i2 Technologies Placement Paper

Company: i2 Technologies

1. convert 0.9375 to binary (c)
a) 0.0111
b) 0.1011
c) 0.1111
d) none

2.( 1a00 * 10b )/ 1010 = 100 (b)
a) a=0, b=0
b) a=0, b=1
c)
d)none

3. in 32 bit memory machine 24 bits for mantissa and 8 bits for exponent. to increase the range of floating point.
a) more than 32 bit is to be there.
b) increase 1 bit for mantissa and decrease 1 bit for exponent
c) increase 1 bit for exponent and decrease one bit for mantissa

5. in C “X ? Y : Z ” is equal to
a) if (X==0) Y ;else Z
b) if (X!=0) Y ;else Z
c) if (X==0) Y ; Z
d)

6. foo()
int foo(int a, int b){
if (a&b) return 1;
return 0;
a) if either a or b are zero returns always 0
b) if both a & b are non zero returns always 1
c) if both a and b are negative…….

7. typedef struct nt{——–} node-type
node-type *p
a) p =( nodetype *)malloc( size of (node-type))

8. function gives some error what changes as to be made
void ( int a,int b){
int t; t=a; a=b; b=t;
a)define void as int and write return tt
b)change everywhere a to *a and b to *b

9. which of the following is incorrect
a) if a and b are defined as int arrays then (a==b) can never be true
b) parameters are passed to functions only by values
c) defining functions in nested loops

10. include
void swap(int*,int*);
main()
int arr[8]={36,8,97,0,161,164,3,9}
for (int i=0; i<7; i++) for (int j=i+1; j<8;j++) if(arr void swap(int*x,int*y) int temp; static int cnt=0; temp= *x; *x=*y; *y=temp; cnt++; cnt = ?
a) 7
b) 15
c)1
d)

11. int main(){ FILE *fp; fp=fopen(“test.dat”,”w”); fprintf(fp,’hellon”); fclose(fp); fp=fopen (“test.dat”,”w”); fprintf (fp, “world”); fclose(fp); return 0; if text.dat file is already present after compiling and execution how many bytes does the file occupy
a) 0 bytes
b)5 bytes
c)11 bytes d)data is insufficient

12. f1(int*x,intflag) int *y; *y=*x+3; switch(flag){ case 0:. break; case 1: *x=*y; break; case 2: break; return(*y) main() *x=5; i=f1(x,0); j=f1(x,1); printf(“%………”,i,j,*x); what is the output?
a) 8 8 8
b) 5 8 8
c) 8 5 8