A user-defined data type can be created with ___________
Answer: D
Correct answer: all of the mentioned
All three create user-defined types. struct aggregates members of different types; enum names a set of integer constants; and typedef gives a new name to an existing type, which is how struct node * becomes NODEPTR. A union belongs on the same list.
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String handling functions such as strcmp(), strcpy() etc can be used with enumerated types.
Answer: B
Correct answer: FALSE
An enumerated value is an integer, not a string, so passing one to strcmp or strcpy would pass a small integer where a char * is expected - a type error, and undefined behaviour if it compiled. The names exist only in the source; nothing of them survives into the program.
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Can we have enum containing enum with the same constant?
#include "stdio.h"
enum Yenum{
C, CPP, Java
};
enum Xenum{
C, CPP, Java, Yenum
};
int main()
{
enum Xenum var;
printf("%d",sizeof(var));
return 0;
}
Answer: A
Correct answer: Enum can contain enum but duplicate constants are not allowed
The two enumerations both declare C, CPP and Java, and enumerators occupy the ordinary identifier namespace at file scope - so the second set are redefinitions and the program will not compile. Naming one enumeration as a member of another is not the difficulty; the repeated constant names are.
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What is the output of this program?
#include <stdio.h>
struct result{
char sub[20];
int marks;
};
void main()
{
struct result res[] = { {"Maths",100},
{"Science",90},
{"English",85}
};
printf("%s ", res[1].sub);
printf("%d", (*(res+2)).marks);
}
Answer: B
Correct answer: Science 85
res[1].sub is the second element’s subject, Science. (*(res+2)).marks is the same as res[2].marks - the third element - which is 85. Both subscripts count from 0.
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What will be the output of the following code?
#include <stdio.h>
enum week { Mon, Tue, Wed };
int main()
{
enum week today;
today = Wed;
Wed = 5;
printf("%d", today);
return 0;
}Answer: A
Correct answer: error
Wed is an enumerator, a constant, so Wed = 5; is an assignment to something that cannot be assigned to. The compiler rejects it. Assigning to the variable, today = 5;, would be allowed.
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Consider the following C declaration
struct {
short s[5];
union {
float y;
long z;
}u;
}t;
Assume that the objects of the type short, float and long occupy 2 bytes, 4 bytes and 8 bytes, respectively. The memory requirement for variable t, ignoring alignment consideration, is
Answer: B
Correct answer: 18 bytes
The array contributes \(5 \times 2 = 10\) bytes, and the union takes the size of its largest member, \(\max(4, 8) = 8\). The structure sums them, giving 18 bytes. The union’s members overlap, which is why the float costs nothing extra beyond the long.
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Which of the following cannot be a structure member?
Answer: A
Correct answer: Function
A structure member must be an object with storage, and a function is not - so a function cannot be a member of a C structure. What can be a member is a pointer to a function, which is how C programs implement dispatch tables and something resembling methods. Arrays and nested structures are ordinary members.
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The size of a union is determined by the size of the
Answer: D
Correct answer: Biggest member in the union
A union’s members all occupy the same storage, so its size is that of its largest member - enough to hold whichever one is in use. A structure, by contrast, gives every member its own space and so is at least the sum of them.
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What is the correct syntax for accessing a member of the ith structure in an array of structures?
Assuming:
struct temp
{
int b;
}s[50];
Answer: D
Correct answer: s[i].b;
s is an array of structures, so subscript first to select the element and then use the dot to select its member: s[i].b. The other forms subscript the member or put the operators the wrong way round.
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Pick the best statement for the below program snippet:
struct {int a[2];} arr[] = {1,2};
Answer: C
Correct answer: No compile error and it’ll create array arr of 1 element. Each of the element of arr contain a struct field of int array of 2 elements. arr[0]. a[0] would be 1 and arr[0].a[1] would be 2.
Note the missing inner braces: {1,2} is a flat list, so the values are used to fill the aggregate in order. Both go into the first structure - a[0] = 1 and a[1] = 2 - and nothing is left for a second element, so arr has just one. Writing {{1},{2}} would have given two elements instead.
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