问题
Is there a difference in ++i and i++ in a for loop? Is it simply a syntax thing?
回答1:
a++ is known as postfix.
add 1 to a, returns the old value.
++a is known as prefix.
add 1 to a, returns the new value.
C#:
string[] items = {"a","b","c","d"};
int i = 0;
foreach (string item in items)
{
Console.WriteLine(++i);
}
Console.WriteLine("");
i = 0;
foreach (string item in items)
{
Console.WriteLine(i++);
}
Output:
1
2
3
4
0
1
2
3
foreach and while loops depend on which increment type you use. With for loops like below it makes no difference as you're not using the return value of i:
for (int i = 0; i < 5; i++) { Console.Write(i);}
Console.WriteLine("");
for (int i = 0; i < 5; ++i) { Console.Write(i); }
0 1 2 3 4
0 1 2 3 4
If the value as evaluated is used then the type of increment becomes significant:
int n = 0;
for (int i = 0; n < 5; n = i++) { }
回答2:
Pre-increment ++i increments the value of i and evaluates to the new incremented value.
int i = 3;
int preIncrementResult = ++i;
Assert( preIncrementResult == 4 );
Assert( i == 4 );
Post-increment i++ increments the value of i and evaluates to the original non-incremented value.
int i = 3;
int postIncrementResult = i++;
Assert( postIncrementtResult == 3 );
Assert( i == 4 );
In C++, the pre-increment is usually preferred where you can use either.
This is because if you use post-increment, it can require the compiler to have to generate code that creates an extra temporary variable. This is because both the previous and new values of the variable being incremented need to be held somewhere because they may be needed elsewhere in the expression being evaluated.
So, in C++ at least, there can be a performance difference which guides your choice of which to use.
This is mainly only a problem when the variable being incremented is a user defined type with an overridden ++ operator. For primitive types (int, etc) there's no performance difference. But, it's worth sticking to the pre-increment operator as a guideline unless the post-increment operator is definitely what's required.
There's some more discussion here:
https://web.archive.org/web/20170405054235/http://en.allexperts.com/q/C-1040/Increment-operators.htm
In C++ if you're using STL, then you may be using for loops with iterators. These mainly have overridden ++ operators, so sticking to pre-increment is a good idea. Compilers get smarter all the time though, and newer ones may be able to perform optimizations that mean there's no performance difference - especially if the type being incremented is defined inline in header file (as STL implementations often are) so that the compiler can see how the method is implemented and can then know what optimizations are safe to perform. Even so, it's probably still worth sticking to pre-increment because loops get executed lots of times and this means a small performance penalty could soon get amplified.
In other languages such as C# where the ++ operator can't be overloaded there is no performance difference. Used in a loop to advance the loop variable, the pre and post increment operators are equivalent.
Correction: overloading ++ in C# is allowed. It seems though, that compared to C++, in c# you cannot overload the pre and post versions independently. So, I would assume that if the result of calling ++ in C# is not assigned to a variable or used as part of a complex expression, then the compiler would reduce the pre and post versions of ++ down to code that performs equivalently.
回答3:
In C# there is no difference when used in a for loop.
for (int i = 0; i < 10; i++) { Console.WriteLine(i); }
outputs the same thing as
for (int i = 0; i < 10; ++i) { Console.WriteLine(i); }
As others have pointed out, when used in general i++ and ++i have a subtle yet significant difference:
int i = 0;
Console.WriteLine(i++); // Prints 0
int j = 0;
Console.WriteLine(++j); // Prints 1
i++ reads the value of i then increments it.
++i increments the value of i then reads it.
回答4:
The question is:
Is there a difference in ++i and i++ in a for loop?
The answer is: No.
Why does each and every other answer have to go into detailed explanations about pre and post incrementing when this is not even asked?
