Type Cast operator and Number System with Examples in C
Type cast :
- It is a concept of converting one data type values into another data type.
- in ‘c’ programming language, “type castings” are classified into two types. i.e
-
implicit type casting
-
Explicit type casting
- when we are converting lower data types value into higher data types. Then it is called. Implicit type casting process
- Implicit type casting is under the control of compiler
- As a programmer, it is not required to consider implicit type casting process.
- When we are converting higher data types values into lower data type then it is called Explicit type casting.
- As a programmer it is mandatary to handle explicit type casting process because data overflow will occur
syntax:
datatype1 variable1=value;
Datatype2 variable2;
variable2=(Data type2) variable1;
Ex:1 int i=45;
float f;
f=1; //f=(float)i;
Ex:-2 double d=123456.789;
long intl;
l=(long int)d;
{
float f;
f=7.6;
if(f==7.6) // if(float==double)4B==8B by default
printf(“welcome”);
else
printf(“Hello”);
o/p: Hello.
- By default , any type of real values is double that why, when we are comparing the float data with double then automatically condition became false.
void main()
{
float f;
f=3.8;
if(f==3.8f) || if(float=float) 4B=4B
printf(“welcome”);
else
printf(“Hello”);
o/p : Welcome
Void main()
{
int i;
float f;
i=6;
f=6.0;
if(f==i)
printf(“welcome”);
else
printf(“Hello”);
}
o/p: Welcome
When we are comparing integer value with float then both are became equal, if fractional part is not available in float i.e “.0”
void main()
{
float f;
f=3.0;
printf(“welcome”);
else
printf(“Hello”);
o/p: welcome
{
flaot f;
f=9.5;
if(f==9.5)
printf(“welcome”);
else
printf(“Hello”);
}
o/p: welcome
Number system :-
- Always Number system will decide that how the numeric values require to hold in a memory.
- Number systems are classified into four types.
- decimal number,
- Octal Number system.
- Hexadecimal
- Binary.
- The Base value of decimal is 10 and the range 0 to 9.
- When we require to print the decimal value, then we required to use %d format specifier
- Decimal can be represent –ve and =ve format data also.
- The base value of octal number system is 8 and the range is 0 to 7
- when we require to print the octal format data, then we require to use %0 format specifier.
- If any numberic value is started with Zero, it indicates octal number.
ex:- 013, 0474,0878
- The base value of hexadecimal is 16 and the range 0-9ABCDEF
- When we require to print octal data, then require to use %x %h %p and %x format specifier
- When we require to represent any hexadecimal value, then we require to use “ OX “ as a prefix
ex: 0X12, 0X-A/3, 0XABC4
10 012 0X4
8
1X1X8
0X2
= 8X2
83 0 1 2 3 0X53
8
3X0+8
2X1+8
1X2+8
0X3
64+16+3
Error
0181 error
Error
0981
range(0-7)
161 0241 0XA1
291 0443 0X123
Binary Number System :-
- The base value of binary Number system is 2 and the range 0,1
- When the memory will represented, always it holds in the form of binary.
- when we are representing the data in the binary complete data will be represented in the forms 1’s and ’0’s
- If there no any format specifiers are available to print binary representation on console
| Decimal |
Binary |
| 1 |
0001 |
| 2 |
0010 |
| 3 |
0011 |
| 4 |
0100 |
| 5 |
0110 |
| 6 |
0110 |
| 7 |
1000 |
| 8 |
1000 |
| 9 |
1001 |
|
|
35
0010 0011 (32+2+1)
61
0011 1101 (32+16+8+4+1)
127
0111 1111(64+32+16+8+4+2+1)
255
1111 1111
32767
0111 1111 1111 1111
65535
1111 1111 1111 1111
| No.of bits rep |
No.of combinations |
max value |
max |
| 1 |
2(21) |
1(21-1) |
01 |
| 2 |
4(22) |
3(22-1) |
11 |
| 3 |
8(23) |
7(23-1) |
0111 |
| 4 |
16(24) |
15(24-1) |
1111 |
| 5 |
32(25) |
31(25-1) |
0001 111 |
| 6 |
64(26) |
63(26-1) |
0011 1111 |
| 7 |
128(27) |
127(27-1) |
0111 1111 |
| 8 |
256(28) |
255/-1(28-1) |
1111 1111 |
| 9 |
65536(216) |
65535/-1(216- |
1111 1111 1111 1111 |
| 10 |
|
|
|
| 8bits - 1 Byte1024B - 1KB
1024KB - 1MB
1024MB - 1GB
1024GB- 1TB |
|
Int i:-
On Dos based compiler the size of int is 2 bytes and the range from -32768 to + 32767
For representing any integer value, we require 16 bit combination i.e 65536 representations.
Among those all combinations, 50% will represent negative data and remaining 50% will represents +ve data.
| 8bits - 1 Byte1024B - 1KB
1024KB - 1MB
1024MB - 1GB
1024GB- 1TB |
- Left side most signigicant bit is called signed bit is called signed bit
- Always signed bit will decides the return value of binary representation
- If sign bit is “0” and remaining b all bits are Zero’s then it gives minimum value of +ve sign ie ‘0’
- if sign bit is ‘0’ and remaining all bits are 1’s then it gives maximum value of +ve sign ie 32767
- If the sign bit is ‘1’ and remaining all bits are 0’s then it gives minimum value of –ve sign i.e 32768
- If the sign bit is ‘1’ and remaining all bits are 1’s then it gives maximum value of –ve sign ie -1
int i;
i=32767+1;
32467
0111 1111 1111 1111
1
0000 0000 0000 0001
1 1 1 1 1 1 1 1 1 1 1 1 1 1
10000 0 0 0 0 0 0 0 0 0 0
Unsigned integer of U:
The size of unsigned int is 4 bytes and the range from 0 to 65,535
For representing any unsigned integer value, we require 16 bit
combination i.e 65536 representations.
- Among those all representations all gives +ve data only, because in unsigned type , signed bit is not available
- When we are working with integer and unsigned integer always format specifier will decide the return value of binary representation.
| %d |
Binary Representations |
%U |
| 0
32767
-32768
-1 |
0000 0000 0000 0000
01111 1111 1111 1111
1000 0000 0000 0000
1111 1111 1111 1111 |
0
32767
32768
65535 |
Character of ch:-
- The size of characters is 1 byte and the range from -128 to +127
- For representing any character we require 8 bit combination i. e 256 representations.
- Among those all representations, 50% will gives +ve data and remaining 50% will give –ve data.
0000 0000
0
0111 1111
127
1000 0000
-128
1111 1111
-1
Unsigned character of ch :-
The size of unsigned character is 1 byte and the range from 0 to 255
For representing any unsigned character we require 8 bit combination i.e 256 combinations and all the representation gives +ve data only.
0000 0000
0
0111 1111
1270000
1000 0000
128
1111 1111
255
255 -128
(-) 127 + 127
128 -1
When we are working with char & unsigned char, always data type will decides the return value of binary representation.
| printf(“%d %U”, 65535,-1); |
-1 |
65535 |
| printf(“%d %U”,32768,-32768); |
-32768 |
32768 |
| printf(“%U %O %X ”,-1,-1,-1); |
65535, |
|
| printf(“%d %U”, 32768,-32768); |
-32768 |
32768 |
| printf(“%u %O %X”,-1,-1,-1); |
65535 |
0177777 |
For representing any octal data we require to take 3 binary bits only, for representing any hexadecimal value, we require to take 4 binary bits.
printf(“%d %0 %u”, 327684,327684,327684);
o/p: -32768 0100000
