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What will be the output of the following C code?
#include <stdio.h>
    void foo(const int *);
    int main()
    {
        const int i = 10;
        printf("%d ", i);
        foo(&i);
        printf("%d", i);
 
    }
    void foo(const int *i)
    {
        *i = 20;
    }
  • a)
    Compile time error
  • b)
    10 20
  • c)
    Undefined value
  • d)
    10
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What will be the output of the following C code?#include <stdio.h&g...
Cannot change a const type value.
Output:
$ cc pgm1.c
pgm1.c: In function ‘foo’:
pgm1.c:13: error: assignment of read-only location ‘*i’
Free Test
Community Answer
What will be the output of the following C code?#include <stdio.h&g...
Explanation:
- Compile time error: The code will throw a compile time error because of the attempt to modify a constant variable `i` inside the `foo` function.
- const int i = 10; This line declares a constant integer `i` with a value of 10.
- printf("%d ", i); This will print the value of `i`, which is 10.
- foo(&i); The address of `i` is passed to the `foo` function which expects a pointer to a constant integer.
- void foo(const int *i) This function takes a constant integer pointer as an argument.
- *i = 20; This line tries to change the value of `i` to 20 which is not allowed since `i` is a constant variable.
- printf("%d", i); This line will print the value of `i` which remains 10 since the `foo` function was not able to modify it.
Therefore, the code will not compile due to the attempt to modify a constant variable, resulting in a compile time error.
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What will be the output of the following C code?#include <stdio.h> void foo(const int *); int main() { const int i = 10; printf("%d ", i); foo(&i); printf("%d", i); } void foo(const int *i) { *i = 20; }a)Compile time errorb)10 20c)Undefined valued)10Correct answer is option 'A'. Can you explain this answer?
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