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[C++][Question]Increments

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wunder312355
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Posted: Mon May 25, 2009 6:45 pm Post subject: [C++][Question]Increments

I am having trouble understanding what is the difference between

Code:

cout << ++i << endl;

and

Code:

cout << i++ << endl;

Can anybody properly explain this?

dnsi0
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Posted: Mon May 25, 2009 6:50 pm Post subject: Re: [C++][Question]Increments

thefun25 wrote:

I am having trouble understanding what is the difference between

Code:

cout << ++i << endl;

and

Code:

cout << i++ << endl;

Can anybody properly explain this?

It simply means whether the variable will be increased by 1 BEFORE OR AFTER the instruction is executed.

hcavolsdsadgadsg
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Posted: Mon May 25, 2009 8:09 pm Post subject:

pre vs post increment, it works exactly like expected.

manc
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Posted: Mon May 25, 2009 10:05 pm Post subject:

http://www.cplusplus.com/doc/tutorial/operators/ wrote:

Increase and decrease (++, --)
Shortening even more some expressions, the increase operator (++) and the decrease operator (--) increase or reduce by one the value stored in a variable. They are equivalent to +=1 and to -=1, respectively. Thus:

c++;
c+=1;
c=c+1;

are all equivalent in its functionality: the three of them increase by one the value of c.

In the early C compilers, the three previous expressions probably produced different executable code depending on which one was used. Nowadays, this type of code optimization is generally done automatically by the compiler, thus the three expressions should produce exactly the same executable code.

A characteristic of this operator is that it can be used both as a prefix and as a suffix. That means that it can be written either before the variable identifier (++a) or after it (a++). Although in simple expressions like a++ or ++a both have exactly the same meaning, in other expressions in which the result of the increase or decrease operation is evaluated as a value in an outer expression they may have an important difference in their meaning: In the case that the increase operator is used as a prefix (++a) the value is increased before the result of the expression is evaluated and therefore the increased value is considered in the outer expression; in case that it is used as a suffix (a++) the value stored in a is increased after being evaluated and therefore the value stored before the increase operation is evaluated in the outer expression. Notice the difference:

Example 1
B=3;
A=++B;
// A contains 4, B contains 4

Example 2
B=3;
A=B++;
// A contains 3, B contains 4

In Example 1, B is increased before its value is copied to A. While in Example 2, the value of B is copied to A and then B is increased.

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Jani
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Posted: Tue May 26, 2009 11:40 am Post subject:

Remember that the pre increment might be faster than the post increment :)

oib111
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Posted: Tue May 26, 2009 1:42 pm Post subject:

It just means that it will increment before or after. This is most easily demonstrated in for loops.

Code:

for(int i = 0; i<10; i++) {
cout<<i;
}

//...

for(int i = 0; i<10; ++i) {
cout<<i;
}

In the first loop, it will display i's value, then increase it's value, while in the second one it will increase i's value and then display it.
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hcavolsdsadgadsg
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Posted: Tue May 26, 2009 3:46 pm Post subject:

Jani wrote:

Remember that the pre increment might be faster than the post increment Smile

Maybe if your compiler is hilariously bad at it's job...

talkerzero
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Posted: Sat Jun 06, 2009 11:17 am Post subject:

Quick example..

Code:

j = 0;
string[j++] = 'x';

Code:

j = 0;
string[++j] = 'x';

The above code will assign 'x' to string[j], when j = 0, then increment j. The below code will increment j to 1, then assign 'x' to string[j] when j now = 1.

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