Answer / p.muthukumar

b=a-b;
a=a-b;
b=a+b;

Answer / 3uggy3oy

Answer #16 is totally wrong it fails
when x>y and many other situation.

Answer / alex

When swapping integers or characters, use:
a^=b;
b^=a;
a^=b;
When swapping floats or doubles use:
a+=b;
b=a-b;
a=a-b;
When swapping strings use:
a=(char*)((int)a^(int)b);
b=(char*)((int)a^(int)b);
a=(char*)((int)a^(int)b);
This works by modifying the address values of the pointer.

Answer / shubham

NOne of the answer is correct except the 2ND one.....Please
don't give wrong answers.

Answer / hanmanth reddy

a=a+b-(b=a);

Answer / bhaskar.mantrala

void main()
{
int a,b;
clrscr();
printf("\n Enter the values of a and b ");
scanf(" %d %d ", &a,&b);
a=a*b;
b=a/b;
a=a/b;
printf("\n \n After swapping ----> a = %d \t b = %d",a,b);
getch();
}

Answer / trinath somarouthu

using X-OR
#define SWAP(x,y) x^=y^=x^=y

x = x ^ y --> x^=y -- (1)
y = y ^ x --> y^=x -- (2)
x = x ^ y --> x^=y -- (3)

(3) in (2) --> y^=x^=y -- (4)
(4) in (1) --> x^=y^=x^=y -- :-)

all togeather, he single line code

#define SWAP(x,y) x^=y^=x^=y

Answer / sunil kharat

a^=b^=a^=b;

try it...
it works. one line solution

Answer / sujata

i think first is right.

Answer / perl guru

($a,$b)=($b,$a)
....this is how we do in PERL....Ha Ha Ha pretty cool !!!!

And BELIEVE me it works not only for NUMBERS but also for
STRINGS, ARRAYS or any Data Structure or any garbage
value....this is mine Challenge.....

None of the any programming language can come close to PERL
in this much of SIMPLICITY and ROBUSTNESS....just one line
does the Magic....

Again I would say the Question itself is very silly one "How
to swap two variables, without using third variable
?"....and I see alot of stupid Answers posted here....

Everyone posting the answer is assuming that Question is
about swapping INTEGER NUMBERS....I would like to ask what
if I provide FLOATING POINT NUMBERS, NEGATIVE NUMBERS, REAL
NUMBERS, and surely it does not work for STRINGS....

That is why I say "Perl is immensely powerful. If you think
something can't be done, the problem is likely to be it is
beyond your ability, not that of Perl."

Welcome to the world of PERL....its more precious than PEARL....

main() { { unsigned int bit=256; printf("%d", bit); } { unsigned int bit=512; printf("%d", bit); } } a. 256, 256 b. 512, 512 c. 256, 512 d. Compile error

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There were 10 records stored in “somefile.dat” but the following program printed 11 names. What went wrong? void main() { struct student { char name[30], rollno[6]; }stud; FILE *fp = fopen(“somefile.dat”,”r”); while(!feof(fp)) { fread(&stud, sizeof(stud), 1 , fp); puts(stud.name); } }

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void ( * abc( int, void ( *def) () ) ) ();

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main() { static char names[5][20]={"pascal","ada","cobol","fortran","perl"}; int i; char *t; t=names[3]; names[3]=names[4]; names[4]=t; for (i=0;i<=4;i++) printf("%s",names[i]); }

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Write a Program in 'C' To Insert a Unique Number Only. (Hint: Just Like a Primary Key Numbers In Database.) Please Some One Suggest Me a Better Solution for This question ??

0 Answers   Home,

How will u find whether a linked list has a loop or not?

8 Answers   Microsoft,

main() { extern int i; i=20; printf("%d",sizeof(i)); }

2 Answers  

Is this code legal? int *ptr; ptr = (int *) 0x400;

1 Answers  

How do you write a program which produces its own source code as its output?

7 Answers  

int i,j; for(i=0;i<=10;i++) { j+=5; assert(i<5); }

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How to count a sum, when the numbers are read from stdin and stored into a structure?

1 Answers  

main() { static int a[3][3]={1,2,3,4,5,6,7,8,9}; int i,j; static *p[]={a,a+1,a+2}; for(i=0;i<3;i++) { for(j=0;j<3;j++) printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j), *(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i)); } }

1 Answers