Expressions Questions and Answers

Q: What is the output of this C code? int main() { int a = 1, b = 1, c; c = a++ + b; printf("%d, %d", a, b); }

Correct answer: a = 2, b = 1c = a++ + b; uses the old value of a, so c becomes 2, and the side effect then makes a equal to 2. b is only read, so it stays 1. The printed values are those of a and b, not of c.

Q: 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++); }

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.

Q: For which of the following, “PI++;” code will fail?

Correct answer: #define PI 3.14A #define is textual substitution, so PI++; expands to 3.14++; - and a constant is not an lvalue, so it cannot be incremented. The other two are variables: float PI can be incremented arithmetically, and char *PI can be incremented as a pointer.

Q: What is the output of this C code? int main() { int a = 10, b = 10; if (a = 5) b--; printf("%d, %d", a, b--); }

Correct answer: a = 5, b = 9if (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.

Q: What is the output of this C code? int main() { int i = 0; int j = i++ + i; printf("%d\n", j); }

Correct answer: 1With i starting at 0, i++ contributes 0 and leaves i as 1, and the second operand then reads 1, giving \(0 + 1 = 1\). Strictly this expression reads i in one operand while modifying it in the other with no sequence point between, which the standard leaves undefined; 1 is what mainstream compilers produce, but the expression should be split into two statements.

Q: What is the output of this C code? int main() { int i = 2; int j = ++i + i; printf("%d\n", j); }

Correct answer: 6++i takes i to 3 and contributes 3, and the second operand then reads 3, giving \(3 + 3 = 6\). As with any expression that both modifies and reads the same variable without an intervening sequence point, this is formally undefined behaviour; 6 is simply what the common compilers give.

Q: Comment on the output of this C code? int main() { int i = 2; i = i++ + i; printf("%d\n", i); }

Correct answer: = operator is not a sequence pointThe point of the example is that = does not introduce a sequence point. In i = i++ + i; the compiler is free to order the read of i, the increment, and the store in any way it likes, so the result is undefined. Sequence points in C come from &&, ||, the comma operator, the conditional ?:, and the end of a full expression - not from assignment.

Q: What is the output of this C code? int main() { int i = 0; int x = i++, y = ++i; printf("%d % d\n", x, y); return 0; }

Correct answer: 0, 2The comma separating two declarators is a sequence point, so the two initialisations happen strictly in order and nothing here is undefined. x = i++ takes the old value 0 and leaves i at 1; then y = ++i makes i 2 and gives 2.

Q: What is the output of this C code? int main() { int i = 10; int *p = &i; printf("%d\n", *p++); }

Correct answer: 10++ binds tighter than *, but it is a post-increment, so *p++ means “dereference the current p, then advance p”. The value printed is therefore i itself, 10; the pointer moves on afterwards, to an address that is no longer valid.

61.

What is the difference between the following 2 codes?

//Program 1
    int main()
    {
        int d, a = 1, b = 2;
        d =  a++ + ++b;
        printf("%d %d %d", d, a, b);
    }

 //Program 2
    int main()
    {
        int d, a = 1, b = 2;
        d =  a++ + ++b;
        printf("%d %d %d", d, a, b);
    }

A.No difference as space doesn’t make any difference, values of a, b, d are same in both the case
B.No difference as space doesn’t make any difference, values of a, b, d are different
C.Program 1 has syntax error, program 2 is not
D.Program 2 has syntax error, program 1 is not

Answer: A

Correct answer: No difference as space doesn’t make any difference, values of a, b, d are same in both the case

The two programs are character for character the same, so naturally they produce the same values: a++ contributes 1 and leaves a as 2, ++b makes b 3 and contributes 3, so d is 4 and the output is 4 2 3. The point the question is driving at is that whitespace between tokens is insignificant - though it is worth knowing the limit of that rule: a+++++b would not compile, because the compiler takes the longest possible token at each step and reads it as a++ ++ +b.