This for-loop:
for (int i = 0; // Initialization
i < 5; // Condition
i++) // Increment
{
Output(i);
}
Would translate to this code without using loops:
int i = 0; // Initialization
loopStart:
if (i < 5) // Condition
{
Output(i);
i++ or ++i; // Increment
goto loopStart;
}
Now does it matter if you put i++ or ++i as increment here? No it does not as the return value of the increment operation is insignificant. i will be incremented AFTER the code's execution that is inside the for loop body.
回答5:
Since you ask about the difference in a loop, i guess you mean
for(int i=0; i<10; i++)
...;
In that case, you have no difference in most languages: The loop behaves the same regardless of whether you write i++ and ++i. In C++, you can write your own versions of the ++ operators, and you can define separate meanings for them, if the i is of a user defined type (your own class, for example).
The reason why it doesn't matter above is because you don't use the value of i++. Another thing is when you do
for(int i=0, a = 0; i<10; a = i++)
...;
Now, there is a difference, because as others point out, i++ means increment, but evaluate to the previous value, but ++i means increment, but evaluate to i (thus it would evaluate to the new value). In the above case, a is assigned the previous value of i, while i is incremented.
回答6:
As this code shows (see the dissambled MSIL in the comments), the C# 3 compiler makes no distinction between i++ and ++i in a for loop. If the value of i++ or ++i were being taken, there would definitely be a difference (this was compiled in Visutal Studio 2008 / Release Build):
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace PreOrPostIncrement
{
class Program
{
static int SomethingToIncrement;
static void Main(string[] args)
{
PreIncrement(1000);
PostIncrement(1000);
Console.WriteLine("SomethingToIncrement={0}", SomethingToIncrement);
}
static void PreIncrement(int count)
{
/*
.method private hidebysig static void PreIncrement(int32 count) cil managed
{
// Code size 25 (0x19)
.maxstack 2
.locals init ([0] int32 i)
IL_0000: ldc.i4.0
IL_0001: stloc.0
IL_0002: br.s IL_0014
IL_0004: ldsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0009: ldc.i4.1
IL_000a: add
IL_000b: stsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0010: ldloc.0
IL_0011: ldc.i4.1
IL_0012: add
IL_0013: stloc.0
IL_0014: ldloc.0
IL_0015: ldarg.0
IL_0016: blt.s IL_0004
IL_0018: ret
} // end of method Program::PreIncrement
*/
for (int i = 0; i < count; ++i)
{
++SomethingToIncrement;
}
}
static void PostIncrement(int count)
{
/*
.method private hidebysig static void PostIncrement(int32 count) cil managed
{
// Code size 25 (0x19)
.maxstack 2
.locals init ([0] int32 i)
IL_0000: ldc.i4.0
IL_0001: stloc.0
IL_0002: br.s IL_0014
IL_0004: ldsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0009: ldc.i4.1
IL_000a: add
IL_000b: stsfld int32 PreOrPostIncrement.Program::SomethingToIncrement
IL_0010: ldloc.0
IL_0011: ldc.i4.1
IL_0012: add
IL_0013: stloc.0
IL_0014: ldloc.0
IL_0015: ldarg.0
IL_0016: blt.s IL_0004
IL_0018: ret
} // end of method Program::PostIncrement
*/
for (int i = 0; i < count; i++)
{
SomethingToIncrement++;
}
}
}
}
回答7:
One (++i) is preincrement, one (i++) is postincrement. The difference is in what value is immediately returned from the expression.
// Psuedocode
int i = 0;
print i++; // Prints 0
print i; // Prints 1
int j = 0;
print ++j; // Prints 1
print j; // Prints 1
Edit: Woops, entirely ignored the loop side of things. There's no actual difference in for loops when it's the 'step' portion (for(...; ...; )), but it can come into play in other cases.
回答8:
There is no difference if you are not using the value after increment in the loop.
for (int i = 0; i < 4; ++i){
cout<<i;
}
for (int i = 0; i < 4; i++){
cout<<i;
}
Both the loops will print 0123.
