


在一個偶然的機會,有機會目睹朋友的肚皮舞表演,特拍照留念.
但,由於天色昏暗,距離又遠,又受限於器材...
只能透過高ISO的方式避免手震..但拍出來雜訊也就可想而知了..
拍舞者真的不容易,舞者的動感的韻味真的很難掌握.
不過有了這次經驗,我相信我下次一定可以拍得更好些.
只是....下次的機會不知道何時才會來到....
苦.澀.酸.甜...生活的筆記,生命的回憶
Return by value
Return by value is the simplest and safest return type to use. When a value is returned by value, a copy of that value is returned to the caller. As with pass by value, you can return by value literals (eg. 5), variables (eg. x), or expressions (eg. x+1), which makes return by value very flexible.
Another advantage of return by value is that you can return variables (or expressions) that involve local variables declared within the function. Because the variables are evaluated before the function goes out of scope, and a copy of the value is returned to the caller, there are no problems when the variable goes out of scope at the end of the function.
1.
int
DoubleValue(
int
nX)
2.
{
3.
int
nValue = nX * 2;
4.
return
nValue;
// A copy of nValue will be returned here
5.
}
// nValue goes out of scope here
Return by value is the most appropriate when returning variables that were declared inside the function, or for returning function arguments that were passed by value. However, like pass by value, return by value is slow for structs and large classes.
Return by reference
Just like with pass by reference, values returned by reference must be variables (you can not return a reference to a literal or an expression). When a variable is returned by reference, a reference to the variable is passed back to the caller. The caller can then use this reference to continue modifying the variable, which can be useful at times. Return by reference is also fast, which can be useful when returning structs and classes.
However, returning by reference has one additional downside that pass by reference doesn’t — you can not return local variables to the function by reference. Consider the following example:
1.
int
& DoubleValue(
int
nX)
2.
{
3.
int
nValue = nX * 2;
4.
return
nValue;
// return a reference to nValue here
5.
}
// nValue goes out of scope here
See the problem here? The function is trying to return a reference to a value that is going to go out of scope when the function returns. This would mean the caller receives a reference to garbage. Fortunately, your compiler will give you an error if you try to do this.
Return by reference is typically used to return arguments passed by reference to the function back to the caller. In the following example, we return (by reference) an element of an array that was passed to our function by reference:
01.
// This struct holds an array of 25 integers
02.
struct
FixedArray25
03.
{
04.
int
anValue[25];
05.
};
06.
07.
// Returns a reference to the nIndex element of rArray
08.
int
& Value(FixedArray25 &rArray,
int
nIndex)
09.
{
10.
return
rArray.anValue[nIndex];
11.
}
12.
13.
int
main()
14.
{
15.
FixedArray25 sMyArray;
16.
17.
// Set the 10th element of sMyArray to the value 5
18.
Value(sMyArray, 10) = 5;
19.
20.
cout << sMyArray.anValue[10] << endl;
21.
return
0;
22.
}
This prints:
5
When we call Value(sMyArray, 10)
, Value() returns a reference to the 10th element of the array inside sMyArray. main() then uses this reference to assign that element the value 5.
Although this is somewhat of a contrived example (because you could access sMyArray.anValue directly), once you learn about classes you will find a lot more uses for returning values by reference.
Return by address
Returning by address involves returning the address of a variable to the caller. Just like pass by address, return by address can only return the address of a variable, not a literal or an expression. Like return by reference, return by address is fast. However, as with return by reference, return by address can not return local variables:
1.
int
* DoubleValue(
int
nX)
2.
{
3.
int
nValue = nX * 2;
4.
return
&nValue;
// return nValue by address here
5.
}
// nValue goes out of scope here
As you can see here, nValue goes out of scope just after its address is returned to the caller. The end result is that the caller ends up with the address of non-allocated memory, which will cause lots of problems if used. This is one of the most common programming mistakes that new programmers make. Many newer compilers will give a warning (not an error) if the programmer tries to return a local variable by address — however, there are quite a few ways to trick the compiler into letting you do something illegal without generating a warning, so the burden is on the programmer to ensure the address they are returning will be to a valid variable after the function returns.
Return by address is often used to return newly allocated memory to the caller:
01.
int
* AllocateArray(
int
nSize)
02.
{
03.
return
new
int
[nSize];
04.
}
05.
06.
int
main()
07.
{
08.
int
*pnArray = AllocateArray(25);
09.
// do stuff with pnArray
10.
11.
delete
[] pnArray;
12.
return
0;
13.
}
Conclusion
Most of the time, return by value will be sufficient for your needs. It’s also the most flexible and safest way to return information to the caller. However, return by reference or address can also be useful, particularly when working with dynamically allocated classes or structs. When using return by reference or address, make sure you are not returning a reference to, or the address of, a variable that will go out of scope when the function returns!