Which of the following statements correct about k used in the below statement?char ****k;a.k is a pointer to a pointer to a pointer to a charb.k is a pointer to a pointer to a pointer to a pointer to a charc.k is a pointer to a char pointerd.k is a pointer to a pointer to a char
Question
Which of the following statements correct about k used in the below statement?char ****k;a.k is a pointer to a pointer to a pointer to a charb.k is a pointer to a pointer to a pointer to a pointer to a charc.k is a pointer to a char pointerd.k is a pointer to a pointer to a char
Solution
To determine which statement is correct about the variable "k" in the given code, we need to analyze the declaration "char ****k;".
Let's break down the declaration step by step:
- "char ****k;" - This line declares a variable named "k" of type "char ****".
Now, let's analyze each statement and compare it with the declaration:
a. "k is a pointer to a pointer to a pointer to a char" - This statement matches the declaration "char ****k;", as "k" is indeed a pointer to a pointer to a pointer to a char. Therefore, statement a is correct.
b. "k is a pointer to a pointer to a pointer to a pointer to a char" - This statement does not match the declaration "char ****k;". The declaration specifies that "k" is a pointer to a pointer to a pointer to a char, not a pointer to a pointer to a pointer to a pointer to a char. Therefore, statement b is incorrect.
c. "k is a pointer to a char pointer" - This statement does not match the declaration "char ****k;". The declaration specifies that "k" is a pointer to a pointer to a pointer to a char, not a pointer to a char pointer. Therefore, statement c is incorrect.
d. "k is a pointer to a pointer to a char" - This statement matches the declaration "char ****k;", as "k" is indeed a pointer to a pointer to a char. Therefore, statement d is correct.
In conclusion, the correct statements about "k" are:
- Statement a: "k is a pointer to a pointer to a pointer to a char"
- Statement d: "k is a pointer to a pointer to a char"
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