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115) Is the following statement a declaration/definition.
Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.
116). What is the output for the program given below
typedef enum errorType{warning, error,
exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One means that it is a enumerator
constant with value 1. The another use is that it is a type name (due to
typedef) for enum errorType. Given a situation the compiler cannot distinguish
the meaning of error to know in what sense the error is used:
error
g1;
g1=error;
// which
error it refers in each case?
When the compiler can distinguish between usages then it will not issue
error (in pure technical terms, names can only be overloaded in different
namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer’s
convenience.
117) typedef struct error{int
warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at
any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as
in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by
the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct
keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.
Note
This code is given here to just explain the concept behind. In real programming
don’t use such overloading of names. It reduces the readability of
the code. Possible doesn’t mean that we should use it!
118) #ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something
is not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.
119) #if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
0 0
Explanation
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.
120). What is the output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==*
arr2D)&&(* arr2D == arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N dimensional
arrays are made up of (N-1) dimensional arrays.
arr2D
is made up of a 3 single arrays that contains 3 integers each .
arr2D
arr2D[1]
The name arr2D refers to the beginning of all the 3 arrays. *arr2D
refers to the start of the first 1D array (of 3 integers) that is the same
address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t
change the value/meaning. Again arr2D[0] is the another way of telling *(arr2D
+ 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is true(1) and the same is printed.
C Aptitude questions : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18