Which is not a proper way of array declaration ?
int a(25) ;
C uses square brackets for arrays, never parentheses, so int a(25); is the improper declaration — option 2.
What the compiler makes of it. Parentheses after an identifier mean a function, not an array. int a(25); is therefore parsed as a function declaration, and since 25 is a constant rather than a type name it is rejected as a syntax error. The mistake usually comes from BASIC or FORTRAN, where A(25) is the normal way to subscript an array.
| Declaration | Verdict |
|---|---|
int num[6] = {…6 values…} | ✓ Size stated, and the initialiser matches |
int a(25); | ✗ Parentheses are for functions |
int n[] = {…} | ✓ Size inferred from the initialiser |
float press[] = {…} | ✓ Same rule; negative values are fine |
Why options 3 and 4 are legal despite the empty brackets. When an array is declared with an initialiser, the size may be omitted and the compiler counts the elements — six in option 3, four in option 4. This is not merely permitted but preferred, since it cannot go out of step with the list. Writing the size explicitly and then changing the initialiser is the classic way to create an off-by-one bug; using sizeof(n)/sizeof(n[0]) to obtain the count keeps everything consistent.
The empty brackets are allowed only with an initialiser. int n[]; on its own is incomplete — the compiler has no way to know how much storage to reserve. That is the one restriction on the form.
What the correct declaration establishes. int num[6] reserves six contiguous int-sized slots, indexed 0 to 5. C performs no bounds checking whatever, so num[6] compiles happily and reads whatever memory lies beyond the array — the source of a large share of real-world security defects. The array name itself decays to a pointer to the first element in almost every expression, which is why an array passed to a function arrives as a pointer and its size must be passed alongside.
Hence, the improper declaration is int a(25);
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