CareerCup

x = x++ + ++y; (1)
y = ++x + ++y; (2)

In (1) this is the following operations to be done.
i) ++y
ii) x = x+y
iii) x ++

In (2) this is the following operations to be done.
i) x ++
ii) y++
iii) y = x + y

What I can tell from the C standard, that in x = x++ + ... the value of x is getting modified twice within one sequence point. So it is Undefined Behaviour and the answer would vart depending upon the implementation of the compiler.

when x = 1, y = 1, we got x = 5, y = 8 after that

What language?

Undefined, will depend on compiler to compiler

x=5 and y=8

5,8 with following compiler.
i686-apple-darwin11-llvm-gcc-4.2 (GCC) 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2335.15.00)

Its UB (Undefined-Bahvior). value of x is being modified without any sequence-point in between. There is no way you can be sure of the value of x.

I wrote the code and tested no matter what you do in both the cases x & y will be incremented first as they are closely bind to "++" and after they are incremented then only they are added, refer wiki for operator precedence.

if x = y = 0 initially, both cases will have answer 2,
if x = 1, y =2 then both cases will have answer 5.

#include<stdio.h>
#include<math.h>
#include<conio.h>
int main()
{
    int x=0,y=0,z;
    z = x++ + ++y;
     printf("%d   %d  %d  \n",x,y,z);x=z;
    y = ++x + ++y;
    printf("%d   %d   \n",x,y);
    getch();
    return(0);
}

output 1 1 1
2 4

x will become x+y+3 & y will be x+2y+5

#include<stdio.h>

int main()
{
    int x=5,y=15;
    x= x++ + ++y;
    y = ++x + ++y;
    printf("%d %d",x,y);
    return 0;
}

Output: 23 40

#include <stdio.h>
main()
{
    int x=0,y=0;
    x= x++ + ++y;
    printf("x: %d\ty:%d\n", x,y);
    y= ++x+ ++y;
    printf("x: %d\ty:%d\n", x,y);
}

Output:
x: 2 y:1
x: 3 y:5

. main ()
{
int x=5;
x=x----1;
printf ("%d", x);
}
Output?

int x=2;y=3;
x=x++ +y++;
y=++x +++y;

x+(y--)-x+(x++)-(y--)

#include<stdio.h> int main(){
int i,x;
int i=printf("c" "++"); for(x=0;x<=i;x++){ printf("%x ",x);
}
return 0;
}

x=(x+1)+y;

y=x+y;

bluej pattern

{
int y = 2;
y = (y++)/(y++);
printf("%d", y++);
}

Cisco gives hike once in 2-3 years.

New-Joinees only get hike after 2-3 years, so take 100% hike at the time of joining .. . otherwise don't cry after joining. :)

int x = 1;
int y = x++;
int z = ++y;
y = z++;
z = ++z;
what will be the value of z?

Void main()
{
Int x=5;
x=x----1;
Of("℅d",x);
}
Will the value of x be 6 or its an error??

same problem comes with:
int x=2,y;
void main()
{
y=++x + ++x + x++; //gives output 12
y=++x + ++x + x++; //gives output 17

}


kindly explain

main()

{ int x, y, flag=0;

for(x=0; x<3; x++)

for(y=x+1; y<5; y++)

flag += x*y;

}

#include<stdio.h>

int main ()
{
int x=2, y=4;
int z=x+x+x+x+(x++);
printf("x=%d \n y=%d \n z=%d",x,y,z);
return 0;
}

OUTPUT:
x=3
y=4
z=10

#include<stdio.h>

int main ()
{
int x=2, y=4;
int z=(x++)+x+x+x+x;
printf("x=%d \n y=%d \n z=%d",x,y,z);
return 0;
}

OUTPUT:
x=3
y=4
z=14

Please someone explain the code. Why the output are different to each other?

#include<stdio.h>

int main ()
{
    int x=2, y=4;
    int z=x+x+x+x+(x++);
    printf("x=%d \n y=%d \n z=%d",x,y,z);
    return 0;
}

OUTPUT:
      x=3
      y=4
      z=10

#include<stdio.h>

int main ()
{
    int x=2, y=4;
    int z=(x++)+x+x+x+x;
    printf("x=%d \n y=%d \n z=%d",x,y,z);
    return 0;
}

OUTPUT: 
       x=3
       y=4
       z=14

Please someone explain the code. Why the output are different to each other?

