What is the output of the following code?1234567891011#include <stdio.h>int main() { char c = 'K'; char* ptr = &c; (*ptr)++; printf("%c ", *ptr); (*ptr)--; printf("%c", *ptr); return 0;}
Question
What is the output of the following code?1234567891011#include <stdio.h>int main() { char c = 'K'; char* ptr = &c; (*ptr)++; printf("%c ", *ptr); (*ptr)--; printf("%c", *ptr); return 0;}
Solution
The output of the code will be "L K".
Here's the step by step explanation:
- The character variable 'c' is initialized with the value 'K'.
- A pointer 'ptr' is created that points to the address of 'c'.
- The value at the address that 'ptr' points to (which is 'c') is incremented by 1. Since 'c' was 'K', after incrementing it becomes 'L'.
- The new value of 'c' ('L') is printed.
- The value at the address that 'ptr' points to is decremented by 1, so it becomes 'K' again.
- The value of 'c' ('K') is printed again.
So, the output is "L K".
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