a 'C' doubt!

ramandeepsingh900

whats the output ??? explain it too!
int main()
{
int x=1, y;
y=++x + ++x + ++x;
printf("%d\n",y);
return 0;
}

dheeraj

int x=1, y;
y=++x + ++x + ++x;

evaluate it from left to right
++x=2
++x + ++x= 2+3 /*it should have to be
*/but in compilation stage it is resolved as
++x + ++x = 3+3
                          means for those first two ++x,compiler assigns only one register for saving value
consider 'a' as a register in which CPU going to stores value
++x means it will save 2 in register a;
++x + ++x; it will increment the value in the same register so value in register a becomes 3
++x + ++x = 6;
a     +     a = 6;
3     +     3 = 6
now ++x + ++x + ++x; now for futher preincrements it will allocate new register for ++x suppose it as b;
in register representation it becomes
a+a+b;
but b=++a;
3+3+4=10

if you consider one more preincrement it will become
++x + ++x + ++x ++x ; it will allocate new register for next ++x suppose it as c;
but c=++b;
a+a+b+c

3+3+4+5=15;

for further details use --save-temps while compiling the code and try to understand the '.S' file which is in assembly

himanshulkce

sir i didn't get that if compiler is incrementing value in register so compiler should increment value according to last change that is four so answer should be ..... 4+4+4
why it is 3+3+4???
why first compiler has changed value than it is not changing when it is increased to four???

spatlou

RCS file: BinarySearch.c,v
Working file: BinarySearch.c
head: 1.4
branch:
locks: strict
	root: 1.4
access list:
symbolic names:
keyword substitution: kv
total revisions: 4;	selected revisions: 4
description:
This is the program that will sort the given array.
So, the pre-requisite for binary search is met.
----------------------------
revision 1.4	locked by: root;
date: 2015/02/07 07:28:34;  author: root;  state: Exp;  lines: +5 -0
checking for the element at mid location.
----------------------------
revision 1.3
date: 2015/02/07 07:26:46;  author: root;  state: Exp;  lines: +4 -1
Declared key.
Took the key input valu to find in the array.
----------------------------
revision 1.2
date: 2015/02/07 07:19:36;  author: root;  state: Exp;  lines: +5 -1
Declared variables l, h, m.
Initialized all these variables.
----------------------------
revision 1.1
date: 2015/02/07 07:13:50;  author: root;  state: Exp;
Initial revision
=============================================================================

ankurg92

Answer should be 4+4+4 = 12

It is an example of Pre- Increment
++x means the value of x is increased before termination of statement.
It should be noted that a variable can contain only one value in a single statement.

Above statement has 3 Pre- Increments .So, value of all three x=4
So, y=4+4+4=12

atulpant063

Answer should be 3+3+4=10.

This is simple implementation of Pre-increment operator.

What pre-increment operator does is , it first increases the value of operator in the memory and uses it after the end of an expression ( it can be full C statement or an expression within the statement ). Here [++x + ++x + ++x ] is evaluated as [ ( ++x + ++x ) + ++x ]. First ++x increments value of x as 2 in memory ( BUT NOT USED) . Then the second ++x increases the value of x to 3 in memory. Now this value will be used for first addition in the expression hence value of x i.e used is 3. Now third ++x increments value of x to 4 in memory. Now since statement ends hence it will use the value of x as 4. So final answer would be 6+4 =10.

And it is always not necessary that a variable can have only one value in a single statement.

dhannjayv1996

char *ptr = "hello Emblogic"

it works without malloc. but how pointer is getting memory without malloc?

deepak4u106

Hello Dhannajay

Because we directly initialising the value of pointer assigning memory to it.

Even if we store value from coming argument from the function in a pointer does not require memory allocation.

a 'C' doubt!

Howdy, Stranger!

Categories