Are logical operators in C evaluated with short-circuit evaluation?
Answer: A
Correct answer: True
Both && and || short-circuit, and the standard requires it rather than leaving it to the compiler. If the left operand of && is false, or of || is true, the result is already determined and the right operand is not evaluated at all - side effects in it simply do not happen.
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What is the output of this C code?
int main()
{
int a = 10, b = 10;
if (a = 5)
b--;
printf("%d, %d", a, b--);
}
Answer: C
Correct answer: a = 5, b = 9
if (a = 5) assigns 5 to a, and the value of that assignment, 5, is non-zero and so true. The body runs, taking b from 10 to 9. The printf then shows a as 5 and, because b-- is a post-decrement, the value passed is the current 9 - the decrement to 8 happens after the value has been handed over, and nothing prints afterwards.
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Which of the following data type will throw an error on modulus operation(%)?
Answer: C
Correct answer: float
% requires integer operands. char, short and int are all integer types and are promoted to int before the operation, so they are fine; float is not, and using it gives a compile-time error. Use fmod() for a floating-point remainder.
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What is the output of this C code?
int main()
{
int a = 10;
if (a == a--)
printf("TRUE 1\t");
a = 10;
if (a == --a)
printf("TRUE 2\t");
}
Answer: C
Correct answer: TRUE 1 TRUE 2
Both comparisons read a in one operand while modifying it in the other, with no sequence point between, so both are formally undefined behaviour. 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. What the question is really illustrating is that a == a-- and a == --a cannot be reasoned about at all - not that one is true and the other false.
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What is the output of this C code?
int main()
{
int i = 1;
if (i++ && (i == 1))
printf("Yes\n");
else
printf("No\n");
}
Answer: B
Correct answer: No
i++ yields the old value 1, which is true, so the right operand is evaluated - but by then the side effect has happened and i is 2. i == 1 is therefore false, the whole condition is false, and the program prints No. The && is a sequence point, so the increment is guaranteed complete before the comparison.
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A function cannot be defined inside another function
Answer: A
Correct answer: True
C has no nested functions: every function is defined at file scope, and a function definition inside another is a syntax error. (GCC offers nested functions as a non-standard extension, but that is not C.) What can be nested is a block, and a block may declare its own variables.
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What is the output of this C code?
int main()
{
int x = 1, y = 0, z = 3;
x > y ? printf("%d", z) : return z;
}
Answer: C
Correct answer: Compile time error
The conditional operator needs expressions in both branches, and return z is a statement. x > y ? printf("%d", z) : return z; is therefore a syntax error. Written properly it would be if (x > y) printf("%d", z); else return z;.
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What is the output of this C code?
void main()
{
int x = 97;
int y = sizeof(x++);
printf("x is %d", x);
}
Answer: A
Correct answer: x is 97
sizeof is evaluated at compile time from the type of its operand, and its operand is not executed. So x++ never runs, y becomes sizeof(int), and x is still 97. (Variable-length arrays are the one exception where the operand is evaluated.)
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What is the output of this C code?
void main()
{
int a = -5;
int k = (a++, ++a);
printf("%d\n", k);
}
Answer: D
Correct answer: -3
The comma operator evaluates its left operand, discards the result, then evaluates the right operand and yields that. So a++ takes a to -4, then ++a takes it to -3 and produces -3. k is therefore -3. There is a sequence point at the comma, so this is well defined.
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What will be the output of the following program?
#include <stdio.h>
void fun();
int main()
{
fun();
return 0;
}
void fun()
{
auto int i = 1;
register char a = 'D';
static int p = 0;
printf("%d %d %ld", i, a, p);
}
Answer: D
Correct answer: 1 68 0
Each declaration in fun behaves as its storage class says:
auto int i = 1; prints as 1.register char a = 'D'; is printed with %d, not %c, so it comes out as the character’s code - 68 in ASCII.static int p = 0; prints as 0.So the output is 1 68 0. The trap is the format specifier: a char is an integer type, and %d shows its numeric value.
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