1. Input Output Formatting

 

Access Specifiers

• %d              : signed int
• %u              : unsigned int
• %s               : string
• %c               : char
• %f               : float
• %lf              : double
• %o              : octal
• %x  or %p : hexadecimal
• %e or %E  : scientific notation
• %hd            : short int
• %ld             : long int

Range of data types

The memory size of the basic data types may change according to 32 or 64-bit operating system.
Let's see the basic data types. Its size is given according to 32-bit architecture.

Data Types	Memory Size	Range
char	1 byte	−128 to 127
signed char	1 byte	−128 to 127
unsigned char	1 byte	0 to 255
short	2 byte	−32,768 to 32,767
signed short	2 byte	−32,768 to 32,767
unsigned short	2 byte	0 to 65,535
int	2 byte	−32,768 to 32,767
signed int	2 byte	−32,768 to 32,767
unsigned int	2 byte	0 to 65,535
short int	2 byte	−32,768 to 32,767
signed short int	2 byte	−32,768 to 32,767
unsigned short int	2 byte	0 to 65,535
long int	4 byte	-2,147,483,648 to 2,147,483,647
signed long int	4 byte	-2,147,483,648 to 2,147,483,647
unsigned long int	4 byte	0 to 4,294,967,295
float	4 byte
double	8 byte
long double	10 byte

The memory size of int for 64-bit architecture is 4 bytes and that of long int is 8 bytes.

1. What is the output of the following C code?
#include <stdio.h>
int main()
{
    double num = 12345.6789;
    printf("%e", num);
}

1.234568e+04

2. What is the output of the following C code?
#include <stdio.h>
int main()
{
    double num = 0.000123456789;
    printf("%e", num);
}

1.234568e-04

3. What is the output of the following C code for the following inputs
1. 12 12
2. 123 123
3. 1234 1234
#include <stdio.h>
int main()
{
    int a, b;
    scanf("%2d %3d", &a, &b);
    printf("%d %d", a, b);
}
1. 12 12
2. 12 3
3. 12 34

Explanation: Here, a can hold maximum 2 characters and b can hold maximum 3 characters. Also if a gets more than 2 characters as input, then the extra characters will be taken as input by b, and other characters will be ignored.

4. What is the output of the following C code for the following inputs
1. 12 12
2. 123 123
3. 1234 1234
#include <stdio.h>
int main()
{
    int a, b;
    scanf("%d %d", &a, &b);
    printf("a=%4d b=%3d", a, b);
}
1. a=    12 b=  12
2. a=  123 b=123
3. a=1234 b=1234

Explanation: %4d in printf means minimum 4 characters needs to be printed and %3d means minimum 3 characters needs to be printed.If variable doesn't have sufficient characters, then whitespace will be printed as required characters.

5. What is the output of the following C code for the following inputs
1. 1.2 1.2
2. 1.23 1.23
3. 12.34 12.34
#include <stdio.h>
int main()
{
    float a, b;
    scanf("%3f %4f", &a, &b);
    printf("%f %f", a, b);
}
1. 1.200000 1.200000
2. 1.200000 3.000000
3. 12.000000 34.000000

Explanation: Here, a can hold maximum 3 characters and b can hold maximum 4 characters including decimal point. Also if a gets more than 3 characters as input, then the extra characters will be taken as input by b, and other characters will be ignored.

6. What is the output of the following C code for the following inputs
1. 1.2 1.2
2. 1.23 1.23
3. 12.34 12.34
#include <stdio.h>
int main()
{
    float a, b;
    scanf("%f %f", &a, &b);
    printf("a=%4.1f b=%7.2f", a, b);
}
1. a=  1.2 b=      1.20
2. a=  1.2 b=      1.23
3. a=12.3 b=    12.34

Explanation: %4.1 means minimum 4 characters are needs to be printed but maximum 1 character will be printed after decimal. If variable doesn't have sufficient characters, then whitespace will be printed as required characters.

7. What is the output of the following C code for the following inputs
1. abc
2. abcde
3. abcdefg
#include <stdio.h>
int main()
{
    char str[5];
    scanf("%s", str);
    printf("str=%s", str);
}
1. str=abc
2. str=abcde
3. str=abcdefg

Explanation: Because of dynamic memory allocation, new array is created because of more characters.

8. What is the output of the following C code for the following inputs
1. a
2. abc
3. abcd
#include <stdio.h>
int main()
{
    char str[10];
    scanf("%3s", str);
    printf("%s", str);
}
1. a
2. abc
3. abc

Explanation: %3s means str can hold maximum 3 characters only.

9. What is the output of the following C code for the following inputs
1. a
2. abc
3. abcd
#include <stdio.h>
int main()
{
    char str[5];
    scanf("%s", str);
    printf("str=%3s", str);
}
1. str=    a
2. str=abc
3. str=abcd

Explanation: %3s in printf means minimum 3 characters needs to be printed. If variable doesn't have sufficient characters, then whitespace will be printed as required characters.

10. What is the output of the following C code for the following inputs
1. a
2. abc
3. abcd
#include <stdio.h>
int main()
{
    char str[10];
    scanf("%s", str);
    printf("%.3s", str);
}
1. a
2. abc
3. abc

Explanation: %0.3s in printf means minimum 0 characters and maximum 3 characters from the input string needs to be printed. If variable doesn't have sufficient characters, then whitespace will be printed as required characters.

11. What is the output of the following C code for the following inputs
1. a
2. abc
3. abcd
#include <stdio.h>
int main()
{
    char str[10];
    scanf("%s", str);
    printf("str=%8.3s", str);
}
1. str=             a
2. str=         abc
3. str=         abc

Explanation: %8.3s in printf means minimum 8 characters and maximum 3 characters from the input string needs to be printed. If variable doesn't have sufficient characters, then whitespace will be printed as required characters.

Supression Character (%*d)

• If used with scanf(), then ignores the current value and takes the next value as input.
• If used with printf(), then prints some whitespaces and gives value of next variable till the last variable.

1. What is the output of the following C code with input: 20 25 30 35?
#include <stdio.h>
int main()
{
    int a, b, c, d;
    scanf("%*d %d %d %d", &a, &b, &c, &d);
    printf("%d %d %d %d", a, b, c, d);
}

25 30 35 0

2. What is the output of the following C code with input: 20 25 30 35?
#include <stdio.h>
int main()
{
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    printf("%d %*d %d %d", a, b, c, d);
}

20           30 35 35