What are the Types of Arrays?
Answer: D
Correct answer: All the above
An array can have elements of any complete type: the built-in types int, float, char, double and long, and user-defined ones such as structures and enumerations. There is no restriction to primitives, so all the options qualify.
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What is Array?
Answer: B
Correct answer: Collection of similar type of elements
An array is a collection of elements of the same type, stored consecutively and reached by index. The uniform element size is what makes the index arithmetic work; a collection of mixed types is a structure, not an array.
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The information about an array used in a program will be stored in
Answer: D
Correct answer: Dope vector
The dope vector is the descriptor an implementation keeps for an array: base address, element size, number of dimensions and the bounds of each. It is what address computation and bounds checking work from. A symbol table holds compile-time names and types; an activation record holds one call’s locals.
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The const feature can be applied to
Answer: D
Correct answer: All of these
const can qualify an ordinary identifier (const int n = 10;), an array (const char msg[] = "hi";), and an array argument (void f(const int a[])). The last is the most useful of the three: it documents, and lets the compiler enforce, that the function will not modify the caller’s data.
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In C, the name of an array used in an expression evaluates to ________.
Answer: B
Correct answer: the address of the first element of the array
An array name decays to a pointer to its first element in almost every expression, which is why passing an array to a function passes only that address. The exceptions are the operands of sizeof and of the address-of operator.
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Pick the correct answers
if x is an one dimensional array, then
Answer: D
Correct answer: both (b) & (c)
(c) is the definition of subscripting: x[i] is *(x + i). (b) follows from it - &x[i] is x + i, so dereferencing it gives the same element back. (a) is false: &x[i] equals x + i, not x + i - 1, which would be the previous element.
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Are the expressions arr and &arr same for an array of 10 integers?
Answer: B
Correct answer: No
They hold the same address but they are not the same expression, because their types differ: arr is int * and &arr is int (*)[10]. The difference shows the moment you do arithmetic - arr+1 advances 4 bytes, &arr+1 advances 40.
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Predict the output of below code:
int main()
{
char a[2][3][3] = {'g','e','e','k','s','f','o',
'r','g','e','e','k','s'};
printf("%s ", **a);
getchar();
return 0;
}
Answer: C
Correct answer: geeksforgeeks
a is char[2][3][3] - 18 bytes in one contiguous block. The 13 characters of the initialiser fill it in order and the remaining 5 bytes are set to zero, because a partly initialised array has the rest zero-filled. **a is the address of the first byte, and %s prints from there to the first NUL - which is byte 14. So the whole run of 13 characters comes out: geeksforgeeks.
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Let x be an array. Which of the following are illegal
Answer: A
Correct answer: x++
An array name is not a modifiable lvalue - it names fixed storage, so it cannot be assigned to or incremented, and x++ is rejected. x+1 is fine: the name decays to a pointer and pointer arithmetic gives the address of the second element. x[2] is ordinary subscripting.
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Size of the array need not be specified, when
Answer: A
Correct answer: All of the above
The size may be omitted in all three cases. In a definition with an initialiser, int a[] = {1,2,3};, the compiler counts the initialisers. In a formal parameter, void f(int a[]), the array decays to a pointer and the size is meaningless. And in a declaration, extern int a[];, the definition elsewhere supplies the size.
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