Programming in C - I/O Operations
8. The following statement printf( ‘%f ’, 9/5) ; prints
- Option : D
- Explanation : This statement is undefined behavior because the argument has to be of type double and 9 / 5 is of type int. Following program yields an output of 1.8 int main() { printf("%f",9/5.0); return 0; }
9. The following program fragment unsigned i = 1; int j = -4; printf ("%u ", i + j); prints
- Option : C
- Explanation : In the computer I used to execute this program. The output was 4294967293. That's because in my system, sizeof (int ) is 4 bytes (32 bits), and negative numbers are represented in 2's complement form. This means -4 will be represented as 11111111 11111111 11111111 11111100 (i.e. 30 one's followed by 2 zeroes). Note that this number is 232 - 1 - 3. Before j gets added to 1, it will be converted to an unsigned integer. So, i + j is essentially adding 1 to 232- 1 - 3 which gives 4294967293.
- Option : C
- Explanation : Let size of ( int ) = 1. So, -4 will be stored as 1 1 1 1 1 1 0 0. Since we are adding unsigned and signed integers, the signed gets converted to unsigned. So, i + j will become 1 1 1 1 1 1 0 1. We are trying to print this as an unsigned integer. So, what is printed will be 28 - 1 - 2. So, log (x + 3) = 8 (i.e.. 8 * sizeof ( int)}.
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