111)
void pascal f(int i,int j,int k)
{
printf(“%d
%d %d”,i, j, k);
}
void cdecl f(int
i,int j,int k)
{
printf(“%d
%d %d”,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf("
%d",i);
}
Answer:
10
11 12 13
12
11 10 13
Explanation:
Pascal
argument passing mechanism forces the arguments to be called from left to
right. cdecl is the normal C argument passing mechanism where the arguments are
passed from right to left.
112).
What is the output of the program given below
main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice
the semicolon at the end of the for loop. THe initial value of the i is set to
0. The inner loop executes to increment the value from 0 to 127 (the positive
range of char) and then it rotates to the negative value of -128. The condition
in the for loop fails and so comes out of the for loop. It prints the current
value of i that is -128.
113)
main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
infinite loop
Explanation
The
difference between the previous question and this one is that the char is
declared to be unsigned. So the i++ can never yield negative value and i>=0
never becomes false so that it can come out of the for loop.
114)
main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
Behavior is
implementation dependent.
Explanation:
The
detail if the char is signed/unsigned by default is implementation dependent.
If the implementation treats the char to be signed by default the program will
print -128 and terminate. On the other hand if it considers char to be unsigned
by default, it goes to infinite loop.
Rule:
You
can write programs that have implementation dependent behavior. But dont write
programs that depend on such behavior.
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 .
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.
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