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if 01111111 represents the delete key in computer keyboard, and 01001000 represents capital letter 'P' on the keyboard. How does the delete key reduces other characters to null character using 8-bits binary representation?

2007-01-31 20:06:35 · 4 answers · asked by Chimex 1 in Computers & Internet Programming & Design

4 answers

delete key does not make previous character in the buffer null. once delete key is pressed then the application running it gets to handle the event. the application may or may not choose to capture and handle this event. then it is passed to the windows system. thus finally the system handles the delete key.
eg. if a delete key is pressed when cursor was in a text box then the app holding the text box handles the event and sends it to the system. since, gui controls are a part of windows api, the api finally gets to handle the key input. the system detects that the key is delete. it checks the cursor position. it then checks if any action needs to be taken. if there is a character in the cursor position and the control is not read-only it removes the character from the buffer where the test inside the textbox was stored. finally, it removed the character from the input buffer.

2007-01-31 22:55:08 · answer #1 · answered by cipher 3 · 0 0

i think it's because the actual code only sends the computer a comand to delete it's last acion on the document or whatever you're using. o's and 1's dont acually do anything its more of a language commanding the computer to do something

2007-01-31 20:16:22 · answer #2 · answered by Anonymous · 0 0

maximum folk of the names i exploit have no magnitude; I only p.c.. them at random from names i think of the mother and dad could have named them. the only cases characters have significant names are whilst the characters call themselves by using names of their very own choosing.

2016-11-23 19:36:25 · answer #3 · answered by wehrly 4 · 0 0

I never really thought of that. I'm curious as well. Great question!

2007-01-31 20:15:03 · answer #4 · answered by Jay S 5 · 0 0

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