Flexible storage for names: Will the following C code allocate enough space for a variable-length name and print it correctly? #include #include struct emp { int len; char name[1]; }; int main() { char newname[] = "Rahul"; struct emp *p = (struct emp *) malloc(sizeof(struct emp) - 1 + strlen(newname) + 1); p->len = strlen(newname); strcpy(p->name, newname); printf("%d %s ", p->len, p->name); return 0; }

C Programming Structures, Unions, Enums Difficulty: Medium
Choose an option
  • A
    Yes on many compilers (the classic “struct hack”), but the C99+ idiom is to use char name[] as a flexible array member.
  • B
    No; this never works on any conforming implementation.
  • C
    Yes, but only if malloc.h is included; stdlib.h will break it.
  • D
    Yes, provided printf uses %ls for the name.
  • E
    No; because name[1] can only hold a single character.

Answer

Correct Answer: Yes on many compilers (the classic “struct hack”), but the C99+ idiom is to use char name[] as a flexible array member.

Explanation

Introduction / Context: This code demonstrates the pre-C99 “struct hack,” where a trailing one-element array is over-allocated to store variable-length data. Modern C (C99 and later) provides flexible array members, which are safer and standard-conforming.

Given Data / Assumptions:

  • Struct contains `int len;` and `char name[1];`.
  • We allocate `sizeof(struct emp) - 1 + strlen(newname) + 1` bytes.
  • We copy the string and print length and contents.

Concept / Approach: The allocation subtracts the initial 1 byte already counted in `sizeof(struct emp)` and then adds the desired string length plus the null terminator, effectively creating room for the whole string inline after the struct header. This works on many compilers, but the standard approach is: struct emp { int len; char name[]; }; p = malloc(sizeof *p + strlen(newname) + 1); This avoids undefined-behavior concerns related to accessing beyond the declared array bounds and is well supported by C99+.

Step-by-Step Solution: 1) Compute needed bytes: header + string length + 1 for terminator. 2) Allocate and set `p->len`. 3) Copy the string: `strcpy(p->name, newname);`. 4) Print: `printf("%d %s\", p->len, p->name);`. 5) Prefer modern pattern with `char name[]` flexible member and `#include <stdlib.h>` for malloc/free.

Verification / Alternative check: Compiles and runs on many systems; replacing with a flexible array member preserves intent and conformance while keeping logic and output identical.

Why Other Options Are Wrong: Option B: It does work widely; the risk is standard conformance, not universal failure.
Option C: `malloc.h` is obsolete; `stdlib.h` is correct and portable.
Option D: `%ls` is for wide strings, which is unrelated.
Option E: The over-allocation provides more space than the nominal `name[1]` declaration.

Common Pitfalls: Forgetting to add space for the null terminator; using `malloc.h` instead of `stdlib.h`; not checking the return of `malloc`; assuming strict aliasing or padding rules that may differ across platforms.

Final Answer: Yes on many compilers, but prefer the C99 flexible array member idiom for portable, standard-conforming code.

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