Predict the output or
error(s) for the following:
81. Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are allowed
inside the [] is just for more readability. So there is no difference between
the two declarations.
82. What is the subtle error in the following
code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++
p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no address. So p remains
NULL where *p =0 may result in problem (may rise to runtime error “NULL pointer
assignment” and terminate the program).
83. What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs. The check s !=
NULL is for error/exception handling and for that assert shouldn’t be used. A
plain if and the corresponding remedy statement has to be given.
84. What is the hidden bug with the following
statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release version. In assert,
the experssion involves side-effects. So the behavior of the code becomes
different in case of debug version and the release version thus leading to a
subtle bug.
Rule to Remember:
Don’t use expressions that have side-effects in assert statements.
85. void main()
{
int *i = 0x400; // i points to the address 400
*i = 0;
// set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because it points to some
location whose value may not be available for modification. This type of
pointer in which the non-availability of the implementation of the referenced
location is known as 'incomplete type'.
86. #define assert(cond)
if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort())
void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else for if in the assert
macro. Hence nothing is printed.
The solution is to use conditional operator instead of if statement,
#define assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s,
file %s, line %d \n",#cond, __FILE__,__LINE__), abort()))
Note:
However this problem of “matching with nearest else” cannot be solved by the
usual method of placing the if statement inside a block like this,
#define assert(cond) { \
if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort()) \
}
87. Is the following code legal?
struct a
{
int x;
struct a b;
}
Answer:
No
Explanation:
Is it not legal for a structure to contain a member that is of the same
type as in this case. Because this will cause the structure declaration to be
recursive without end.
88. Is the following code legal?
struct a
{
int x;
struct a *b;
}
Answer:
Yes.
Explanation:
*b is a pointer to type struct a and so is legal. The compiler knows, the size
of the pointer to a structure even before the size of the structure
is determined(as you know the pointer to any type is of same size). This type of
structures is known as ‘self-referencing’ structure.
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