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Awk and a global variable


Awk and a global variable

By : Markus Fried
Date : November 22 2020, 04:01 AM
I think the issue was by ths following , Your variable is in single quotes, therefore it isn't expanded. In the awk part replace $numIterations with '"$numIterations"'.
Better alternative: Use awk -v ni="$numIterations" to set an awk variable and use ni inside the awk script. Example for clarity:
code :
$ x=7
$ echo awk 'BEGIN {print 1 * $x}'
awk BEGIN {print 1 * $x}              # this is literally what awk sees
$ awk 'BEGIN {print 1 * $x}'
0                                     # because $x is not set in awk
$ awk -v x="$x" 'BEGIN {print 1 * x}'
7


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How do remove the global variable in my 'debug mode' (Global Variable Scare)

How do remove the global variable in my 'debug mode' (Global Variable Scare)


By : user3191246
Date : March 29 2020, 07:55 AM
should help you out Try to learn a PHP framework. They solve these problems better and in an OOP way.
Anyway, in your example, you can remove a lot of the code:
code :
$debug_mode = TRUE;

function dump() {
  global $debug_mode;

  if ($debug_mode) {
    call_user_func_array('var_dump', func_get_args());
  }
}
Initializing a global variable with integer-casted global variable pointer in C

Initializing a global variable with integer-casted global variable pointer in C


By : Mike
Date : March 29 2020, 07:55 AM
will be helpful for those in need I found an unexpected 'invalid' C code snippet, which I believed a valid one: ,
Which part of C specification makes this erroneous?
code :
char *pstr = str;
why we cannot assign value to variable after defining as global variable in global area space

why we cannot assign value to variable after defining as global variable in global area space


By : luxferre
Date : March 29 2020, 07:55 AM
To fix the issue you can do Only assignmemnt is an instruction and each instruction should appear inside a function body.
In global scope only a special case, initialization through assignment while definition is allowed.
How can I change a local variable assigned to the value of a global one without changing the global variable?

How can I change a local variable assigned to the value of a global one without changing the global variable?


By : Cá Bống Mải Chơi
Date : March 29 2020, 07:55 AM
I wish this help you So this is basically how the code works that I'm using. , The problem here is the following line:
code :
Lvar = Gvar
Lvar = Gvar[:]
global Gvar
Gvar = ["Hello"]

def someFunction():
    Lvar = Gvar[:]
    Lvar.append("World")
    print(Lvar)
    print(Gvar)

someFunction()
Why Global variable does not update with global function within the same Global variable? Android/Java

Why Global variable does not update with global function within the same Global variable? Android/Java


By : 黄雨生
Date : March 29 2020, 07:55 AM
it helps some times In your code, the constant GET_CART_ALL_ITEM is created and initialize only one time, and it takes the current value of ReturnUserRrQuote(). Later It will not trigger the function as the constant already has its value and does not require a new initialization.
If at the beginning of code ReturnUserRrQuote() => "100". Then GET_CART_ALL_ITEM is created and initialize with this value, it will be "store/3/cart/100/type/customer" and not "store/3/cart/" + ReturnUserRrQuote() + "/type/customer". It's due cause at initialization the expression "store/3/cart/" + ReturnUserRrQuote() + "/type/customer" is evaluated and the result is affected to the constant (the expression is not affected to the constant).
code :
public static final const1 = "store/3/cart/";
public static final const2 = "/type/customer";

//whenever you have to obtain your value
String value = const1 + ReturnUserRrQuote() + const2;
//Static function that return the number of times it has been called
public static returnNumber() {
    final int i=1;
    return i++;
 }

public static void main() {
    int a = returnNumber(); //Initialize a
    for (j=0; j<10; j++) {
        System.out.println(a); //print the current value of a
    }
}
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