What will be output of following c code? void main() { struct bitfield { unsigned a:5; unsigned c:5; unsigned b:6; }bit; char *p; struct bitfield *ptr,bit1={1,3,3}; p=&bit1; p++; clrscr(); printf("%d",*p); getch(); }

Correct Answer: Output: 1 1 1 1 1 1 For loop will execute six times Note: continue keyword in do-while loop bring the program its while condition (while(0)) which is always false

Correct Answer: #include "stringh" #include "stdlibh" struct stud { int rollno; int marks; char name[30]; }; int sort_m (struct stud *, struct stud *); int sort_name (struct stud *, struct stud *); int sort_marks (struct stud *, struct stud *); main() { static struct stud ss[] = { { 15, 96, "Akshay" }, { 2, 97, "Madhuri" }, { 8, 85, "Aishvarya" }, { 10, 80, "Sushmita" } }; int x,w; clrscr(); w = sizeof (struct stud); printf ('\nIn order of roll numbers:"); qsort (ss, 4, w, sort_rn); for(x=0; x<4;x++) printf ("\n%d%s%d", ss[x]rollno, ss[x]name,ss[x]marks); printf("\n\nIn order of names:"); qsort(ss, 4, sort_name); for (x=0; x<4;x++) printf("\n%d%s%d",ss[x]rollno, ss[x]name,ss[x]marks); printf("\n\nIn order of marks:"); qsort(ss,4,w,sort_marks); for (x=0;x<4;x++) printf ("\n%d%s%d",ss[x]rollno,ss[x]name,ss[x]marks); } int sort_rn (struct stud *t1, struct stud *t2) { return (t1->rollno-t2->rollno); } int sort_name (struct stud *t1, struct stud *t2) { return (strcmp(t1->name,t2->name)); } int sort_marks (struct stud *t1, struct stud *t2) { return (t2->marks-t1->marks); }

Correct Answer: When a function accepts a variable number of arguments , its prototype cannot provide any information about the number of arguments and type of those variable arguments Hence the compiler cannot warn about the mismatches The programmer must make sure that arguments match or must manually insert explicit typecast

Correct Answer: As against malloc(), calloc() needs two arguments, the number of elements to be allocated and the size of each element For example, p = (int *) calloc (10, sizeof (int)); would allocate space for a 10- integer array Additionally, calloc() would also set each of this element with a value 0 Thus the above call to calloc() is equivalent to: p = (int *) malloc (10 * sizeof (int)); memset (p, 0, 10 * sizeof( int ));

Correct Answer: Yes, using the realloc() function as shown below: main() { int *p; p = ( int *) malloc (20) ; t = p; t = (int *) realloc ( p, 40); if ( t == NULL ) Printf (" Cannot reallocate, leaves previous allocated region unchanged "); else { if ( p ==t ) ; / * the array expanded at the same region */ else { free ( p ); / * deallocate the original array */ p = t; /* set p to newly allocated region */ } } }

Correct Answer: Johnny Depp

Explanation:

Default value of enum constant
GaryOldman = -2 +1 = -1
And default value of enum constant
EdNorton = -1 + 1 = 0
Note: Case expression can be enum constant.

Correct Answer: #include int main( int count,char * argv[] ) { int i; FILE *ptr; char *str; char ch; if( count == 1) { printf( "The syntax of the command is incorrect\n" ); } for( i=1;i2) printf("\nError occurred while procesing : %s\n",argv[i]); } else { if(count>2) { printf("%s\n\n",argv[i]); } while((ch=getc(ptr))!=-1) printf("%c",ch); } fclose(ptr); } return 0; } Save the above file as openc, compile and execute the go to command mode (current working directory) and write: open xyc (xyc any file present in that directory) To run the open command in all directories and drive you will have to give the path of current working directory in command mode Write: C:tc\bin>PATH c:\tc\bin Now press enter key Now your open command will work in all directory and drive

Correct Answer: It is also called as post tested loop It is used when it is necessary to execute the loop at least one time Syntax: do { Loop body } while (Expression); Example: int main(){ int num,i=0; do{ printf("To enter press 1\n"); printf("To exit press 2"); scanf("%d",&num); ++i; switch(num){ case 1:printf("You are welcome\n");break; default : exit(0); } } while(i<=10); return 0; } Output: 3 3 4 4

Correct Answer: #include int main(){ int a=5,b=10; a=b+a; b=a-b; a=a-b; printf("a= %d b= %d",a,b); }

Correct Answer: #include int main(){ int k,r; long int i=0l,j=1,f; //Taking maximum numbers form user printf("Enter the number range:"); scanf("%d",&r); printf("FIBONACCI SERIES: "); printf("%ld %ld",i,j); //printing firts two values for(k=2;k