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Unsigned char c;
for ( c=0;c!=256;c++2)
printf("%d",c);
No. of times the loop is executed ?
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Answer Posted / 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)
| Is This Answer Correct ? | 4 Yes | 1 No |
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