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What will be the final value of j in the following C code?
#include <stdio.h>
int main()
{
int i = 0, j = 0;
if (i && (j = i + 10))
//do something
;
}
int main()
{
int i = 0, j = 0;
if (i && (j = i + 10))
//do something
;
}
- a)0
- b)10
- c)Depends on the compiler
- d)Depends on language standard
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
What will be the final value of j in the following C code? #include &...
Understanding the C Code
The given C code includes a conditional statement that utilizes logical operators and assignment. Here's a breakdown of how it works:
Initial Variables
- The variables `i` and `j` are both initialized to `0`.
Conditional Statement Analysis
- The condition `if (i && (j = i + 10))` is evaluated:
- The logical AND operator (`&&`) checks if both operands are true.
- The first operand is `i`, which is `0`. In C, `0` is considered `false`.
Short-Circuit Evaluation
- Since the first operand (`i`) is `false`, the second operand `(j = i + 10)` is not evaluated due to short-circuit evaluation.
- This means `j` is never assigned a new value; it remains `0`.
Final Value of j
- Because of the short-circuit evaluation, the assignment `j = i + 10` does not occur.
- Therefore, the final value of `j` is `0`.
Conclusion
- The correct answer is option **A**: `0`. This is due to the behavior of the logical operators in C and the way they handle short-circuit evaluation. If `i` is false, the second part of the expression is skipped, leaving `j` unchanged.
The given C code includes a conditional statement that utilizes logical operators and assignment. Here's a breakdown of how it works:
Initial Variables
- The variables `i` and `j` are both initialized to `0`.
Conditional Statement Analysis
- The condition `if (i && (j = i + 10))` is evaluated:
- The logical AND operator (`&&`) checks if both operands are true.
- The first operand is `i`, which is `0`. In C, `0` is considered `false`.
Short-Circuit Evaluation
- Since the first operand (`i`) is `false`, the second operand `(j = i + 10)` is not evaluated due to short-circuit evaluation.
- This means `j` is never assigned a new value; it remains `0`.
Final Value of j
- Because of the short-circuit evaluation, the assignment `j = i + 10` does not occur.
- Therefore, the final value of `j` is `0`.
Conclusion
- The correct answer is option **A**: `0`. This is due to the behavior of the logical operators in C and the way they handle short-circuit evaluation. If `i` is false, the second part of the expression is skipped, leaving `j` unchanged.
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