why the range of an unsigned integer is double almost than
the signed integer.
Answer / srinivasu
By default int is declared as unsigned integer.It's range is
0 to 65535.Signed integer range is -32767 to 32768.
A variable declared as Unsigned int doesn't holds a negative
value whereas a signed integer does.So signed and unsigned
integer doesn't differ in the range, but in the kind of
value it holds.
| Is This Answer Correct ? | 3 Yes | 0 No |
why java is platform independent?
main() { static int a[3][3]={1,2,3,4,5,6,7,8,9}; int i,j; static *p[]={a,a+1,a+2}; for(i=0;i<3;i++) { for(j=0;j<3;j++) printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j), *(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i)); } }
There were 10 records stored in “somefile.dat” but the following program printed 11 names. What went wrong? void main() { struct student { char name[30], rollno[6]; }stud; FILE *fp = fopen(“somefile.dat”,”r”); while(!feof(fp)) { fread(&stud, sizeof(stud), 1 , fp); puts(stud.name); } }
int main() { int x=10; printf("x=%d, count of earlier print=%d", x,printf("x=%d, y=%d",x,--x)); getch(); } ================================================== returns error>> ld returned 1 exit status =================================================== Does it have something to do with printf() inside another printf().
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]); }
main() { int i=5,j=6,z; printf("%d",i+++j); }
Write a Program in 'C' To Insert a Unique Number Only. (Hint: Just Like a Primary Key Numbers In Database.) Please Some One Suggest Me a Better Solution for This question ??
{ int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }
posted by surbhi just now main() { float a = 5.375; char *p; int i; p=(char*)&a; for(i=0;i<=3;i++) printf("%02x",(unsigned char) p[i]); } how is the output of this program is :: 0000ac40 please let me know y this output has come
Which version do you prefer of the following two, 1) printf(“%s”,str); // or the more curt one 2) printf(str);
void main() { int x,y=2,z; z=(z*=2)+(x=y=z); printf("%d",z); }
#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(); }