But the difference comes when you uses the value after increment/decrement in your loop as below:
Pre Increment Loop:
for (int i = 0,k=0; i < 4; k=++i){
cout<<i<<" ";
cout<<k<<" ";
}
Output: 0 0 1 1 2 2 3 3
Post Increment Loop:
for (int i = 0, k=0; i < 4; k=i++){
cout<<i<<" ";
cout<<k<<" ";
}
Output: 0 0 1 0 2 1 3 2
I hope the difference is clear by comparing the output. Point to note here is the increment/decrement is always performed at the end of the for loop and hence the results can be explained.
回答9:
Here is a Java-Sample and the Byte-Code, post- and preIncrement show no difference in Bytecode:
public class PreOrPostIncrement {
static int somethingToIncrement = 0;
public static void main(String[] args) {
final int rounds = 1000;
postIncrement(rounds);
preIncrement(rounds);
}
private static void postIncrement(final int rounds) {
for (int i = 0; i < rounds; i++) {
somethingToIncrement++;
}
}
private static void preIncrement(final int rounds) {
for (int i = 0; i < rounds; ++i) {
++somethingToIncrement;
}
}
}
And now for the byte-code (javap -private -c PreOrPostIncrement):
public class PreOrPostIncrement extends java.lang.Object{
static int somethingToIncrement;
static {};
Code:
0: iconst_0
1: putstatic #10; //Field somethingToIncrement:I
4: return
public PreOrPostIncrement();
Code:
0: aload_0
1: invokespecial #15; //Method java/lang/Object."<init>":()V
4: return
public static void main(java.lang.String[]);
Code:
0: sipush 1000
3: istore_1
4: sipush 1000
7: invokestatic #21; //Method postIncrement:(I)V
10: sipush 1000
13: invokestatic #25; //Method preIncrement:(I)V
16: return
private static void postIncrement(int);
Code:
0: iconst_0
1: istore_1
2: goto 16
5: getstatic #10; //Field somethingToIncrement:I
8: iconst_1
9: iadd
10: putstatic #10; //Field somethingToIncrement:I
13: iinc 1, 1
16: iload_1
17: iload_0
18: if_icmplt 5
21: return
private static void preIncrement(int);
Code:
0: iconst_0
1: istore_1
2: goto 16
5: getstatic #10; //Field somethingToIncrement:I
8: iconst_1
9: iadd
10: putstatic #10; //Field somethingToIncrement:I
13: iinc 1, 1
16: iload_1
17: iload_0
18: if_icmplt 5
21: return
}
回答10:
Yes, there is. The difference is in the return value. The return value of "++i" will be the value after incrementing i. The return of "i++" will be the value before incrementing. This means that code that looks like the following:
int a = 0;
int b = ++a; // a is incremented and the result after incrementing is saved to b.
int c = a++; // a is incremented again and the result before incremening is saved to c.
Therefore, a would be 2, and b and c would each be 1.
I could rewrite the code like this:
int a = 0;
// ++a;
a = a + 1; // incrementing first.
b = a; // setting second.
// a++;
c = a; // setting first.
a = a + 1; // incrementing second.
回答11:
There is no actual difference in both cases 'i' will be incremented by 1.
But there is a difference when you use it in an expression, for example:
int i = 1;
int a = ++i;
// i is incremented by one and then assigned to a.
// Both i and a are now 2.
int b = i++;
// i is assigned to b and then incremented by one.
// b is now 2, and i is now 3
回答12:
There is more to ++i and i++ than loops and performance differences. ++i returns a l-value and i++ returns an r-value. Based on this, there are many things you can do to ( ++i ) but not to ( i++ ).
1- It is illegal to take the address of post increment result. Compiler won't even allow you.
2- Only constant references to post increment can exist, i.e., of the form const T&.
3- You cannot apply another post increment or decrement to the result of i++, i.e., there is no such thing as I++++. This would be parsed as ( i ++ ) ++ which is illegal.
4- When overloading pre-/post-increment and decrement operators, programmers are encouraged to define post- increment/decrement operators like:
T& operator ++ ( )
{
// logical increment
return *this;
}
const T operator ++ ( int )
{
T temp( *this );
++*this;
return temp;
}
回答13:
It boggles my mind why so may people write the increment expression in for-loop as i++.
