C declarations: interpret the following declaration char *scr;

Introduction / Context:This item tests basic pointer declarations in C. The expression char *name; is one of the most common idioms and denotes a pointer that can reference a character or the first element of a character array representing a C string.

Given Data / Assumptions:

• Declaration under analysis: char *scr;
• No qualifiers (const, volatile) or storage specifiers are present.
• We are only classifying the type, not its initialization or lifetime.

Concept / Approach:

In C, the * binds to the identifier as part of the declarator, so char *scr; defines scr as “pointer to char”. This pointer can hold the address of a single char object or the first element of a char buffer that is terminated with '\\0' to represent a string. The declaration does not allocate storage for characters; it only allocates space for the pointer itself.

Step-by-Step Solution:

Verification / Alternative check:

sizeof(scr) equals the size of a pointer on the platform, not the size of a character buffer. Dereferencing *scr yields type char, which confirms the one-level indirection.

Why Other Options Are Wrong:

Pointer to pointer — Would require char **scr.

Function pointer — No function declarator () exists.

Member of function pointer — Not meaningful terminology in C type system.

Common Pitfalls:

• Assuming char *scr allocates storage for a full string; it does not allocate the characters themselves.
• Forgetting const when pointing to string literals to avoid modifying read-only storage.

Final Answer:

scr is a pointer to char.