Answer / subbu

instead of "c++2" if there is "c++" it will not give any
errors. output will be continuosly for infinite number of
times.

Answer / shalabh

This will be an infinite loop as the value of unsigned char
cannot exceed 255....c++2 can be errornous..but the value
of unsigned char cannot go beyond 255

Answer / manishsoni

It produce compile time error becoz the statement c++2 is
not allowed here.

if this statement is written as c+=2,then it will give us an
infinite loop between
0--->254----->0---->254...
because it check simply c!=256(mean c is not equal to
256,mean it simply it didnot print the value at 256,that
mean it did not print 256...after 254...and print 0)
------------------------------------------------------------
No. of times the loop is executed ?
------------------------------------------------------------
loop will execute 128 times becoz:
the loop is increased each time by 2 and loop is execute
254 time without zero so
254/2=127
127+1(zero)=128;
so the loop is execute 128 times....
BY:ManisH SonI(MoNu)

Answer / sundeep

First of all the code returns compile time error....bcoz
of "c++2",and c!=256 becoz unsigned char takes just 255
bytes..........and exceeding that results in compile time
error........

Answer / manish soni

IT SHOW ACTUAL O/P.
main()
{
int c;
int i=1;
for ( c=0;c!=256;c+=2)
{
printf("%d=%d\n",i,c);
i++;
}
getch();
}

MANISH SONI
[GYAN CORPORATION,CHANDIGARH]

Answer / prabu

infinite loop executed

Answer / manishsoni

The program is as:
------------------------------------------------------------
#include<stdio.h>
#include<conio.h>
int main()
{
unsigned char c;
for(c=0;c!=256;c++2)
{
printf("%d",c);
printf("\n");
}

getch();
}

Answer / sanjeev

this loop is executed at 254 times.

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