placement c programming quiz with solution

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’.

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’.

3)

main()

{

 int  i, n;

 char *x = “girl”;

 n = strlen(x);

 *x = x[n];

 for(i=0; i

   {

printf(“%s\n”,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 (‘\0’) to the first location. Now the string becomes “\0irl” . 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] = ‘\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.

4) int i,j;

for(i=0;i<=10;i++)

{

j+=5;

assert(i<5);

}

Answer: 

Runtime error: Abnormal program termination. 

assert failed (i<5),>, 

Explanation:

asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,

#undef NDEBUG

and this will disable all the assertions from the source code. Assertion

is a good debugging tool to make use of.  

  

5) main()

{

int i=-1;

+i;

printf("i = %d, +i = %d \n",i,+i);

}

Answer:

 i = -1, +i = -1

Explanation:

Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

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