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i have a simple function, in which i need to divide two integers. But the casting does not work.
I can't understand what is wrong in my code:
double new=0.0;
if(N>0) new = double(Ns)/double(N);
The error-message at the place double new; is (error:expected unqualified-id) and at the place new=double(Ns)/double(N)
and at
new is a reserved keyword in C++. Pick another name for your variable:
double double_new=0.0;
if(N>0) double_new = double(Ns)/double(N);
new is a reserved keyword in C++. You cannot have objects named new.
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auto data = new char[480][640][3]();
char data = new char[480][640][3]();
First works.
Second doesnt.
Why? Isn't auto supposed to just replace itself with the type of the initializer?
Because the type isn't char. The type is char(*)[640][3] and the declaration would be written as
char (*data)[640][3] = new char[480][640][3]();
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In C++, please explain meaning of string = "" in vector declaration -> vector<string = "">v;
Assuming these are std::vector and std::string, it is syntactically invalid. It won't compile.
Otherwise, you need to provide more context like: what are vector, string and v? Are they objects? Objects of what type? Are they macros? Are they classes? Are they templates? It could do anything. Here it prints Hello, World, here it calculates pi, here it creates a std::vector of std::string, and here it doesn't compile.
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Why is g++ giving an error like this?
blahblah.h:80:10: error: decomposition declaration not permitted in this context
float[NUM_OUTPUTS] output_buffer;
(Already solved, but creating this because there's no good google hits for this error text, and the error message is inscrutable.)
In C++ declarations, the array size goes after the variable name, not after the type:
float output_buffer[NUM_OUTPUTS];
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I am using the following code to check if all the elements in a list are the same:
def sameItem(myList):
return all(x==myList[0] for x in myList)
However, in my test case:
myL1 = ['dog','cat','dog']
sameItem(myL1)
returns True. Shouldn't it be False? Or did I have a bug in the sameItem() function?
Also, I am using Jupyter Notebook, could it cause any problem is this scenario?
Thanks!
Your method should be correct and works for me. As an alternative, you can try this method to double check, which is a one line that does the same thing
return myList[1:] == myList[:-1]
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I think I saw something on using this to truncate test as a filter, but I've seen to no idea how to use it. Using as xx|do_trucate(20) gives the following:
TemplateAssertionError: no filter named 'do_truncate'
What is the correct usage?
Doh, from the spec I saw
do_trucate
http://code.nabla.net/doc/jinja2/api/jinja2/jinja2.filters.html
But in reality, its just truncate