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1. Start line with an ENDLOCAL command.

2. Add an ampersand and a SET global = local variable.

3. Repeat step 2 for additional variables.

Notes:

Works vice versa with the SETLOCAL command.

Works in nested arrangements of SETLOCAL and ENDLOCAL.

Local and global variables can have same names.

Spaces are permitted after a SETLOCAL or ENDLOCAL but might get added to a variable's value if put after a SET command so recommend avoiding spaces before the second and subsequent ampersands.

Spaces are permitted but not required after the ampersand and before a SET command.

The first line of this example shows where spaces are allowed, the last line shows where spaces can be omitted.

SETLOCAL & SET LTEMP=%TEMP%& SET PATH=%PATH%

SET LPATH=C:\BAT;%PATH%

SET L1=valueL1

SET P1=valueP1

ENDLOCAL&SET G1=%L1%&SET P1=%P1%&SET PATH=%LPATH%

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Q: In a DOS batch file how do you copy a local variable to a global variable?
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What is the difference between local and global variable of C language?

4m Sai. Gloabal Variable is variable which is declared before main (). int a=10; main() { } here a is a global variable. where ever u r using this variable ... we'll get the value as 10. Local Variable is a variable which is declared inside the main(). int a=10; main() { int a=5; ........ ....... } here a=5 is a local variable and a=10 is called as global variable. then if u want to get a value... printf("%d",a); then result 'll be a=5; B'cos compiler gives main preference to local variables.. if any local declareation is not there then it 'll prefer global variable.


What is a variable that is used within a function?

If the variable is declared within the function body, it is a local variable, one that is local to the function. Local variables fall from scope when the function returns, they are only accessible within the function. However, local variables can be returned by value, which creates an automatic variable that is returned to the caller. If the caller does not store the return value, the automatic variable falls from scope when the expression containing the function call ends. However, the expression may evaluate the return value without storing it. Note that functions cannot return local variables by reference since the local variable falls from scope when the function returns. If the variable is passed as an argument to the function, then the variable is a parameter of the function. Arguments may be passed by value or by reference, depending upon the function signature. Passing by value means the function parameter is a copy of the argument (if the argument is an object, the object's copy constructor is invoked automatically). Thus any changes made to the parameter within the function are not reflected in the argument that was originally passed, and the parameter will fall from scope when the function returns. However, the value of the parameter can be returned as previously explained. Passing by reference means the function parameter refers directly to the argument that was passed. Thus any changes made to the parameter are reflected in the argument. Parameters that are declared as constant references assure the caller that the reference's immutable members will not be altered by the function. If the parameter is a non-const reference but the caller does not wish changes to be reflected in the argument, the caller should pass a copy of the argument instead.


When a reference variable is passed to any function than whether it pass address or a copy of a variable?

You cannot arbitrarily determine what is passed to a function from outside of the function. If the function expects a reference, then the memory address of the variable will be automatically passed to the function. If the function expects a value, then a copy of the variable's value is automatically passed instead. In other words, it is the function signature that determines what is passed, not how you call the function.


How can one copy local disk C to external hard disk E with DOS Commands?

copy c:\,e: or copy c:,e:


How can a pointer create a perfect copy of a variable?

In order to copy a variable you must copy its value and store that value in another variable, thus creating an independent copy of the value. That is, after the copy, changing one value has no effect upon the other: void f (int x) { int x = y; assert (x == y); // assert the values are the same assert (&x != &y); // assert the variables occupy different addresses (independent variables) } To achieve the same thing with a pointer you must create an independent variable to store the copied value: void g (int x) { int* p = malloc (sizeof (int)); // allocate memory to hold the value *p = x; // copy the value to the allocated memory assert (*p == x); // assert the values are the same assert (p != &x); // assert the variables occupy different addresses (independent variables) // ... free (p); // release the allocated memory } Note that the variable referred to by p is anonymous (has no name). To refer to the value we must dereference the pointer (*p). Copying a value via a pointer is achieved similarly: void h (int* p) { int x = *p; // copy the value referred to by p assert (x == *p); // assert the values are the same assert (&x != p); // assert the variables occupy different addresses (independent variables) } void i (int* p) { int* p2 = malloc (sizeof (int)); // allocate memory to hold the value *p2 = *p; // copy the value to the allocated memory assert (*p2 == *p); // assert the values are the same assert (p2 != p); // assert the variables occupy different addresses (independent variables) // ... free (p2); // release the allocated memory }

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