CODEWITHSUNDEEP
cincrement
912M0FR34K
912M0FR34K

Reputation: 121

Can someone tell me how does this work ? (C programming)

int m, n, j;
n=16;
j=15;
m = n++ -j+10;
printf("%d", m);

Output: 11.

Here, first, the old value of n is given to m and then it is incremented so the new value i get is 17 and then the expression is solved i.e. j+10 = 25 then the new value of n is subtracted by 25 i.e 17-25. Am i right ? but the answer doesn't match the output 11. Then how does this work ? And also, i am new to programming and started learning C. Which book will you suggest is the best for me ? As I've no programming experience. Thank you.

Upvotes: 1

Views: 293

Answers (10)

user2713461
user2713461

Reputation: 397

In the expression m = n++ -j+10;
The compiler treats the expression as m= n++ ((-j)+10)
As the intialized values of n and j are n = 16 and j = 15. We have m = 16++ ((-15)+10). We get output as 11.
After the expression is executed n will be incremented.

Upvotes: 0

Shiplu Mokaddim
Shiplu Mokaddim

Reputation: 57690

m = n++ -j+10; is same as 

m = n -j+10;
n = n + 1; // m is 11.

If it was ++n It would be 

n = n + 1;
m = n -j+10; //m is 12.

then the expression is solved i.e. j+10 = 25

No. It would be -j+10 = -5

My suggestion is, dont write complex expression unless you are completely sure what you are writing.

Upvotes: 4

hkulekci
hkulekci

Reputation: 1942

for m, first of all calculate n - j + 10 and assign it to m. After that n++ is executed.

at the end n = 17, m = 11

Upvotes: 0

joaogoncalv_es
joaogoncalv_es

Reputation: 309

n++ is post incrementation. It only increments the value of n after doing: m = n++ -j+10;

++n is pre incrementation. It increments the value of n before calculing m. m = ++n -j+10;

Upvotes: -1

Kidi
Kidi

Reputation: 61

in fact the post increment operation is done on n after the operation... you have 16-15+10 = 11 but if you print n you should have 17.

to begin, you can read some book on basics but this example is not simple; it include the precedence of operator which can be tricky.

begin simple... it's quite simple to write unreadable code in c. http://www.cs.cf.ac.uk/Dave/C/node4.html

hope it helps

Upvotes: 1

JimmyG
JimmyG

Reputation: 131

In answer to your question about a good book - you probably want to consider learning C++ instead of just plain old C, since C++ is a superset of C. And for C++ you need to get Bjarne Stroustrup's 'The C++ Programming Language'. It's easy enough to read and will last a long time on your bookshelf as a good reference.

Upvotes: 0

yurib
yurib

Reputation: 8147

n++ first returns the value of n and then increments it.
so, the actual computation that takes place is m = 16 - 15 + 10 which is 11

i think what you want is:

m = (n+1) - (j+10);

the use of the ++ operator is to increment the value of n for future use after you use it's current value to compute m.

Upvotes: 0

borrible
borrible

Reputation: 17376

You're making two incorrect interpretations.

Firstly, as indicated in other answers, n++ only increments n after the entire expression has been evaluated.

Secondly, you have -j+10. This is not equal to -(j+10), so it is wrong to say that j+10 is 25 and you are looking a something - 25. Another way of view -j+10 is 10-j.

Upvotes: 0

mattjgalloway
mattjgalloway

Reputation: 34912

You've got a few things wrong there.

  1. n++ will increment n and return the original result, so you've then got m = 16 ....

  2. -j so you've got m = 16 - 15 ....

  3. +10 so you've got m = 16 - 15 + 10.

Now the last time I did maths that would come out as m = 11 like you're seeing.

If you wanted it to be m = 17 - (15 + 10) then you wanted:

int m, n, j;
n=16;
j=15;
m = ++n -(j+10);
printf("%d", m);

Upvotes: 1

Graham Borland
Graham Borland

Reputation: 60721

n is incremented after its value is used in the expression which ends up in m.

Upvotes: -1

Related Questions