Printf() Quiz in C Set-2

Printf() Quiz in C Set-2 | The printf() function in C is used for output formatting. It is used to display information required by the user and also prints the value of the variables.

It formats the output, like the width of the output, the sign of the output e.t.c. We will learn those formatting using printf() C.

If you finding difficulties to solve these questions then first you should learn these:-

• Output formatting using printf() in C
• Input formatting using scanf() in C

Printf() Quiz in C

Find the output of the programs given in printf Quiz in C Set-2

Q1) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%d+%d=%d",1,2,3);
  return 0;
}

a) 123
b) 1+2=3
c) 1+2=garbage_value
d) garbage_value + garbage_value = garbage_value

View Answer

Answer:- b) 1+2=3

Q2) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%d+%d=%d",1,2);
  return 0;
}

a) 1+2=garbage_value
b) garbage_value + garbage_value = garbage_value
c) 123
d) 1+2=3

View Answer

Answer:- a) 1+2=garbage_value

Q3) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%d+%d=%d",1,2,3,4);
  return 0;
}

a) garbage_value + garbage_value = garbage_value
b) 123
c) 1+2=3
d) 1+2=garbage_value

View Answer

Answer:- c) 1+2=3

Q4) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%d+%d=%d",'1','2','3');
  return 0;
}

a) 49+50=51
b) None of these
c) error
d) 1+2=3

View Answer

Answer:- a) 49+50=51

Q5) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%c+%c=%c",'1','2','3');
  return 0;
}

a) 1+2=3
b) None of these
c) 49+50=51
d) error

View Answer

Answer:- a) 1+2=3

Q6) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%c+%c=%c",65,66,67);
  return 0;
}

a) A+B=C
b) error
c) None of these
d) 65+66=67

View Answer

Answer:- a) A+B=C

Q7) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%c+%c=%c",a,b,c);
  return 0;
}

a) 65+66=67
b) A+B=C
c) None of these
d) error

View Answer

Answer:- d) error

The compiler creates a, b, and c as variables not as constants. The variables must be declared first before using it in the program.

Q8) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%d+%d=%d",'a','b','a'+'b');
  return 0;
}

a) a + b = a+b
b) a + b = c
c) 97+98=195
d) error

View Answer

Answer:- c) 97+98=195

Q9) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%d+%d=%D",1,2,3);
  return 0;
}

a) None of these
b) error
c) 1+2=%D
d) 1+2=3

View Answer

Answer:- c) 1+2=%D

C language is case sensitive, %d, and %D are different. %d is conversion specifier but %D is treated as simple text or string.

Q10) Find the output of the given C program.

#include<stdio.h>
int main()
{
  printf("%d%%d=%d",9,2,9%2);
  return 0;
}

a) 9%2 = 1
b) 9%d = 2
c) 9%d = 1
d) 9%2 = 2

View Answer

Answer:- b) 9%d = 2

In C language, to print ‘%’ we use escape character %% So, in this program %%d converted to string %d. The 9 will be used for the first %d, so before assignment operator, it will display 9%d. Now, after assignment operator %d is there, since 9 is already used and the next constant is 2. Finally, it gives the output 9%2=2, Note that the last constant (9%2) was never used by the program because there was only two %d to display variable but there are three constants (9, 2, 9%2)

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