Part1 | Part2 | ; Part3 | ; Part4
97) void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
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Explanation:
The int ret(int ret), ie., the function name and the argument name
can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is
passed, after the first expression the value in ret will be 6, as ret is
integer hence the value stored in ret will have implicit type conversion
from float to int. The ret is returned in main() it is printed after and
preincrement.
98) void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will
be counted from 0 till the null character. Hence the 'I' will hold the value
equal to 5, after the pre-increment in the printf statement, the 6 will be
printed.
99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence
printing a null character returns 1 which makes the if
statement true, thus "Ok here" is printed.
101) void main()
{
void *v;
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int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can
be done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such
pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later
point of time.
102) void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of
its declaration.
So expressions such as i = i++ are valid statements. The i, j and k
are automatic variables and so they contain some garbage value.
Garbage in is garbage out (GIGO).
103) void main()
{
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.
104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
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The inner printf executes first to print some garbage value. The
printf returns no of characters printed and this value also cannot be
predicted. Still the outer printf prints something and so returns a
non-zero value. So it encounters the break statement and comes
out of the while statement.
104) main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the
expression i-- >=0 will always be true, leading to an infinite loop.
105) #include
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition
reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.
106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
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Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10
108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i
becomes 0 it comes out of while loop. Due to post-increment on i
the value of i while printing is 1.
109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on
the expression and now the while loop is, while(i--!=0) which is
false and so breaks out of while loop. The value –1 is printed due
to the post-decrement operator.
113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2); printf("%lf\n",f%g); printf("%lf\n",fmod(f,g)); } Answer: Line no 5: Error: Lvalue required Line no 6: Cannot apply leftshift to float Line no 7: Cannot apply mod to float Explanation: Enumeration constants cannot be modified, so you cannot apply ++. Bit-wise operators and % operators cannot be applied on float values. fmod() is to find the modulus values for floats as % operator is for ints. http://www.exforsys.com Page 35 6/26/2004 35 110) main() { int i=10; void pascal f(int,int,int); f(i++,i++,i++); printf(" %d",i); } void pascal f(integer :i,integer:j,integer :k) { write(i,j,k); } Answer: Compiler error: unknown type integer Compiler error: undeclared function write Explanation: Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions. 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
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-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.
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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;
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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];
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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.
121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented
in memory”);
}
Answer
You can answer this if you know how values are represented
in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on
0 to produce all ones to fill the space for an integer. –1 is
represented in unsigned value as all 1’s and so both are
equal.
122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
arr2D
arr2D[1]
arr2D[2]
arr2D[3]
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printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help
understand this.
123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is
required.
125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++)
and ++*p++. Parenthesis just works as a visual clue for the
reader to see which expression is first evaluated.
126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
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}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions
that takes no arguments and returns the type int. By the
assignment ptr[0] = aaa; it means that the first function pointer in
the array is initialized with the address of the function aaa.
Similarly, the other two array elements also get initialized with the
addresses of the functions bbb and ccc. Since ptr[2] contains the
address of the function ccc, the call to the function ptr[2]() is same
as calling ccc(). So it results in printing "bye".
127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of
higher precedence than = operator. In the inner expression, ++i is
equal to 6 yielding true(1). Hence the result.
128) main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”.
Since this string becomes the format string for printf and ASCII
value of 65 is ‘A’, the same gets printed.
129) void ( * abc( int, void ( *def) () ) ) ();
Answer::
abc is a ptr to a function which takes 2 parameters .(a). an
integer variable.(b). a ptrto a funtion which returns void. the
return type of the function is void.
Explanation:
Apply the clock-wise rule to find the result.
130) main()
{
while (strcmp(“some”,”some\0”))
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printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference.
So “some” and “some\0” are equivalent. So, strcmp returns 0
(false) hence breaking out of the while loop.
131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not
appended automatically to the string. Since str1 doesn’t have null
termination, it treats whatever the values that are in the following
positions as part of the string until it randomly reaches a ‘\0’. So
str1 and str2 are not the same, hence the result.
132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}
Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and
hence it cannot appear on the left hand side of an
assignment operation.
133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas
calloc returns the allocated memory space initialized to zeros.
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134) void main()
{
static int i;
while(i<=10) (i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the
conditional operator evaluates to false, executing i--. This
continues till the integer value rotates to positive value (32767).
The while condition becomes false and hence, comes out of the
while loop, printing the i value.
135) main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}
Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else
statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant
char )
*a='F' : illegal
a="Hi" : legal
2. 'const' applies to 'a' rather than to the value of a (constant
pointer to char )
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*a='F' : legal
a="Hi" : illegal
3. Same as 1.
137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner
expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is
1. Hence the result.
138) main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}
Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth
value of the expression is not known. j is not equal to zero itself
means that the expression’s truth value is 1. Because it is followed
by || and true || (anything) => true where (anything) will not be
evaluated. So the remaining expression is not evaluated and so the
value of i remains the same.
Similarly when && operator is involved in an expression, when any
of the operands become false, the whole expression’s truth value
becomes false and hence the remaining expression will not be
evaluated.
false && (anything) => false where (anything) will not be
evaluated.
139) main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register
variables.
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140) main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.
141) main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern
has no use in resolving it.
142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes
through f1 and f2 ultimately affects only the value of a.
143) main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d",
sizeof(p), sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
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46
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the
array and it is not the same as the sizeof the pointer variable. Here
the sizeof(a) where a is the character array and the size of the
array is 5 because the space necessary for the terminating NULL
character should also be taken into account.
144) #define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The
macro expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) /
sizeof(int) => 10.
145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the
pointers can be passed. So the argument is equivalent to int *
array (this is one of the very few places where [] and * usage are
equivalent). The return statement becomes, sizeof(int *)/
sizeof(int) that happens to be equal in this case.
146) main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++) { for(j=0;j<3;j++) printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j), http://www.exforsys.com Page 47 6/26/2004 47 *(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i)); } } Answer: 1 1 1 1 2 4 2 4 3 7 3 7 4 2 4 2 5 5 5 5 6 8 6 8 7 3 7 3 8 6 8 6 9 9 9 9 Explanation: *(*(p+i)+j) is equivalent to p[i][j]. 147) main() { void swap(); int x=10,y=8; swap(&x,&y); printf("x=%d y=%d",x,y); } void swap(int *a, int *b) { *a ^= *b, *b ^= *a, *a ^= *b; } Answer: x=10 y=8 Explanation: Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement. Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments. This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows, void swap() int *a, int *b { *a ^= *b, *b ^= *a, *a ^= *b; } where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments. 148) main() { int i = 257; int *iPtr = &i; printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) ); http://www.exforsys.com Page 48 6/26/2004 48 } Answer: 1 1 Explanation: The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed. 149) main() { int i = 258; int *iPtr = &i; printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) ); } Answer: 2 1 Explanation: The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010. 150) main() { int i=300; char *ptr = &i; *++ptr=2; printf("%d",i); } Answer: 556 Explanation: The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.
Part1 | Part2 | ; Part3 | ; Part4
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