Wednesday, March 9, 2011

C programming Language Technical Interview Questions and Answers

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

8.       # include
aaa() {
  printf("hi");
 }
bbb(){
 printf("hello");
 }
ccc(){
 printf("TechPreparation.com");
 }
main()
{
  int (*ptr[3])();
  ptr[0]=aaa;
  ptr[1]=bbb;
  ptr[2]=ccc;
  ptr[2]();
}

Answer:
TechPreparation.com

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.
9.       #include
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.
10.       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.
11.     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.
12.       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
13.       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.
14.       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.
15. 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.
16. 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

C programming Language Technical Interview Questions and Solution

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

1.     struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
 struct aaa abc,def,ghi,jkl;
 int x=100;
 abc.i=0;abc.prev=&jkl;
 abc.next=&def;
 def.i=1;def.prev=&abc;def.next=&ghi;
 ghi.i=2;ghi.prev=&def;
 ghi.next=&jkl;
 jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
 x=abc.next->next->prev->next->i;
 printf("%d",x);
}

Answer:
2

Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
2.       struct point
 {
 int x;
 int y;
 };
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);

           
Answer:
origin is(0,0)
origin is(0,0)

Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point  is globally declared x & y are initialized as zeroes
3.       main()
{
 int i=_l_abc(10);
             printf("%d\n",--i);
}
int _l_abc(int i)
{
 return(i++);
}

Answer:
9

Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.
4.       main()
{
 char *p;
 int *q;
 long *r;
 p=q=r=0;
 p++;
 q++;
 r++;
 printf("%p...%p...%p",p,q,r);
}

Answer:
0001...0002...0004

Explanation:
++ operator  when applied to pointers increments address according to their corresponding data-types.
5.      main()
{
 char c=' ',x,convert(z);
 getc(c);
 if((c>='a') && (c<='z'))  x=convert(c);  printf("%c",x);
}
convert(z)
{
  return z-32;
}

Answer:
Compiler error

Explanation:
declaration of convert and format of getc() are wrong.
6.      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. 
7.       # include 
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.

100+ C programming Technical Interview Questions With Answers

63. main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}

Answer:
1==1 is TRUE

Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".
64. main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}

Answer:
2000 is a leap year

Explanation:
An ordinary program to check if leap year or not.
65. #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}

Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.
66. int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}

Answer:
30,20,10

Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.
67. 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 upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.
68. 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.
69. #include
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
70. #include
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 pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.
71. #include
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}

Answer:
hello 5

Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.
72. main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}

Answer:
11

Explanation:
the expression i+++j is treated as (i++ + j)

100+ C Technical Interview Questions With Solution

Predict the output or error(s) for the following:
44. main()
{
extern out;
printf("%d", out);
}
int out=100;

Answer:
100

Explanation:
This is the correct way of writing the previous program.
45. main()
{
show();
}
void show()
{
printf("I'm the greatest");
}

Answer:
Compier error: Type mismatch in redeclaration of show.

Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
46. main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}

Answer:
100, 100, 100, 2
114, 104, 102, 3
47. main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; p =" a;" j="0;">
Answer:
Compiler error: lvalue required.

Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
48. main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}

Answer:
111
222
333
344
49. 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’.
50. 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’.
51. 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.

C Technical Programming questions and solution

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

25. main()
{
printf("%p",main);
}

Answer:
Some address will be printed.

Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.
26. main()
{
clrscr();
}
clrscr();

Answer:
No output/error

Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).
27. enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}

Answer:
0..1..2

Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
28. 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
29. 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.
30. main()
{
char *p;
p="Hello";
printf("%c\n",*&*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.
31. 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.
32. 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.
33. 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
34. void main()
{
int i=5;
printf("%d",i+++++i);
}

Answer:
Compiler Error

Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
35. #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}

Answer:
Compiler Error: Constant expression required in function main.

Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

C Technical Interview Questions With Answers | C Programming Language Questions With Solution

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

7
. 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.
8. 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.
9. 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.
10. main()
{
printf("%x",-1<<4);>
}

Answer:
fff0

Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.
11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}

Answer:
Compiler Error : Type mismatch in redeclaration of function display

Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.
12. main()
{
int c=- -2;
printf("c=%d",c);
}

Answer:
c=2;

Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}

Answer:
sizeof(i)=1

Explanation:
Since the #define replaces the string int by the macro char

14. main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}

Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

C Technical Interview Questions With Solution | C Programming Language Questions With Answers

Predict the output or error(s) for the following:
1. 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".
2. 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:
mmmm
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].
3. 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 (== , >, <, <=, >=,!= ) .
4. 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.
5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) j="0;j<5;j++){">}

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.
6. 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 .

Style /Convention of writing Program in C++ programming language

C++ Programming Style Guidelines


1 Introduction
o 1.1 Layout of the Recommendations
o 1.2 Recommendations Importance
2 General Recommendations
3 Naming Conventions
o 3.1 General
o 3.2 Specific
4 Files
o 4.1 Source Files
o 4.2 Include Files and Include Statements
5 Statements
o 5.1 Types
o 5.2 Variables
o 5.3 Loops
o 5.4 Conditionals
o 5.5 Miscellaneous
6 Layout and Comments
o 6.1 Layout
o 6.2 White space
o 6.3 Comments
7 References

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Calculating Free Bytes (Wasted ) On Disk-Data Structures

Calculating Wasted Bytes On Disk

When a file gets stored on the disk, at a time DOS allocates one cluster for it. A cluster is nothing but a group of sectors. However, since all file sizes cannot be expected to be a multiple of 512 bytes, when a file gets stored often part of the cluster remains unoccupied. This space goes waste unless the file size grows to occupy these wasted bytes. The
following program finds out how much space is wasted for all files in all the directories of the current drive.
#include 
#include 
#include 
#include 
#include 
unsigned bytes_per_cluster ;
unsigned long wasted_bytes ;
unsigned long num_files = 0 ;
main( )
{
int ptr = 0, flag = 0, first = 0 ;
struct ffblk f[50] ;
struct dfree free ;
/* get cluster information and calculate bytes per cluster */
getdfree ( 0, &free ) ;
bytes_per_cluster = free.df_bsec * free.df_sclus ;
chdir ( "\\" ) ;
/* check out files in root directory first */
cal_waste( ) ;
/* loop until all directories scanned */
while ( ptr != -1 )
{
/* should I do a findfirst or a findnext? */
if ( first == 0 )
flag = findfirst ( "*.*", &f[ptr], FA_DIREC ) ;
else
flag = findnext ( &f[ptr] ) ;
while ( flag == 0 )
{
/* make sure its a directory and skip over . & .. entries */
if ( f[ptr].ff_attrib == FA_DIREC && f[ptr].ff_name[0] != '.' )
{
flag = chdir ( f[ptr].ff_name ) ; /* try changing directories */
if ( flag == 0 ) /* did change dir work? */
{
cal_waste( ) ;
first = 0 ; /* set for findfirst on next pass */
break ;
}
}
flag = findnext ( &f[ptr] ) ; /* search for more dirs */
}
if ( flag != 0 || ptr == 49 ) /* didn't find any more dirs */
{
ptr-- ;
chdir ( ".." ) ; /* go back one level */
first = 1 ; /* set to findnext on next pass */
}
else
ptr++ ;
}
printf ( "There are %lu bytes wasted in %lu files.\n", wasted_bytes,
num_files ) ;
}
cal_waste( )
{
int flag = 0 ;
long full_cluster ;
struct ffblk ff ;
/* look for all file types */
flag = findfirst ( "*.*", &ff, FA_RDONLY | FA_HIDDEN | FA_SYSTEM | FA_ARCH
) ;
while ( flag == 0 )
{
num_files++ ;
full_cluster = ff.ff_fsize / bytes_per_cluster * bytes_per_cluster ;
wasted_bytes += bytes_per_cluster - ( ff.ff_fsize - full_cluster ) ;
flag = findnext ( &ff ) ;
}
}

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