This question already has answers here: Closed 11 years ago. P开发者_高级运维ossible Duplicate: When all does comma operator not act as a comma operator?
In the following type of code is there a sequence point between each variable construction, or is the result undefined?
The line below is inside a for loop. If the method fails, it needs to break. CATEGORY_1 is an enum. I added this enum as a new parameter to AddToList method with a default value. If you see closely be
MDN claims that: The comma operator evaluates 开发者_StackOverflowboth of its operands (from left to right)
I came across this code snippet 1 int return_printChar1() { // code // oops! no return statement } int return_printChar2()
I have read in a lot of places but I really can\'t understand the specified behavior in conditionals.
I am reading Bjarne Stroustrup\'s Programming : Principles and Practice Using C++ In the drill section for Chapter 2 it talks about various ways to look at typing errors when compiling the hello_worl
If you see this code, class A{ public: A(int a):var(a){} int var开发者_StackOverflow; }; int f(A obj) { return obj.var;
I have been programming in C and C++ for a few years and now I\'m just now taking a college course in it and our book had a function like this for an example:
is it possible to construct variadic arguments for function by overloading operator comma of the argument? i want to see an example how to do so.., maybe something like this:
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