what is the output of following program ?
void main()
{
int i=5;
printf("%d %d %d %d %d ",i++,i--,++i,--i,i);
}
Answers were Sorted based on User's Feedback
Answer / sameeksha agrawal
ya i can tell u abt d ans ..the reasn behind it x++ is
example of post increament in which frstly value of any
variable is prnt thn oprtr r prfrmed and the oprtions r
perfrmd frm right side...
Is This Answer Correct ? | 7 Yes | 2 No |
Answer / sandeep gupta
Friends, actually the output is compiler dependent. Never
give any specific answer in this type of questions because C
does not provide any format in evaluation order of postfix
and prefix expressions when passed to any functions like
printf() eg. fun1(i++,i) may pass fun1(5,6) in some
compilers and fun1(5,5) in another. So it totally depends on
what compiler we're using.
Is This Answer Correct ? | 4 Yes | 0 No |
Answer / jalp
Here if i do it manually then i got 44545 but compiler shows
me : 45545
Can any body show the stack process that how it execute
internally,
And also reply through mail.
Thanks.
Is This Answer Correct ? | 6 Yes | 5 No |
Answer / jalp
That i know .. I want to know the stack process , how it
internally works .. if you elaborate through step then
please explain it ..
Thanks.
Is This Answer Correct ? | 2 Yes | 1 No |
Answer / kavsac
Guys,
There is something, I wanna add on. The above result occurs only in windows, in Unix its 56656
Is This Answer Correct ? | 2 Yes | 1 No |
Answer / shenbagam
in c its left to right operation will be perfomed by
compiler..... so take this
int i=5;
printf("%d %d %d %d %d ",i++,i--,++i,--i,i);
i=5;
--i=4; after tis the value of i=4 only;
because --i will decrement the value of i, and then return the decremented value.
++i=5; after this the value of i=5;
because ++i will increment the value of i, and then return the incremented value.
i--=5; after this the value of i=4;
because will decrement the value of i, but return the pre-defind value of i. so i=5 before the step na?.....
i++=4; after this the value of i=5;
because will increment the value of i, but return the pre-defined value of i.so i=4 before the step na?...
so only the result will be like 45545 this.......
int i=5;
printf("%d %d %d %d %d ",i++,i--,++i,--i,i);
after the step if u print i; the value must be 5
only......... so the doubt will be cleared aha?............
all the best:):):):).
Is This Answer Correct ? | 4 Yes | 3 No |
Answer / sameeksha agrawal
no Kavsac u r wrng try again its answer...
Is This Answer Correct ? | 0 Yes | 0 No |
Answer / sameeksha agrawal
sry u r totally wrng cz c is structrd progrmmmng languaga
Is This Answer Correct ? | 0 Yes | 0 No |
Answer / sindhu
Answer:after compilation i got 45545.... but if i consider it from left to right in manual my answer is 55454...
i want clear xplanation for tis program...
Is This Answer Correct ? | 0 Yes | 0 No |
union u { union u { int i; int j; }a[10]; int b[10]; }u; main() { printf("\n%d", sizeof(u)); printf(" %d", sizeof(u.a)); // printf("%d", sizeof(u.a[4].i)); } a. 4, 4, 4 b. 40, 4, 4 c. 1, 100, 1 d. 40 400 4
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main() { int x=5; for(;x!=0;x--) { printf("x=%d\n", x--); } } a. 5, 4, 3, 2,1 b. 4, 3, 2, 1, 0 c. 5, 3, 1 d. none of the above
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