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What will be the output of the program? #include<stdio.h> int main() { unsigned int i = 65536; / Assume 2 byte integer/ while(i != 0) printf("%d",++i); printf("\n"); return 0; } ?->(Show Answer!)
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1. What will be the output of the program? #include<stdio.h> int main() { unsigned int i = 65536; / Assume 2 byte integer/ while(i != 0) printf("%d",++i); printf("\n"); return 0; }





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  • By: guest on 01 Jun 2017 06.01 pm
    Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535. Step 1:unsigned int i = 65536; here variable i becomes '0'(zero). because unsigned int varies from 0 to 65535. Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed. Hence there is no output. Note: Don't forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.
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