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Representing Data Types in
C language
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HOW CAN YOU SAY THAT A
VARIABLE IS INTEGER?

Integer - will hold values that are
zero and positive and negative
whole not numbers and should not
contain decimal places.

variables and constants integer data
type, preceded by int.

EXAMPLES:

Const int Dozen = 12;

Int score1, score2, score3;
HOW CAN YOU THAT A VARIABLES IS
CONSTANT FLOAT?
Float- variables and constant float of data type,
on the other hand can store real numbers that
contain fractional parts.
The value of float data type can have a precision
of up to six decimal places and whose value may
range 3.4 x 10 -38 to 3.4 x 10 38
By default, float constant are automatically
treated as double precision during program
execution. To prevent such conversion from taking
place, use the suffix F or after the value of your
constant float.
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EXAMPLES of float

Float salary;

Const float price = 45.50f
HOW CAN YOU TELL IF THE CONSTANT
VARIABLE IS DOUBLE?
Double - variables constant of this data type, can
store values twice the precession of the float data
type, thus the word double is used. It occupies
two times of the memory as compared to that of
float data type.
It occupies two times of the memory as compared
to that of float data type.
Double RATE: 235.34554550;
const double speed=446383.68789450;
double distance;
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HOW CAN SAY THE DECLARED
VARIABLE IS CHARACTER?

Character - variables constant of
this type, usually preceded by char,
can accept a single character
enclosed by a pair of single
quotation marks.
ALSO, any number that is enclose to a pair of
single and double quotation mark is considered
data type char.
Complete the following table by indicating
the correct variable/constant, data type or
declarations for each item.
Variable/Constant
data type
declarations
1.
a=17;
int
Int a=17;
2.
ANS=‘y’ const
char
char cons ANS = ‘y’;
3.
Price =533144.75
double
double price=5333144.75;
4.
Status [15];
char
5.
X=1.9E-150
double
char status [15]’;
double X=1.9E-150
6.
Bunos=50 const;
int
cons int bunos = 50;
7.
Level=‘6’;
char
char level = ‘6’;
Create a program that will allow you
to enter test scores in four subject
and compute the average using
declaration of basic data type.
#include<stdio.h>
main()
{
int score1, score2,score3,score4;
Double average, sum;
printf(“Please type your test score 1 to 4 separated by spaces:”);
scanf(“%d%d%d%d”, &score1,&score2,&score3,&score4);
sum=score1+score2+score3+score4;
average=sum/4;
printf(“Your average scoe is %.2lf”, average);
}
End of the C data types
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