31. Why doesn't the following statement work?
char str[ ] = "Hello" ;
strcat ( str, '!' ) ;
Answer : The string function strcat( ) concatenates strings and not a character. The basic difference between a string and a character is that a string is a collection of characters, represented by an array of characters whereas a character is a single character. To make the above statement work writes the statement as shown below:
strcat ( str, "!" ) ;
32. What is the benefit of using #define to declare a constant?
Using the #define method of declaring a constant enables you to declare a constant in one place and use it throughout your program. This helps make your programs more maintainable, because you need to maintain only the #define statement and not several instances of individual constants throughout your program.
For instance, if your program used the value of pi (approximately 3.14159) several times, you might want to declare a constant for pi as follows: #define PI 3.14159
Using the #define method of declaring a constant is probably the most familiar way of declaring constants to traditional C programmers. Besides being the most common method of declaring constants, it also takes up the least memory.
Constants defined in this manner are simply placed directly into your source code, with no variable space allocated in memory. Unfortunately, this is one reason why most debuggers cannot inspect constants created using the #define method.
33. What is the purpose of main( ) function ?
The function main( ) invokes other functions within it.It is the first function to be called when the program starts execution.
· It is the starting function
· It returns an int value to the environment that called the program
· Recursive call is allowed for main( ) also.
· It is a user-defined function
· Program execution ends when the closing brace of the function main( ) is reached.
· It has two arguments 1)argument count and 2) argument vector (represents strings passed).
· Any user-defined name can also be used as parameters for main( ) instead of argc and argv
34. How can I search for data in a linked list?
Unfortunately, the only way to search a linked list is with a linear search, because the only way a linked list’s members can be accessed is sequentially.
Sometimes it is quicker to take the data from a linked list and store it in a different data structure so that searches can be more efficient.
35. Why should we assign NULL to the elements (pointer) after freeing them?
This is paranoia based on long experience. After a pointer has been freed, you can no longer use the pointed-to data. The pointer is said to dangle; it doesn’t point at anything useful.
If you NULL out or zero out a pointer immediately after freeing it, your program can no longer get in trouble by using that pointer. True, you might go indirect on the null pointer instead, but that’s something your debugger might be able to help you with immediately.
Also, there still might be copies of the pointer that refer to the memory that has been deallocated; that’s the nature of C. Zeroing out pointers after freeing them won’t solve all problems.
36. What is a null pointer assignment error? What are bus errors, memory faults, and core dumps?
These are all serious errors, symptoms of a wild pointer or subscript. Null pointer assignment is a message you might get when an MS-DOS program finishes executing. Some such programs can arrange for a small amount of memory to be available “where the NULL pointer points to (so to speak). If the program tries to write to that area, it will overwrite the data put there by the compiler.
When the program is done, code generated by the compiler examines that area. If that data has been changed, the compiler-generated code complains with null pointer assignment. This message carries only enough information to get you worried. There’s no way to tell, just from a null pointer assignment message, what part of your program is responsible for the error. Some debuggers, and some compilers, can give you more help in finding the problem.
Bus error: core dumped and Memory fault: core dumped are messages you might see from a program running under UNIX. They’re more programmer friendly. Both mean that a pointer or an array subscript was wildly out of bounds. You can get these messages on a read or on a write. They aren’t restricted to null pointer problems. The core dumped part of the message is telling you about a file, called core, that has just been written in your current directory. This is a dump of everything on the stack and in the heap at the time the program was running. With the help of a debugger, you can use the core dump to find where the bad pointer was used. That might not tell you why the pointer was bad, but it’s a step in the right direction. If you don’t have write permission in the current directory, you won’t get a core file, or the core dumped message
37. Predict the output or error(s) for the following programmes:
void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer : Compiler error: Cannot modify a constant value.
Explanation: p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
38. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer :
mmm
aaaa
nnnn
Explanation: s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/ displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
39. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer : I hate U
Explanation: For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb: Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .
40. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer : 5 4 3 2 1
Explanation: When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.
41. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer : 2 2 2 2 2 2 3 4 6 5
Explanation: Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.
42. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer : Linker Error : Undefined symbol '_i'
Explanation: extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .
43. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer : 0 0 1 3 1
Explanation: Logical operations always give a result of 1 or 0. And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.
44. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}
Answer : 1 2
Explanation: The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.
45. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer : Three
Explanation: The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.
