Strange answer after executing n++

Question

Why is the value of int d 25 and not 26 after executing the following code snippet?

int n = 20;
int d = n++ + 5;

Console.WriteLine(d);

possible duplicate of [What is the difference between ++i and i++](http://stackoverflow.com/questions/24853/what-is-the-difference-between-i-and-i) — Sam Leach, Oct 15 '13 at 12:18
See also: https://communities.coverity.com/blogs/development-testing-blog/2013/09/24/c-c-and-c — Rowland Shaw, Oct 15 '13 at 12:20

score 11 · Accepted Answer · answered Oct 15 '13 at 12:19

11

n++ is the "post-increment operator", which only increments the value after its initial value has been used in the surrounding expression.

Your code is equivalent to:

int d = n + 5;
n = n + 1;

To increment the value before its value gets used, use ++n, the pre-increment operator.

answered Oct 15 '13 at 12:19

RichieHindle

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+1 for breaking out the equivalent surgarless syntax. – Mike Perrenoud Oct 15 '13 at 12:21

score 3 · Answer 2 · answered Oct 15 '13 at 12:18

Because you need to use ++n to use the incremented value in that expression.

See, in the expression tree it's not incrementing n and then using that value in the addition because n++ returns the value of n but increments it for the next expression used.

However, ++n will actually return the incremented value of n for this expression.

Therefore, n++ + 5 yields 25 whereas ++n + 5 yields 26.

score 2 · Answer 3 · answered Oct 15 '13 at 12:19

2

n++ means execute the adition after the operation, so first d will equal to n+5 and then n will get raised.

answered Oct 15 '13 at 12:19

Giannis Paraskevopoulos

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Neel · Answer 4 · 2013-10-15T12:24:21.683

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Because n++ will first assign the value and after the completion of iteration it will increment thats the reason its giving 25

Hence,

int d= n++ +  5;

is interpreted as

int d = n +  5;

edited Oct 15 '13 at 12:24

answered Oct 15 '13 at 12:19

Neel

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score 1 · Answer 5 · answered Oct 15 '13 at 12:20

1

Because of you are using Postfix express

int d = n++ + 5;

where compiler first assign value to d, but in following

int d = ++n + 5;

You will got d's value 26

answered Oct 15 '13 at 12:20

Manwal

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score 1 · Answer 6 · answered Oct 15 '13 at 12:20

++:post increment operator.

The result of the postﬁx ++ operator is the value of the operand. After the result is obtained, the value of the operand is incremented

Hence,

int d= n++ +  5;

is interpreted as

int d = n +  5;

after execution of the above interpretaion. n is incremented by 1.

Strange answer after executing n++

6 Answers6