aptitude c programs with answers

main(int argc, char **argv)
{
 printf("enter the character");
 getchar();
 sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
 return num1+num2;
}

Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

82)

	# include <stdio.h>
int one_d[]={1,2,3};
main()
{
 int *ptr; 
 ptr=one_d;
 ptr+=3;
 printf("%d",*ptr);
}

Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.

83)

	# include<stdio.h>
aaa() {
  printf("hi");
 }
bbb(){
 printf("hello");
 }
ccc(){
 printf("bye");
 }
main()
{
  int (*ptr[3])();
  ptr[0]=aaa;
  ptr[1]=bbb;
  ptr[2]=ccc;
  ptr[2]();
}

Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

85)

	#include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}

Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.

86)

	main()
{
 int i =0;j=0;
 if(i && j++)
   	printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}

Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.

87)

	main()
{
 int i;
 i = abc();
 printf("%d",i);
}
abc()
{
 _AX = 1000;
}

Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

88)

	int i;
        	main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
             	printf("%d--",t--);
            	}
	// If the inputs are 0,1,2,3 find the o/p

Answer:
4–0
3–1
2–2
Explanation:
Let us assume some x= scanf(“%d”,&i)-t the values during execution
will be,
t i x
4 0 -4
3 1 -2
2 2 0

89)

	main(){
  int a= 0;int b = 20;char x =1;char y =10;
  if(a,b,x,y)
        printf("hello");
 }

Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, “hello” will be printed.

90)

	main(){
 unsigned int i;
 for(i=1;i>-2;i--)
        		printf("c aptitude");
}

Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn’t match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

91) In the following pgm add a stmt in the function fun such that the address of
‘a’ gets stored in ‘j’.

main(){
  int * j;
  void fun(int **);
  fun(&j);
 }
 void fun(int **k) {
  int a =0;
  /* add a stmt here*/
 }

Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.

92) What are the following notations of defining functions known as?

i.      int abc(int a,float b)
        		{
                	/* some code */
 }
ii.    int abc(a,b)
        int a; float b;
        		{
          		/* some code*/
        		}

Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation

93)

	main()
{
char *p;
p="%d\n";
           	p++;
           	p++;
           	printf(p-2,300);
}

Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to ‘%d\n’ and 300 is printed.

94)

	main(){
 char a[100];
 a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
 abc(a);
}
abc(char a[]){
 a++; 
   	 printf("%c",*a);
 a++;
 printf("%c",*a);
}

Explanation:
The base address is modified only in function and as a result a points to ‘b’ then after incrementing to ‘c’ so bc will be printed.

95)

	func(a,b)
int a,b;
{
 return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
 }

Answer:
The value if process is 0 !
Explanation:
The function ‘process’ has 3 parameters – 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function ‘func’. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function ‘process’.

96)

	void main()
{
	static int i=5;
	if(--i){
		main();
		printf("%d ",i);
	}
}

Answer:
0 0 0 0
Explanation:
The variable “I” is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

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
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;
	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:
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<conio.h>
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
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.

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.

If you have questions, please ask below

4 Comments


  1. priyanka says:

    please help me to prepare C and C++ for my forth coming interviews. I am a begginer in those feilds.

  2. Anu says:

    i need learn from the begining of c so pls guide me

  3. nik says:

    C Program for finding the Multiplication of Two INT no. of 40 digits.

  4. maliga says:

    additon program in c with only two end of statements

Leave a Reply

If you have any questions headover to our forums

You can use these XHTML tags: <a href="" title=""> <abbr title=""> <acronym title=""> <blockquote cite=""> <code> <em> <strong>