Given a spherical surface, write bump-mapping procedure to
generate the bumpy surface of an orange
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what will be the output of this program? void main() { int a[]={5,10,15}; int i=0,num; num=a[++i] + ++i +(++i); printf("%d",num); }
#define a 10 int main() { printf("%d..",a); foo(); printf("%d..",a); return 0; } void foo() { #undef a #define a 50 }
Given a list of numbers ( fixed list) Now given any other list, how can you efficiently find out if there is any element in the second list that is an element of the first list (fixed list)
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why the range of an unsigned integer is double almost than the signed integer.
print numbers till we want without using loops or condition statements like specifically(for,do while, while swiches, if etc)!
void main() { static int i; while(i<=10) (i>2)?i++:i--; printf(“%d”, i); }
main() { char c=' ',x,convert(z); getc(c); if((c>='a') && (c<='z')) x=convert(c); printf("%c",x); } convert(z) { return z-32; }
main() { int i; clrscr(); for(i=0;i<5;i++) { printf("%d\n", 1L << i); } } a. 5, 4, 3, 2, 1 b. 0, 1, 2, 3, 4 c. 0, 1, 2, 4, 8 d. 1, 2, 4, 8, 16
main() { char *p = “ayqm”; char c; c = ++*p++; printf(“%c”,c); }
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); }
PROG. TO PRODUCE 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
main() { int i; float *pf; pf = (float *)&i; *pf = 100.00; printf("\n %d", i); } a. Runtime error. b. 100 c. Some Integer not 100 d. None of the above