for c = 2, value of c after c <<= 1;
Answer: D
Correct answer: c = 4;
<<= shifts left and assigns. Shifting left by one position doubles the value, because each bit moves to the next higher power of two:
So c becomes 4.
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What will be the output of the following program?
#include <stdio.h>
int main()
{
static unsigned int a = 23;
register unsigned char c = 'R';
auto long unsigned q = 345L;
static long signed p = 345L;
printf("a = %u c = %c", a ,c);
printf("\nq = %ld p = %ld", q, p);
return 0;
}
Answer: A
Correct answer: a=23 c=R q = 345 p = 345
Each variable prints its own value, and the storage classes affect only lifetime and placement, not the values: a is 23, and c printed with %c shows the character R. The two long variables both hold 345 and print with %ld as 345. The only real trap is the case of the letters in the options - the format string writes a lower-case a.
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Relational operators cannot be used on:
Answer: A
Correct answer: structure
Relational operators work on scalars - arithmetic types and pointers - because comparison needs a single value to order. A struct is an aggregate of several members, so s1 < s2 has no meaning and is rejected; even == is not defined for structures, since padding bytes would make a byte-wise comparison unreliable. Comparing them means comparing their members one by one.
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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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