What is the output of this C code?
int main()
{
int a = 20;
double b = 15.6;
int c;
c = a + b;
printf("%d", c);
}
Answer: A
Correct answer: 35
a + b promotes the int to double, giving \(20 + 15.6 = 35.6\). Assigning that to an int truncates toward zero rather than rounding, so c becomes 35. Rounding would have given 36 - the distractor.
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What is the output of this C code?
int main()
{
int a = 1, b = 1, d = 1;
printf("%d, %d, %d", ++a + ++a+a++, a++ + ++b, ++d + d++ + a++);
}
Answer: A
Correct answer: 15, 4, 5
++a + ++a + a++ modifies a three times in one expression, and the three printf arguments may be evaluated in any order - the standard does not fix it. The expression modifies a variable more than once, and reads it as well, with no sequence point in between. The C standard makes that undefined behaviour, so the answer listed is what one common compiler happens to produce - another compiler, or another optimisation level, may give something else, and no program should depend on it. The lesson to take away is not the number but the rule: one variable, one modification per expression.
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What will be the output of the following code snippet?
Y = 5;
if (! Y > 10)
X = Y + 3;
else
X = Y + 10;
printf(" X = %d Y = %d", X, Y);
Answer: A
Correct answer: The program will print X = 15 Y = 5
! binds far tighter than >, so the condition is (!Y) > 10, not !(Y > 10). With Y equal to 5, !Y is 0, and \(0 > 10\) is false. The else branch runs, giving \(X = 5 + 10 = 15\), and Y is never modified, so the output is X = 15 Y = 5.
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Which of the following statement is correct about the code snippet given below?
num = 5;
printf( "%d", ++num++ );
Answer: C
Correct answer: The code will result in L – value required
++num++ groups as ++(num++), because postfix binds tighter than prefix. The result of num++ is a value, not an object, so the prefix ++ has nothing it can modify - the compiler reports that an lvalue is required. The same applies to ++(a + b) or 5++.
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What is the output of this C code?
int main()
{
unsigned int a = 10;
a = ~a;
printf("%d\n", a);
}
Answer: C
Correct answer: -11
~ flips every bit, and for a two’s-complement integer that is exactly \(\sim a = -a - 1\):
The variable is unsigned, so it actually holds 4294967285, but %d reads those bits back as a signed int and prints -11.
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For which of the following situation should the register storage class be used?
Answer: B
Correct answer: For loop counter
register is worth asking for where a variable is accessed very often, and a loop counter is the classic case - it is read and written on every iteration. It cannot be used where the variable’s address is needed, since a register has no address. In practice modern compilers allocate registers better than the hint does, so the keyword is now of historical interest.
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p++ executes faster than p + 1 since
Answer: B
Correct answer: Single machine instruction required for p++
p++ maps to a single increment instruction that updates p in place, whereas p + 1 computes a new value that then has to be stored somewhere. The saving is historical, though: any modern optimising compiler generates identical code for p = p + 1 and p++.
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Comment on the output of this C code?
int main()
{
int i, n, a = 4;
scanf("%d", &n);
for (i = 0; i < n; i++)
a = a * 2;
}
Answer: B
Correct answer: No output
The loop multiplies a by 2 on each of the n passes, so a ends as \(4 \times 2^n\). But the program never prints anything - there is no printf after the loop - so the computed value is simply discarded when main returns. No output.
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Where will the space be allocated for an automatic storage class variable?
Answer: C
Correct answer: In memory
An automatic variable lives on the stack, which is memory. The register class is the one that asks for a CPU register, and even that is only a request. A compiler may of course keep an automatic variable in a register as an optimisation, but the storage class itself specifies memory.
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Which is executed quickly?
Answer: C
Correct answer: Both
On any modern compiler ++p and p++ compile to the same machine code when the result is discarded, as in a for loop’s third clause - the optimiser sees that the old value is unused. The historical preference for ++p matters for C++ iterators, where p++ must copy the object before incrementing it.
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