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What will be output of the following c code? ( according to GCC compiler)
#include<stdio.h>
int main()
{
signed x;
unsigned y;
x = 10 +- 10u + 10u +- 10;
y = x;
if(x==y) printf("%d %d",x,y);
else if(x!=y) printf("%u %u",x,y);
return 0;
}
  • a)
    0 0
  • b)
    65536 -10
  • c)
    0 65536
  • d)
    Compilation error
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
What will be output of the following c code? ( according to GCC compil...
Explanation: Consider on the expression:
x = 10 +- 10u + 10u +- 10;
10: It is signed integer constant.
10u: It is unsigned integer constant.
X: It is signed integer variable.
As we know operators enjoy higher precedence than binary operators. So
x = 10 + (-10u) + 10u + (-10);
= 10 + -10 + 10 + (-10);
= 0
So, Corresponding signed value of unsigned 10u is +10.
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Most Upvoted Answer
What will be output of the following c code? ( according to GCC compil...
Understanding the Code
The code snippet provided involves arithmetic operations between signed and unsigned integers. Let's break it down.
Code Breakdown
- The signed variable `x` is initialized and will hold the result of the expression `10 +- 10u + 10u +- 10`.
- The unsigned variable `y` will store the value of `x`.
Expression Evaluation
- The expression `10 +- 10u + 10u +- 10` can be simplified:
- `10 + (-10u) + 10u + (-10)`
- The `10u` (unsigned) is treated as `10` in terms of value but has an unsigned representation.
- The signed integer `10` and `-10` lead to:
- `10 - 10 = 0`.
- Now, evaluating `0 + 10u + (-10)` (with `10u` being treated as unsigned) results in:
- `0 + 10 + (-10) = 0`.
Assigning and Comparing
- After evaluating the expression, `x = 0` and `y = x` implies `y = 0`.
- The comparison `x == y` will evaluate to `true` since both are `0`.
Output Statement
- Since `x == y`, the code will execute the first `printf` statement:
- `printf("%d %d", x, y);` outputs `0 0`.
Conclusion
Thus, the final output of the program is `0 0`, confirming that option 'A' is correct. The signed and unsigned integer behaviors alongside the arithmetic operation rules lead to this conclusion.
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What will be output of the following c code? ( according to GCC compiler)#include<stdio.h>int main(){signed x;unsigned y;x = 10 +- 10u + 10u +- 10;y = x;if(x==y) printf("%d %d",x,y);else if(x!=y) printf("%u %u",x,y);return 0;}a)0 0b)65536 -10c)0 65536d)Compilation errorCorrect answer is option 'A'. Can you explain this answer?
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