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How to return from multiple layers in the call stack?


By : Lakhvir Singh Gurm
Date : September 16 2020, 08:00 AM
like below fixes the issue Assume we have a game loop, along the lines of this Python pseudocode , Return an error and match on the error in the main function:
code :
#[derive(Debug)]
enum Error {
    SuccessfulExit,
}

fn handle_events() -> Result<(), Error> {
    // for e in get_frame_events() {
        // if e.kind == CloseEvent {
            return Err(Error::SuccessfulExit);
        // }
    // }

    // Ok(())
}

fn gameloop() -> Result<(), Error> {
    loop {
        handle_events()?;
        // handle_physics();
        // render();
    }
}

fn main() {
    match gameloop() {
        Ok(()) | Err(Error::SuccessfulExit) => return,
        Err(e) => eprintln!("Error: {:?}", e),
    }
}


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Relationship between performance and the number of call stack layers?


By : user3088673
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , There should be no correlation with the depth.
While "chained" scopes do need to be traversed, this is related to lexical scopes and not call stacks. The cost of the call frame ("method call") should be constant -- and independent -- from the depth of the frame.

Why does a recursed return call break out of stack without an explicit return statement?


By : N04H
Date : March 29 2020, 07:55 AM
Does that help Things like that only work if accidentally the return value happens to be in the register where the caller expects it. This only works if this is realized by your compiler as a recursive function. Technically it is undefined behavior to use the return value of a function that doesn't provide one.
Edit: On modern architectures the return value of a function for values for which it is possible is passed in a specific hardware register. When you call your function recursively, on the bottom in all cases that hardware register is set to the expect value. If by chance when popping up from recursion that hardware register is never changed, you end up with the correct value.

How to stack multiple layers of conv2d_transpose() of Tensorflow


By : Ayman Kasaby
Date : March 29 2020, 07:55 AM
this one helps. I guess you need to change your 'stride' paramter in conv2d_transpose. conv2d_transpos is like conv2d but input and output are reversed.
For conv2d, the stride and input shape will decide the output shape. For conv2d_transpose, the stride and output shape will decide the input shape. Now your stride is [1 1 1 1], which means output and input of conv2d_transpose is about same (ignoring boundary effect).

R: Replacing several layers in a raster stack with layers from a second raster stack


By : May Sanchez
Date : March 29 2020, 07:55 AM
wish helps you I have a raster stack and need to replace some of the layers it contains with layers from another stack. , This should work:
code :
for(ind in seq(along = l_replace)) r_stack[[l_replace[ind]]][] <- rep_stack[[ind]][]

r_stack
class       : RasterStack 
dimensions  : 10, 10, 100, 4  (nrow, ncol, ncell, nlayers)
resolution  : 1, 1  (x, y)
extent      : 0, 10, 0, 10  (xmin, xmax, ymin, ymax)
coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0 
names       : layer.1, layer.2, layer.3, layer.4 
min values  :    1.00,  -54.25,    2.00,  -54.25 
max values  :   100.0,    51.5,   100.0,    51.5
for(ind in seq(along = l_replace)){
  setValues(r_stack, getValues(rep_stack[[ind]]), layer = l_replace[ind])
} 

Stack multiple images in semitransparent layers with Imagemagick?


By : Tarek Khattab
Date : March 29 2020, 07:55 AM
it helps some times There are several ways to do this in Imagemagick, depending upon what you are really trying to achieve. Using version 6.9.9.12 Q16 Mac OSX.
I have corrected the commands to fix a typo in the arguments of the third example and included Xenoid's suggested weights to make the first example below become equivalent to the other two so that each result is the same as averaging the images.
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