Answer / ram mohan

Ur approach may correct but I have a small dought
Plz tell me output of
Printf("", ++i + --z +i + --i * ++y) ;
@pawanjhi

Answer / pawanjha12

here in first statement

printf("%d",++i + --z + i++ + --i * ++y);

argument is : ++i + --z + i++ + --i * ++y.

first it will maintain stack operation like

++y (now fifth, it will execute and, y=4)(top 4)
--i (now fourth, it will execute and, i=6)
(because, last value of i were 7, once i++ were
executed, now --i will less one value in i)(top 3)
i++ (now third, it will execute and, i=6, it will
as it is, its value will for next stack value.)
(top 2)
--z (now second, it will execute and, z=1)(top 1)
++i (first it will execute and, i=6)(top 0)

now (++i + --z + i++ + --i * ++y)
(6+1+6+6*4)=(37)

it is output, say 37,for this printf("%d",++i + --z + i++
+ --i * ++y);

.................................................

Now Let me go with second statement, that is :

ans=++i + --z + i++ + --i * ++y;

here,
first of all ++y will contain the value of variable y=4
++y=4

after this, --i will less the value of variable i, say now
i = 4,
(--i=4),

after this, i++ will execute and, it will not increase the
value of variable i, right now, so value of i, say now i =
4, as it is.

after this, --z will less the value of variable z, say now
z = 1,
(--z=1)

now, ++i will increase the value of variable i.
Say i = 5.
.......................................................

now value of valiable i in memory is 5.
ans=5+1+5+5*4
ans=5+1+5+20
ans=31..........,
still if you are not getting, so come to me at
pankajace12@gmail.com

how tally is useful?

2 Answers  

---

WHAT WILL BE THE OUTPUT OF THE FOLLOWING QUESTION void main() { int x=4,y=3,z; z=x-- -y; printf("%d%d%d",x,y,z); }

25 Answers   HCL,

---

Display this kind of output on screen. 1 0 1 1 0 1 3. Display this kind of output on screen. 1 1 0 1 0 1 4. Display this kind of output on screen. 1 1 0 1 0 1 5.Display this kind of output on screen. 1 2 3 4 5 6 7 8 9 10

1 Answers  

---

main() { char c; for(c='A';c<='Z';c++) getch(); }

9 Answers  

---

class test { int a; public: test(int b):a(b){} void show(){ cout<<a; } }; void main() { test t1; test t2(5); t1.show(); t2.show(); } }

1 Answers  

---

void main() { int i=1; printf("%d%d%d",i,++i,i++); } Cau u say the output....?

24 Answers   HCL,

---

I'm having trouble with coming up with the correct code. Do I need to put a loop? Please let me know if I'm on the right track and what areas I need to correct. I still don't have a good grasp on this programming stuff. Thanks =) The assignment was to write a program using string functions that accepts a coded value of an item and displays its equivalent tag price. The base of the keys: 0 1 2 3 4 5 6 7 8 9 X C O M P U T E R S Sample I/O Dialogue: Enter coded value: TR.XX Tag Price : 68.00

3 Answers   UCB,

---

Why are memory errors hard to debug?

1 Answers  

---

I'm having trouble with coming up with the correct code. Thank You!! The assignment was to write a program using string functions that accepts a price of an item and displays its coded value. The base of the keys: X C O M P U T E R S 0 1 2 3 4 5 6 7 8 9 Sample I/O Dialogue: Enter Price: 489.50 Coded Value: PRS.UX

0 Answers  

---

what is meant by linking error? how can i solve it? if there is a linking error " unable to open file 'cos.obj'? then what should i do?

1 Answers  

---

Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have already been declared, use a do...while loop to compute the sum of the cubes of the first n whole numbers, and store this value in total . Thus if n equals 4, your code should put 1*1*1 + 2*2*2 + 3*3*3 + 4*4*4 into total . Use no variables other than n , k , and total .

3 Answers  

---

void main() { for(int i=0;i<5;i++); printf("%d",i); } What is the output?..

32 Answers   College School Exams Tests, CTS, HCL, iGate, SmartData,

---