C Aptitude Questions and answers with explanation

C Aptitude 21
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 21.1

 char *Fun1()
{
char temp[ ] = “example";
return temp;
}
char *Fun2()
{
char temp[ ] = {‘e’, ‘x’,’a’,’m’,’p’,’l’,'e'};
return temp;
}
int main()
{
puts(Fun1());
puts(Fun2());
}


Answer: Garbage values.

Explanation: Both the functions suffer from the problem of dangling pointers. In Fun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function Fun2() also suffers from the same problem but the problem can be easily identified in this case.

C aptitude 21.2

   char *strexp()
{
char *temp = "example string";
return temp;
}
int main()
{
puts(strexp);
}

Answer: example string

Explanation: The character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

C aptitude 21.3

     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.

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C Aptitude Questions and answers with explanation

C Aptitude 20
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 20.1

  main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}


Answer: 10

Explanation: The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives up to the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

C aptitude 20.2

   main()
{
int i=-1;
-i;
printf("i = %d, -i = %d n",i,-i);
}


Answer: i = -1, -i = 1

Explanation: -i is executed and this execution doesn’t affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

C aptitude 20.3

     main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}

Answer: Compiler error

Explanation: i is a constant. you cannot change the value of constant

C aptitude 20.4

      main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}

Answer: garbagevalue..1

Explanation: p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer.

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Write a C program to reverse a string using pointers.

C Strings:
Write a C program to reverse a string using pointers.
In this program, we reverse the given string by using the pointers. Here we use the two pointers to reverse the string, strptr holds the address of given string and in loop revptr holds the address of the reversed string.
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* and browse!
*
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***********************************************************/

#include<stdio.h>
int main(){
int i=-1;
char str[100];
char rev[100];
char *strptr = str;
char *revptr = rev;
printf("Enter the string:n");
scanf("%s",str);
while(*strptr)
{
strptr++;
i++;
}
while(i >=0) {
strptr--;
*strptr = *revptr;
revptr++;
--i;
}
printf("nn Reversed string is:%s",rev);
return 0;
}



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C Aptitude Questions and answers with explanation

C Aptitude 18
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 18.1

  main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}



Answer: 2 5 5

Explanation: In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the ‘’ termination character). The third sizeof is similar to the second one.

C aptitude 18.2

  main()
{
char not;
not=!2;
printf("%d",not);
}


Answer: 0
Explanation:! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

C aptitude 18.3

    #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}

Answer: TRUE

Explanation: The input program to the compiler after processing by the preprocessor is,
main()
{
  if(0)
    puts(“NULL”);
  else if(-1)
    puts(“TRUE”);
  else
    puts(“FALSE”);
 }
Preprocessor doesn’t replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence “TRUE” is printed.

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C Aptitude Questions and answers with explanation

C Aptitude 16
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 16.1

 main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}


Answer: g20fy

Explanation: Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

C aptitude 16.2

  main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}

Answer: ck

Explanation:In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

C aptitude 16.3

    main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i {
printf(“%sn”,x);
x++;
}
}

Answer:(blank space)
irl
rl
l

Explanation: Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘’) to the first location. Now the string becomes “irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

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C Aptitude Questions and answers with explanation

C Aptitude 14
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 14.1

  main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}


Answer: Compiler Error

Explanation: in the end of nested structure yy a member have to be declared

C aptitude 14.2

  main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}

Answer: Linker error: undefined symbol ‘i’.

Explanation:extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn’t find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

C aptitude 14.3

     main()
{
printf("%d", out);
}

int out=100;

Answer: Compiler error: undefined symbol out in function main.

Explanation: The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main.

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C Aptitude Questions and answers with explanation

C Aptitude 13
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

 
Predict the output or error(s) for the following:

C aptitude 13.1

  ain()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}


Answer: 1

Explanation: before entering into the for loop the checking condition is “evaluated”. Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

C aptitude 13.2

   main()
{
char s[]={'a','b','c','n','c',''};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}


Answer: M

Explanation:p is pointing to character ‘n’.str1 is pointing to character ‘a’ ++*p meAnswer:”p is pointing to ‘n’ and that is incremented by one.” the ASCII value of ‘n’ is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:”str1 is pointing to ‘a’ that is incremented by 1 and it becomes ‘b’. ASCII value of ‘b’ is 98. both 11 and 98 is added and result is subtracted from 32. i.e. (11+98-32)=77(“M”);

C aptitude 13.3

   main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}

Answer: Compiler Error

Explanation: Initialization should not be done for structure members inside the structure declaration.

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Aptitude Questions and answers with explanation

C Aptitude 11
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

 
Predict the output or error(s) for the following:

C aptitude 11.1

   main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}

Answer: Compiler error: Undefined label ‘here’ in function main

Explanation: Labels have functions scope, in other words The scope of the labels is limited to functions . The label ‘here’ is available in function fun() Hence it is not visible in function main.

C aptitude 11.2

   main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}

Answer: Compiler error: Lvalue required in function main

Explanation:Array names are pointer constants. So it cannot be modified.

C aptitude 11.3

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

Answer: Output Cannot be predicted exactly.

Explanation: Side effects are involved in the evaluation of i

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C Aptitude Questions and answers with explanation

C Aptitude 10
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 10.1

  void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}

Answer: 4..2

Explanation: the second pointer is of char type and not a far pointer

C aptitude 10.2

    main()
{
int i=400,j=300;
printf("%d..%d");
}

Answer: 400..300

Explanation:printf takes the values of the first two assignments of the program. Any number of printf’s may be given. All of them take only the first two values. If more number of assignments given in the program, then printf will take garbage values.

C aptitude 10.3

    main()
{
char *p;
p="Hello";
printf("%cn",*&*p);
}

Answer: H

Explanation: * is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string “Hello”. *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

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C Aptitude Questions and answers with explanation

C Aptitude 7
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
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Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 7.1

  main()
{
printf("nab");
printf("bsi");
printf("rha");
}

Answer: hai

Explanation: n – newline b – backspace r – linefeed

C aptitude 7.2

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

Answer:45545

Explanation:The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

C aptitude 7.3

    #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:64

Explanation: the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

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