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62.

What is the output of this C code?

 int main()
    {
        int a = 1, b = 1, c;
        c = a++ + b;
        printf("%d, %d", a, b);
    }

A.a = 1, b = 1
B.a = 2, b = 1
C.a = 1, b = 2
D.a = 2, b = 2

Answer: B

Correct answer: a = 2, b = 1

c = a++ + b; uses the old value of a, so c becomes 2, and the side effect then makes a equal to 2. b is only read, so it stays 1. The printed values are those of a and b, not of c.

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63.

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++);
    }

A.15, 4, 5
B.9, 6, 9
C.9, 3, 5
D.6, 4, 6

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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64.

For which of the following, “PI++;” code will fail?

A.#define PI 3.14
B.char *PI = “A”;
C.float PI = 3.14;
D.Both (A) and (B)

Answer: A

Correct answer: #define PI 3.14

A #define is textual substitution, so PI++; expands to 3.14++; - and a constant is not an lvalue, so it cannot be incremented. The other two are variables: float PI can be incremented arithmetically, and char *PI can be incremented as a pointer.

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65.

What is the output of this C code?

int main()
    {
        int a = 10, b = 10;
        if (a = 5)
        b--;
        printf("%d, %d", a, b--);
    }

A.a = 10, b = 9
B.a = 10, b = 8
C.a = 5, b = 9
D.a = 5, b = 8

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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66.

What is the output of this C code?

int main()
    {
        int i = 0;
        int j = i++ + i;
        printf("%d\n", j);
    }

A.1
B.2
C.Compile time error

Answer: A

Correct answer: 1

With i starting at 0, i++ contributes 0 and leaves i as 1, and the second operand then reads 1, giving \(0 + 1 = 1\). Strictly this expression reads i in one operand while modifying it in the other with no sequence point between, which the standard leaves undefined; 1 is what mainstream compilers produce, but the expression should be split into two statements.

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67.

What is the output of this C code?

int main()
    {
        int i = 2;
        int j = ++i + i;
        printf("%d\n", j);
    }

A.6
B.5
C.4
D.Compile time error

Answer: A

Correct answer: 6

++i takes i to 3 and contributes 3, and the second operand then reads 3, giving \(3 + 3 = 6\). As with any expression that both modifies and reads the same variable without an intervening sequence point, this is formally undefined behaviour; 6 is simply what the common compilers give.

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68.

Comment on the output of this C code?

int main()
    {
        int i = 2;
        i = i++ + i;
        printf("%d\n", i);
    }

A.= operator is not a sequence point
B.++ operator may return value with or without side effects
C.it can be evaluated as (i++)+i or i+(++i)
D.Both a and b

Answer: A

Correct answer: = operator is not a sequence point

The point of the example is that = does not introduce a sequence point. In i = i++ + i; the compiler is free to order the read of i, the increment, and the store in any way it likes, so the result is undefined. Sequence points in C come from &&, ||, the comma operator, the conditional ?:, and the end of a full expression - not from assignment.

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69.

What is the output of this C code?

int main()
    {
        int i = 0;
        int x = i++, y = ++i;
        printf("%d % d\n", x, y);
        return 0;
    }

A.0, 2
B.0, 1
C.1, 2
D.Undefined

Answer: A

Correct answer: 0, 2

The comma separating two declarators is a sequence point, so the two initialisations happen strictly in order and nothing here is undefined. x = i++ takes the old value 0 and leaves i at 1; then y = ++i makes i 2 and gives 2.

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70.

What is the output of this C code?

int main()
    {
        int i = 10;
        int *p = &i;
        printf("%d\n", *p++);
    }

A.10
B.11
C.Garbage value
D.Address of i

Answer: A

Correct answer: 10

++ binds tighter than *, but it is a post-increment, so *p++ means “dereference the current p, then advance p”. The value printed is therefore i itself, 10; the pointer moves on afterwards, to an address that is no longer valid.

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