In a for-loop, when the 3rd component is a simple increment statement, as in
for (i=0; i<x; i++)
or
for (i=0; i<x; ++i)
there is no difference in the resulting executions.
回答14:
As @Jon B says, there is no difference in a for loop.
But in a while or do...while loop, you could find some differences if you are making a comparison with the ++i or i++
while(i++ < 10) { ... } //compare then increment
while(++i < 10) { ... } //increment then compare
回答15:
In javascript due to the following i++ may be better to use:
var i=1;
alert(i++); // before, 1. current, 1. after, 2.
alert(i); // before, 2. current, 2. after, 2.
alert(++i); // before, 2. current, 3 after, 3.
While arrays (I think all) and some other functions and calls use 0 as a starting point you would have to set i to -1 to make the loop work with the array when using ++i.
When using i++ the following value will use the increased value. You could say i++ is the way humans count, cause you can start with a 0.
回答16:
There can be a difference for loops. This is the practical application of post/pre-increment.
int i = 0;
while(i++ <= 10) {
Console.Write(i);
}
Console.Write(System.Environment.NewLine);
i = 0;
while(++i <= 10) {
Console.Write(i);
}
Console.ReadLine();
While the first one counts to 11 and loops 11 times, the second does not.
Mostly this is rather used in a simple while(x-- > 0 ) ; - - Loop to iterate for example all elements of an array (exempting foreach-constructs here).
回答17:
They both increment the number. ++i is equivalent to i = i + 1.
i++ and ++i are very similar but not exactly the same. Both increment the number, but ++i increments the number before the current expression is evaluated, whereas i++ increments the number after the expression is evaluated.
int i = 3;
int a = i++; // a = 3, i = 4
int b = ++a; // b = 4, a =
Check this link.
回答18:
Yes, there is a difference between ++i and i++ in a for loop, though in unusual use cases; when a loop variable with increment/decrement operator is used in the for block or within the loop test expression, or with one of the loop variables. No it is not simply a syntax thing.
As i in a code means evaluate the expression i and the operator does not mean an evaluation but just an operation;
++imeans increment value ofiby 1 and later evaluatei,i++means evaluateiand later increment value ofiby 1.
So, what are obtained from each two expressions differ because what is evaluated differs in each. All same for --i and i--
For example;
let i = 0
i++ // evaluates to value of i, means evaluates to 0, later increments i by 1, i is now 1
0
i
1
++i // increments i by 1, i is now 2, later evaluates to value of i, means evaluates to 2
2
i
2
In unusual use cases, however next example sounds useful or not does not matter, it shows a difference
for(i=0, j=i; i<10; j=++i){
console.log(j, i)
}
for(i=0, j=i; i<10; j=i++){
console.log(j, i)
}
回答19:
For i's of user-defined types, these operators could (but should not) have meaningfully different sematics in the context of a loop index, and this could (but should not) affect the behavior of the loop described.
Also, in c++ it is generally safest to use the pre-increment form (++i) because it is more easily optimized. (Scott Langham beat me to this tidbit. Curse you, Scott)
回答20:
I dont know for the other languages but in Java ++i is a prefix increment which means: increase i by 1 and then use the new value of i in the expression in which i resides, and i++ is a postfix increment which means the following: use the current value of i in the expression and then increase it by 1. Example:
public static void main(String [] args){
int a = 3;
int b = 5;
System.out.println(++a);
System.out.println(b++);
System.out.println(b);
} and the output is:
- 4
- 5
- 6
回答21:
i++ ; ++i ; both are similar as they are not used in an expression.
class A {
public static void main (String []args) {
int j = 0 ;
int k = 0 ;
++j;
k++;
System.out.println(k+" "+j);
}}
prints out : 1 1
来源：https://stackoverflow.com/questions/484462/difference-between-pre-increment-and-post-increment-in-a-loop