void main()
{
for(int i=0;i<5;i++);
printf("%d",i);
}
What is the output?..
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Answer / kumaran
declaring int i inside for loop is not available in
traditional 'c'
| Is This Answer Correct ? | 80 Yes | 30 No |
Answer / mahfooz
we can not declare loop condition variable in c..we can do
in c++.
| Is This Answer Correct ? | 38 Yes | 20 No |
Answer / samir isakoski
why shoul have a cout if is c++
buth this is c (printf)
and why cannot daclare a int in the for ciclus
| Is This Answer Correct ? | 17 Yes | 11 No |
Answer / divya
no output because the for loop is terminated with a semicolon which means the loop will terminate at once.
| Is This Answer Correct ? | 15 Yes | 12 No |
Answer / shilpi
it will give 6 as a answer because condition is true inside and then it will increase its value by 1 and the terminate from the loop.
| Is This Answer Correct ? | 2 Yes | 0 No |
hi friend i am ricky dobriyal.
if this programe in c then this program produce error
expression syntex error.
if in c++ then it will produce o/p=5
| Is This Answer Correct ? | 10 Yes | 9 No |
Answer / pramod
if it is in C it displays an error "indicating that "i"
cannot be declared in loop and i as undefined symbol"
if it the case of C++ it displays "5" because C++ supports
dynamic declaration and once declared anywhere in the
program(even in loops) can be used anytime in that program
| Is This Answer Correct ? | 2 Yes | 2 No |
Answer / marius
if i have
$x=(11,22,33,44,55);
for($i=0;$i<5;$i++)
echo$x[$i]." ";
and we will print this:
11 22 33 44 55
$i=0 => 11
$i<5 => 44
$i++ => 55
ghigamarius@gmail.com
| Is This Answer Correct ? | 1 Yes | 2 No |
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