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........................
Answers were Sorted based on User's Feedback
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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