What will be the output of the following C code?#includeint main(){int p = 1, q = 2, r = 3, s = 4, x;e = r + s = q * p;printf(“%d, %d ”, x, s);}a)Syntax errorb)5, 2c)7, 2d)7, 4Correct answer is option 'A'. Can you explain this answer?

Question

What will be the output of the following C code?#include<stdio.h>int main(){int p = 1, q = 2, r = 3, s = 4, x;e = r + s = q * p;printf(&ldquo;%d, %d
&rdquo;, x, s);}a)Syntax errorb)5, 2c)7, 2d)7, 4Correct answer is option 'A'. Can you explain this answer?

Answer

Explanation:

- The code provided has a syntax error in the line where the variable 'e' is being assigned a value.
- The expression 'e = r + s = q * p;' is not a valid assignment statement in C.
- In C, assignment is done from right to left, so the expression 'q * p' will be evaluated first and then added to 'r + s'.
- However, the statement 'e = r + s = q * p;' is trying to assign a value to both 'e' and 's' simultaneously, which is not allowed in C syntax.

Therefore, the correct answer is option 'A' which states that there is a syntax error in the code provided.

Answer

The code provided has a syntax error. The line e = r + s = q * p; is invalid in C programming. It is not possible to assign values to multiple variables in a single line like this.

The correct way to assign values to multiple variables would be:
int p = 1, q = 2, r = 3, s = 4, x;
x = r + s;
s = q * p;
printf("%d, %d ", x, s);
However, even with this correction, the output of the code would be 7, 2, not 7, 4. This is because x is uninitialized, and its value is unpredictable. The value of x would depend on the initial state of the memory where x is stored.

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