The code is::::: if(condition)
Printf("Hello");
Else
Printf("World");
What will be the condition in if in such a way that both
Hello and world are printed in a single attempt?????? Single
Attempt in the sense... It must first print "Hello" and it
Must go to else part and print "World"..... No loops,
Recursion are allowed........................
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Answer / gg
#include<stdio.h>
main()
{
if(printf("hello")==0)
printf("hello");
else
printf("world");
}
Think is there any Other.....
| Is This Answer Correct ? | 62 Yes | 10 No |
Answer / abdur rab
#include <stdio.h>
int main ( int argc, char* argv[] )
{
if ( !printf("Hello") )
printf ("Hello");
else printf (" World");
}
| Is This Answer Correct ? | 17 Yes | 6 No |
but goto is not a good sort of programming ............. a
well knowledgeable programmer wont use goto..... according
to me........... plz think of any other logic??????????
| Is This Answer Correct ? | 5 Yes | 5 No |
but yhe printf statement will print one "hello" and one
world.......................... but i said it must enter to
the if part as well as else part controls..............
before a long time baxk itself i tried this method.........
then only i read the question of IBM properly
| Is This Answer Correct ? | 4 Yes | 4 No |
Answer / anudyuti
#include <stdio.h>
int executeboth();
void main()
{
both();
}
int both()
{
static int i=0;
if(i++==0 ? executeboth():1)
{
printf ("Hello");
}
else
{
printf (" World");
}
return 0;
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / srikanth
int main()
{
if(fork())
{
printf("hello");
}
else
{
printf("world");
}
}
| Is This Answer Correct ? | 0 Yes | 0 No |
#include<stdio.h>
void main()
{
if(1)
{
printf("HELLO!");
goto HERE;
}
else
{
HERE:
printf("HAI....");
}
}
| Is This Answer Correct ? | 18 Yes | 19 No |
Answer / subbu
#include<stdio.h>
int main()
{
if(!printf("Hello")
{
printf("Hello");
}
else
printf("World");
}
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / fazil
/* Solution 1: */
int main(int argc, char* argv[])
{
if( argc == 2 || main( 2, NULL ) )
{
printf("Hello ");
}
else
{
printf("World\n");
}
return 0;
}
/* Solution 2 (Only for Unix and Linux): */
int main(int argc, char* argv[])
{
if( !fork() )
{
printf("Hello ");
}
else
{
printf("World\n");
}
return 0;
}
| Is This Answer Correct ? | 4 Yes | 7 No |
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C 