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char str[ ] = "Hello" ;
strcat ( str, '!' ) ;
Answer : The string function strcat( ) concatenates strings and not a character. The basic difference between a string and a character is that a string is a collection of characters, represented by an array of characters whereas a character is a single character. To make the above statement work writes the statement as shown below:
strcat ( str, "!" ) ;
32. What is the benefit of using #define to declare a constant?
Using the #define method of declaring a constant enables you to declare a constant in one place and use it throughout your program. This helps make your programs more maintainable, because you need to maintain only the #define statement and not several instances of individual constants throughout your program.
For instance, if your program used the value of pi (approximately 3.14159) several times, you might want to declare a constant for pi as follows: #define PI 3.14159
Using the #define method of declaring a constant is probably the most familiar way of declaring constants to traditional C programmers. Besides being the most common method of declaring constants, it also takes up the least memory.
Constants defined in this manner are simply placed directly into your source code, with no variable space allocated in memory. Unfortunately, this is one reason why most debuggers cannot inspect constants created using the #define method.
33. What is the purpose of main( ) function ?
The function main( ) invokes other functions within it.It is the first function to be called when the program starts execution.
· It is the starting function
· It returns an int value to the environment that called the program
· Recursive call is allowed for main( ) also.
· It is a user-defined function
· Program execution ends when the closing brace of the function main( ) is reached.
· It has two arguments 1)argument count and 2) argument vector (represents strings passed).
· Any user-defined name can also be used as parameters for main( ) instead of argc and argv
34. How can I search for data in a linked list?
Unfortunately, the only way to search a linked list is with a linear search, because the only way a linked list’s members can be accessed is sequentially.
Sometimes it is quicker to take the data from a linked list and store it in a different data structure so that searches can be more efficient.
35. Why should we assign NULL to the elements (pointer) after freeing them?
This is paranoia based on long experience. After a pointer has been freed, you can no longer use the pointed-to data. The pointer is said to dangle; it doesn’t point at anything useful.
If you NULL out or zero out a pointer immediately after freeing it, your program can no longer get in trouble by using that pointer. True, you might go indirect on the null pointer instead, but that’s something your debugger might be able to help you with immediately.
Also, there still might be copies of the pointer that refer to the memory that has been deallocated; that’s the nature of C. Zeroing out pointers after freeing them won’t solve all problems.
36. What is a null pointer assignment error? What are bus errors, memory faults, and core dumps?
These are all serious errors, symptoms of a wild pointer or subscript. Null pointer assignment is a message you might get when an MS-DOS program finishes executing. Some such programs can arrange for a small amount of memory to be available “where the NULL pointer points to (so to speak). If the program tries to write to that area, it will overwrite the data put there by the compiler.
When the program is done, code generated by the compiler examines that area. If that data has been changed, the compiler-generated code complains with null pointer assignment. This message carries only enough information to get you worried. There’s no way to tell, just from a null pointer assignment message, what part of your program is responsible for the error. Some debuggers, and some compilers, can give you more help in finding the problem.
Bus error: core dumped and Memory fault: core dumped are messages you might see from a program running under UNIX. They’re more programmer friendly. Both mean that a pointer or an array subscript was wildly out of bounds. You can get these messages on a read or on a write. They aren’t restricted to null pointer problems. The core dumped part of the message is telling you about a file, called core, that has just been written in your current directory. This is a dump of everything on the stack and in the heap at the time the program was running. With the help of a debugger, you can use the core dump to find where the bad pointer was used. That might not tell you why the pointer was bad, but it’s a step in the right direction. If you don’t have write permission in the current directory, you won’t get a core file, or the core dumped message
37. Predict the output or error(s) for the following programmes:
void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer : Compiler error: Cannot modify a constant value.
Explanation: p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".
38. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer :
mmm
aaaa
nnnn
Explanation: s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/ displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
39. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer : I hate U
Explanation: For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb: Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .
40. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer : 5 4 3 2 1
Explanation: When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.
41. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer : 2 2 2 2 2 2 3 4 6 5
Explanation: Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.
42. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer : Linker Error : Undefined symbol '_i'
Explanation: extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .
43. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer : 0 0 1 3 1
Explanation: Logical operations always give a result of 1 or 0. And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.
44. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}
Answer : 1 2
Explanation: The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.
45. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer : Three
Explanation: The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.