Arithmetic Operation

What is printed by the following ANSI C program?

#include < stdio.h >
int main(int argc, char *argv[]){ 
 char a = 'P';
 char b = 'x';
 char c = (a  &  b) + '*';
 char d = (a | b) - '-';
 char e = (a ^ b) + '+'; 
 printf("%c %c %c \n", c, d, e); 
 return 0;
}

ASCII encoding for relevant characters is given below

Consider the following C program:

#include < stdio.h >
int main() {
 float sum = 0.0, j = 1.0, i = 2.0;
 while (i / j > 0.0625) {
    j = j + j; 
    sum = sum + i/j;
    printf("%f \n", sum);
 }
 return 0;
}

The number of times variable sum will be printed When the above program is executed is _________ .

Consider the following C program:

 main()
{
float sum= 0.0, j=1.0,i=2.0;
while(i/j>0.001){
    j=j+1;
    sum=sum+i/j;
    printf("%f/n", sum);
  }
}

Consider the following C code segment:

 #include < stdio.h >
main()
{
    int i, j, x;
    scanf("%d", &x);
    i=1; j=1;
    while (i<10) {
            j =j*i;
            i= i+1;
            if(i==x) break;
        }
}

For the program fragment above, which of the following statements about the variables i and j must be true after execution of this program? [ !(exclamation) sign denotes factorial in the answer]

What does the following program do when the input is unsigned 16 bit integer?

 #include < stdio.h >
main(){
  unsigned int num;
  int i;
  scanf("%u", &num);
  for(i=0;i<16;i++){
    printf("%d", (num < < i&1 < < 15)?1:0);
  }
}

Consider the following C program.

 #include < stdio.h >
int main ( ) {
    int m = 10;
    int n, n1;
    n = ++m;
    n1 = m++;
    n--;
    --n1;
    n-=n1;
printf ("%d", n) ;
return 0;
}

The output of the program is ______________.

The attributes of three arithmetic operators in some programming language are given below.

The value of the expression 2-5+1-7*3 in this language is_______ .

Consider the following statements

 #define hypotenuse (a, b) sqrt (a*a+b*b);

The macro call hypotenuse(a+2,b+3);

The for loop

 for (i=0; i<10; ++i)
printf("%d", i&1);

prints

Consider the following program fragment

 i=6720; j=4;
while (i%j)==0
{
    i=i/j;
    j=j+1;
}

On termination j will have the value