Question

a) use for loop to get the ascii code from ‘A’ to ‘Z’ with the following format:

A - - - - ?(a number)

B - - - - ?(another number)

.

.

Z - - - - ?(the ascii code of ‘Z’)

b) verify ‘A’ > ‘z’ or ‘a’ > ‘Z’

c) what will be printed for expression 3344.996 with this format string %012.05f or %12.05f or %-12.05f or %-012.5f?

d) read a float number to a float variable tttt and show the size(bytes) of 0.5 * tttt, 0.5 is a constant.

Answer 1

a)

#include<stdio.h>
void main()
{
char ch;
for(ch='A';ch<='Z';++ch)
printf("%c %d\n",ch,ch);
}

Output

A 65
B 66
C 67
D 68
E 69
F 70
G 71
H 72
I 73
J 74
K 75
L 76
M 77
N 78
O 79
P 80
Q 81
R 82
S 83
T 84
U 85
V 86
W 87
X 88
Y 89
Z 90

b)

#include<stdio.h>
void main()
{

//ascii values of characters will be compared
if('A' > 'z')
printf("A>z\n");
else if('a' > 'Z')
printf("a>Z\n");

}

Output

a>Z

c)

#include<stdio.h>
void main()
{
printf("%012.05f\n",3344.996);
printf("%12.05f\n",3344.996);
printf("%-12.05f\n",3344.996);
printf("%-012.5f\n",3344.996);
}

Output

003344.99600
3344.99600
3344.99600
3344.99600

d)

#include<stdio.h>
void main()
{
float ttt;
scanf("%f",&ttt);
printf("%d",sizeof(0.5*ttt));
}

Output

2.35

8 (since 0.5*ttt gives real number by default any real number is treated as double so size is 8 bytes instead of 4 bytes)