### Find Output Of Program

Question 6 :

What will be the output of the program?

``````#include
int main()
{
float f=43.20;
printf("%e, ", f);
printf("%f, ", f);
printf("%g", f);
return 0;
}``````

A). 4.320000e+01, 43.200001, 43.2
B). 4.3, 43.22, 43.21
C). 4.3e, 43.20f, 43.00
D). Error
Answer : Option A

Explanation :

printf("%e, ", f); Here '%e' specifies the "Scientific Notation" format. So, it prints the 43.20 as 4.320000e+01.
printf("%f, ", f); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 43.20 as 43.200001.
printf("%g, ", f); Here '%g' "Use the shorter of %e or %f". So, it prints the 43.20 as 43.2.

Question 7 :

What will be the output of the program?

``````#include
int main()
{
float a=0.7;
if(a < 0.7f)
printf("C\n");
else
printf("C++\n");
return 0;
}``````

A). C
B). C++
C). Compiler error
D). Non of above
Answer : Option B

Explanation :

if(a < 0.7f) here a is a float variable and 0.7f is a float constant. The float variable a is not less than 0.7f float constant. But both are equal. Hence the if condition is failed and it goes to else it prints 'C++'
Example:

``````#include
int main()
{
float a=0.7;
printf("%.10f %.10f\n",0.7f, a);
return 0;
}``````
Output:
0.6999999881 0.6999999881

Question 8 :

What will be the output of the program?

``````#include
#include
int main()
{
float n=1.54;
printf("%f, %f\n", ceil(n), floor(n));
return 0;
}``````

A). 2.000000, 1.000000
B). 1.500000, 1.500000
C). 1.550000, 2.000000
D). 1.000000, 2.000000
Answer : Option A

Explanation :

ceil(x) round up the given value. It finds the smallest integer not < x.
floor(x) round down the given value. It finds the smallest integer not > x.

printf("%f, %f\n", ceil(n), floor(n)); In this line ceil(1.54) round up the 1.54 to 2 and floor(1.54) round down the 1.54 to 1.

In the printf("%f, %f\n", ceil(n), floor(n)); statement, the format specifier "%f %f" tells output to be float value. Hence it prints 2.000000 and 1.000000.

Question 9 :

What will be the output of the program?

``````#include
int main()
{
float d=2.25;
printf("%e,", d);
printf("%f,", d);
printf("%g,", d);
printf("%lf", d);
return 0;
}``````

A). 2.2, 2.50, 2.50, 2.5
B). 2.2e, 2.25f, 2.00, 2.25
C). 2.250000e+000, 2.250000, 2.25, 2.250000
D). Error
Answer : Option C

Explanation :

printf("%e,", d); Here '%e' specifies the "Scientific Notation" format. So, it prints the 2.25 as 2.250000e+000.

printf("%f,", d); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 2.25 as 2.250000.

printf("%g,", d); Here '%g' "Use the shorter of %e or %f". So, it prints the 2.25 as 2.25.

printf("%lf,", d); Here '%lf' specifies the "Long Double" format. So, it prints the 2.25 as 2.250000.