ISRO2015-69

Question

If n has  3, then the statement a[++n]=n++;

• A. assigns 3 to a[5]
• B. assigns 4 to a[5]
• C. assigns 4 to a[4]
• D. what is assigned is compiler dependent

Comments

This is a classic example of Undefined Behavior (UB) in C.

The statement a[++n] = n++; modifies the variable n twice between two "sequence points" (in this case, the start of the statement and the semicolon at the end).

1. ++n (pre-increment): Increments n before its value is used for the array index.

2. n++ (post-increment): Uses the current value of n for the assignment, and then increments n after.

The C standard does not specify the order in which the left-hand side (a[++n]) and the right-hand side (n++) are evaluated. A compiler is free to do it in any order, leading to different results.

Scenario 1: Compiler evaluates Left-to-Right

1. a[++n]: n (which is 3) becomes 4. The array index is a[4].

2. n++: The value of n (which is now 4) is used. The value 4 is assigned. n is then incremented to 5.

3. Result: a[4] = 4

Scenario 2: Compiler evaluates Right-to-Left

1. n++: The value of n (which is 3) is used. The value 3 will be assigned. n is then incremented to 4.

2. a[++n]: n (which is now 4) becomes 5. The array index is a[5].

3. Result: a[5] = 3

Because different compilers (or even the same compiler with different optimization settings) can produce different results, the behavior is "compiler dependent," which is the practical outcome of undefined behavior.

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This is a classic example of Undefined Behavior (UB) in C.

The statement a[++n] = n++; modifies the variable n twice between two "sequence points" (in this case, the start of the statement and the semicolon at the end).

1. ++n (pre-increment): Increments n before its value is used for the array index.

2. n++ (post-increment): Uses the current value of n for the assignment, and then increments n after.

The C standard does not specify the order in which the left-hand side (a[++n]) and the right-hand side (n++) are evaluated. A compiler is free to do it in any order, leading to different results.

Scenario 1: Compiler evaluates Left-to-Right

1. a[++n]: n (which is 3) becomes 4. The array index is a[4].

2. n++: The value of n (which is now 4) is used. The value 4 is assigned. n is then incremented to 5.

3. Result: a[4] = 4

Scenario 2: Compiler evaluates Right-to-Left

1. n++: The value of n (which is 3) is used. The value 3 will be assigned. n is then incremented to 4.

2. a[++n]: n (which is now 4) becomes 5. The array index is a[5].

3. Result: a[5] = 3

Because different compilers (or even the same compiler with different optimization settings) can produce different results, the behavior is "compiler dependent," which is the practical outcome of undefined behavior. 

so option is D

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The statement a[++n] = n++; modifies the variable n twice between two "sequence points" (in this case, the start of the statement and the semicolon at the end).

1. ++n (pre-increment): Increments n before its value is used for the array index.

2. n++ (post-increment): Uses the current value of n for the assignment, and then increments n after.

The C standard does not specify the order in which the left-hand side (a[++n]) and the right-hand side (n++) are evaluated. A compiler is free to do it in any order, leading to different results.

Scenario 1: Compiler evaluates Left-to-Right

1. a[++n]: n (which is 3) becomes 4. The array index is a[4].

2. n++: The value of n (which is now 4) is used. The value 4 is assigned. n is then incremented to 5.

3. Result: a[4] = 4

Scenario 2: Compiler evaluates Right-to-Left

1. n++: The value of n (which is 3) is used. The value 3 will be assigned. n is then incremented to 4.

2. a[++n]: n (which is now 4) becomes 5. The array index is a[5].

3. Result: a[5] = 3

Because different compilers (or even the same compiler with different optimization settings) can produce different results, the behavior is "compiler dependent," which is the practical outcome of undefined behavior. 

so option is D

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This is a classic example of Undefined Behavior (UB) in C.

The statement a[++n] = n++; modifies the variable n twice between two "sequence points" (in this case, the start of the statement and the semicolon at the end).

1. ++n (pre-increment): Increments n before its value is used for the array index.

2. n++ (post-increment): Uses the current value of n for the assignment, and then increments n after.

The C standard does not specify the order in which the left-hand side (a[++n]) and the right-hand side (n++) are evaluated. A compiler is free to do it in any order, leading to different results.

Scenario 1: Compiler evaluates Left-to-Right

1. a[++n]: n (which is 3) becomes 4. The array index is a[4].

2. n++: The value of n (which is now 4) is used. The value 4 is assigned. n is then incremented to 5.

3. Result: a[4] = 4

Scenario 2: Compiler evaluates Right-to-Left

1. n++: The value of n (which is 3) is used. The value 3 will be assigned. n is then incremented to 4.

2. a[++n]: n (which is now 4) becomes 5. The array index is a[5].

3. Result: a[5] = 3

Because different compilers (or even the same compiler with different optimization settings) can produce different results, the behavior is "compiler dependent," which is the practical outcome of undefined behavior.

Answer 1

It is compiler dependent

Reference :Undefined behaviour

Comments

@manoj sir, codepad give 6 

                 tutorial point give 5

                  ideaone give 5.         why?

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That is what is undefined behavior of C.

i = i++

which modify the same value twice which needn't be allowed.

Similarly here

a[i] = i++

which modify i and use it along the way.

Its all depends on compiler what value is will be assigned .

I think it should be clear . I dont know much about undefined stuffs.

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Final value of i is 6

Answer 2

It is an undefined behaviour since between two sequence points which is here ; at the end of this statement and the other ; which would be present above this statement in the actual code we cannot modify the value of a variable more than once therefore here u r trying to modify the value of n two times so it depends on compiler wither it will evaluate ++n inside array subscript or assign ++n to array index .

Comments

the more explanation in this video about sequence points :https://www.youtube.com/watch?v=hCLsW1qttZ0

Answer 3

Ans is D.It depends on the compiler.Here n is updated twice before the next sequence point is reached.ref:https://en.wikipedia.org/wiki/Sequence_point

Comments

So whenever there are 2 or more modifications on the same variable, it is always compiler dependent?

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No. It is about how and in which order prefix and postfix operator perform its function

Answer 4

Correct Answer - Option 4 : what is assigned is compiler dependent

 EXPLANATION:
This question can be solved in two different ways by different compilers:
 1. First perform n++ then calculate ++n for index value
2. Calculate ++n for index value initially and then find n++. Therefore different compilers can solve this code in one of the above ways. Hence, the correct answer is "option 4".
Increment operation is of two types:
1. Pre increment (++n): Increase the value first & then substitute the value in variable n.
2. Post increment (n++): Initially substitute the value in n & then increase the value of n.

Answer 5

Answer : option C- assigns 4 to a[4]

Answer 6

what is assigned is compiler dependent......