#include<stdio.h>

int main ()
{
    int x=2, y=4;
    int z=x+x+x+x+(x++);
    printf("x=%d \n y=%d \n z=%d",x,y,z);
    return 0;
}

OUTPUT:
x=3
y=4
z=10

#include<stdio.h>

int main ()
{
    int x=2, y=4;
    int z=(x++)+x+x+x+x;
    printf("x=%d \n y=%d \n z=%d",x,y,z);
    return 0;
}

OUTPUT:
x=3
y=4
z=14

Please someone explain the following code above. Why the output of z are different to each oher?

how to declare a variable in a sub Function as "a=b" on c programming ?

what will be the syntax of it?

x=20 y=15 z=22

x+=2;
y=x;
y+=10;
z–=3;
y++;
x––;
z++;
y*=2;
x/=1;
z–=7;

x=1;
z=x++ + x;

if x=3,y=4 then value of=++x+ - - y+x - - +y - -

I have a doubt
if x=3,y=4 then value of=++x+ - - y+x - - +y - -

int x, y = 1, z;
if (x = z = y);
x = 3;
while (y < 4) {
x += ++y;
}

23 39

how does this output come??? can ou provide explanation
int y=5,x;
x=(++y)+(++y);
cout<<x;
output is 14

Int x,y=0;
For(x=1;x<=5;++x)
y=x++;
--y;
What will be the value of x and y?

if intial value of x and y was a and b respectively,then
x=a+b+2 and y=a+2*b+4

If P=5; find d=++p + 5

Find value of x and y


{
int x,y;
x=3,y=1;
if(x--)
y=x+2;
if(--x)
y= --x;
if(--x)
y=x--;
}

Sala

j= --k+2k+(l=k,l++)
the value of k is 20 initially

what is the value of j=--k+2k+(l=k,l++)

x=x×x++×++x+--x-x++×2
Please send the value

if x=5, y=4, c=8
g=c-y ,D=c/y+5 what is the solution of this question

d=2;Y=++d*++d*(d++*d++);

d=2;Y=++d*++d*(d++*d++);

d=2;Y=++d*++d*(d++*d++);

{
d=2;Y=++d*++d*(d++*d++);
}

d=2;Y=++d*++d*(d++*d++);

what will be the output ?
x=10;
x=+x++;

int a=0,x;
x = ++a * --a;
cout<<++a<<" "<<a++<<" "<<x;

output?? and how??

Y+=++Y+Y- - + - -8

PSVM(string args[ ])
Int x=6
x=++×*++×%4
System. out. println(x)

#include <stdio.h>
int main()
{
int i=0;
printf("%d %d %d %d", i,i++,++i,i--);
return 0;
}

Output :- 1,0,1,0
Explanation?

int i=0;
printf("%d %d %d %d", i,i++,++i,i--);

output:- 1,0,1,0
Explamation required.

Int x=2;
Printf("\n%d",++x % x++);
What should be the output??

A=--A +B++ + ++B + --B * C++ (A=12 B=13 C= 11)

#include<stdio.h>
#include<conio.h>
void main()
{
int x=5,y;
y=(++x)+(--x);
printf("%d",y);
}

#include<stdio.h>
#include<conio.h>
void main()
{
int x=5,y;
y=(++x)+(--x);
printf("%d",y);
}
What will be the answer of this???

++8 + ++7* ++0

Using which procedure the following two expressions are solved;

#include<stdio.h>
#include<math.h>
void main()
{
	int z=10,a,b;
	a=++z + ++z + ++z + z++;
	b=z++ + ++z + ++z + z++;
	printf("%d\n",a); // i am getting a=49
	printf("%d\n",b); // b= 47
	printf("%d\n",z); // z=14

getch();
}

y=++x+y+++x+++++y
x=++y+++y+x+++x++

y=++x+y+++x+++++y, x=++y+++y+x+++x++

int x=20,y=35;
x=y++ +x++;
y=++y + ++x;
Printf("%d%d",x,y);

Int a= 25 int b=5

What I can tell from the C standard, that in x = x++ + ... the value of x is getting modified twice within one sequence point. So it is Undefined Behaviour and the answer would vart depending upon the implementation of the compiler.

x+(y--)-x+(x++)-(y--) if x=y=2