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What will be the output of the following C code?
#include <stdio.h>
#define a 10
int main()
{
const int a = 5;
printf("a = %d\n", a);
}
#define a 10
int main()
{
const int a = 5;
printf("a = %d\n", a);
}
- a)a = 5
- b)a = 10
- c)Compilation error
- d)Runtime error
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
What will be the output of the following C code? #include <stdio.h...
Explanation:
Constant Reassignment:
- In the given C code, there are two definitions of the variable 'a'.
- The first definition is done using a macro '#define a 10' which defines 'a' as 10.
- The second definition is a constant integer 'const int a = 5;' which defines 'a' as 5.
- Since 'const int a = 5;' is defined in the local scope of the main() function, it shadows the global definition 'define a 10'.
Printing Value:
- The printf statement inside the main() function prints the value of 'a'.
- Since 'const int a = 5;' is used in the local scope, the value of 'a' printed will be 5.
Compilation Error:
- The code will not give a compilation error as it is syntactically correct.
- However, the output will be 'a = 5' instead of 'a = 10' due to the reassignment of 'a' as a constant integer with a value of 5.
Therefore, the correct output of the code will be:
a = 5
Constant Reassignment:
- In the given C code, there are two definitions of the variable 'a'.
- The first definition is done using a macro '#define a 10' which defines 'a' as 10.
- The second definition is a constant integer 'const int a = 5;' which defines 'a' as 5.
- Since 'const int a = 5;' is defined in the local scope of the main() function, it shadows the global definition 'define a 10'.
Printing Value:
- The printf statement inside the main() function prints the value of 'a'.
- Since 'const int a = 5;' is used in the local scope, the value of 'a' printed will be 5.
Compilation Error:
- The code will not give a compilation error as it is syntactically correct.
- However, the output will be 'a = 5' instead of 'a = 10' due to the reassignment of 'a' as a constant integer with a value of 5.
Therefore, the correct output of the code will be:
a = 5
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Community Answer
What will be the output of the following C code? #include <stdio.h...
The #define substitutes a with 10 without leaving any identifier, which results in Compilation error.
Output:
$ cc pgm3.c
pgm3.c: In function ‘main’:
pgm3.c:5: error: expected identifier or ‘(’ before numeric constant
Output:
$ cc pgm3.c
pgm3.c: In function ‘main’:
pgm3.c:5: error: expected identifier or ‘(’ before numeric constant
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What will be the output of the following C code? #include <stdio.h> #define a 10 int main() { const int a = 5; printf("a = %d\n", a); }a)a = 5b)a = 10c)Compilation errord)Runtime errorCorrect answer is option 'C'. Can you explain this answer?
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