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What will be the output of the program, if a short int is 2 bytes wide? #include<stdio.h> int main() { short int i = 0; for(i<=5 && i>=-1; ++i; i>0) printf("%u,", i); return 0; } ?->(Show Answer!)
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1. What will be the output of the program, if a short int is 2 bytes wide? #include<stdio.h> int main() { short int i = 0; for(i<=5 && i>=-1; ++i; i>0) printf("%u,", i); return 0; }





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  • By: guest on 01 Jun 2017 06.01 pm
    for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression. In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one. Loop condition always get evaluated to true. Also at this point it increases i by one. An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)
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