: 400;

change .subcat to .subcat li

ul, li {
    font-weight: 700;

.subcat li {
    font-weight: 400;
        <ul class="subcat">

Categories : CSS

Related to : CSS style not inherited from class
Does an inherited class realize the interface the class's parent already realized in java?
1) Implementing an interface (Q) is a contract stating that your class contains certain methods. 2) When you make a subclass (R), all methods from the super class (W) are inherited by the subclass. 3) So..what do you think?

Categories : Java
I can't create public inherited class - Something is wrong with constructors C++
Monitor does not have a default constructor. You have to invoke its constructor in the constructor of the derived class. That's because otherwise the Prodaja instance doesn't know how to construct its Monitor part. So you need Prodaja (int monster1): Monitor(monster1) or to define a constructor of Monitor that takes no parameters (or has a default parameter).

Categories : C++
Apply css style of one class into another class
Im not sure i understand you completely but i think this is what you mean. So add a class which will only be applied within another class you would do the folowing .col-sm-2 .x-col-sm-left { /*Style you want here*/ } so now x-col-sm-left can be used in col-sm-2 once you have a parent of col-sm-2.. this will overwrite .col-sm-2{ /*Syle*/ } as for inheritance, you can use the , do defin

Categories : CSS
WPF TreeView style template resetting IsExpanded on style switch
When you take away a template or replace by another some dependency properties will reset to their default value. The reason for that is dependency property precedence. Take a look at this here: Think of it like this: A dependency property may contain many values which are layered. The value in top most layer is always the current value. If you take a st

Categories : Wpf
Use inherited member operator instead of free one
Short answer, you can do this: template <typename T> struct DerivedA : Base<T> { using Base<T>::operator*; }; Long answer: in order to figure out what *d to call, we have to determine all the viable functions (§13.3.2): From the set of candidate functions constructed for a given context (13.3.1), a set of viable functions is chosen, from which the best function wil

Categories : C++
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