Answer / gouri shanker m.ojha

using Dynamic memory allocation,you can create large array size.

example:m=1000000,n=1000000
void enter(int m,int n)
{
int i,j;
long int c=0;


matrix=new long int*[m]; /*m is row size*/
for(i=m-1;i>=0;i--)
{
matrix[i]=new long int[n]; /*n is col size*/
for(j=0;j<n;j++)
{
matrix[i][j]=++c;
}
delete matrix[i];
}
}

Write a program that produces these three columns sequence nos. using loop statement Sequence nos. Squared Squared + 5 1 1 6 2 4 9 3 9 14 4 16 21 5 25 30

1 Answers   GoDB,

---

main() { extern int i; { int i=20; { const volatile unsigned i=30; printf("%d",i); } printf("%d",i); } printf("%d",i); } int i;

1 Answers  

---

What is the output for the following program main() { int arr2D[3][3]; printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) ); }

1 Answers  

---

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

2 Answers  

---

main() { int i=5,j=6,z; printf("%d",i+++j); }

2 Answers  

---

#include<stdio.h> #include<conio.h> void main() { int a=(1,2,3,(1,2,3,4); switch(a) { printf("ans:"); case 1: printf("1");break; case 2: printf("2");break; case 3: printf("1");break; case 4: printf("4");break; printf("end"); } getch(); }

0 Answers  

---

Give a very good method to count the number of ones in a 32 bit number. (caution: looping through testing each bit is not a solution)

7 Answers   Microsoft,

---

void main() { while(1){ if(printf("%d",printf("%d"))) break; else continue; } }

1 Answers  

---

¦void main() ¦{ ¦int i=10,j; ¦ j=i+++i+++i; ¦printf("%d",j); ¦getch(); ¦} ¦ output:-30 but in same question if we write as- ¦void main() ¦{ ¦int i=10; ¦ int j=i+++i+++i; ¦printf("%d",j); ¦getch(); ¦} ¦ output:-33 why output is changed from 30 to 33. Can any body answer...

3 Answers  

---

main() { clrscr(); } clrscr();

2 Answers  

---

write the function. if all the character in string B appear in string A, return true, otherwise return false.

11 Answers   Google,

---

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]); }

2 Answers  

---