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void main()
{
static int i = 5;
if(--i)
{
main();
printf("%d
",i);
}
}
what would be output of the above program and justify your
answer?
}
- 5 Answers
- 31049 Views
- C DAC, CDAC, Infosys, Wipro, I also Faced
- E-Mail Answers
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Answer / samrat
Ans: 0 0 0 0
The first thing you have to remember is that static
variables are initialized only once. The second thing is
that static variables have a life time scope and they retain
their value between function calls.
"i" is first initialized to 5. in the if condition the value
of i is changed to 4. main() is called again and the value
of i is changed to 3 in the if condition and main is called
again. Now the value of i is changed to 2 and main is called
again. Now the value of i is changed to 1 and main is called
again. After this the value of i is changed to "0" and the
block is excited.
As the value of i is now "0", it is printed 4 times for each
of the calls for main(). So the ans will be
0
0
0
0
Thanks,
Samrat
| Is This Answer Correct ? | 67 Yes | 5 No |
Answer / srsabariselvan
0
0
0
0
static variable's value is stored in memory statically upto
end of the program. so if the variable comes out of the
function it retains its value
| Is This Answer Correct ? | 13 Yes | 5 No |
Answer / sri ram
This prog'll not produce any output since the value of i
reduces when it reaches zero if block will not be executed
and the program is terminated....
| Is This Answer Correct ? | 7 Yes | 16 No |
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