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Binary representations
-^ We have already described binary numbers^ –^ On the computer, all integers are stored in binary^ –^ Thus,^ to^
store^ each^ of^ these
numbers,^ we^
must^ store^ the corresponding binary digits (bits):^3
-^ To store a googol (
10010 ), we must store
333 bits: 10010010010011010110100100101100101001100001101111100111…01011000010110010011110000100110001001100111000001011111…10011100010101100111001000000100011100010000100011010011…11100101010101011001001000011000010001010100000101110100…01111000100000000000000000000000000000000000000000000000… 00000000000000000000000000000000000000000000000000000
Adeel Israr | COMSATS Institute of Information Technology, Islamabad
Storage
-^ Do we store as many bits as are necessary?^ –^ You could, but this would be exceedingly difficult to manage •^ Instead, each primitive data type has a fixed amount of storage^ –^ 8 bits are defined as 1 byte^ –^ All data types are an integral number of bytes^ •^ Usually
1 ,^2 ,^4 ,^8 or^16 bytes • Because we use binary, powers of 2 are very common: Exponent^ Decimal
Binary 0 21
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Example
-^ Consider this program:^ #include // Function declarationsint main();// Function definitionsint main() {unsigned short a=42;unsigned int
b=207500;unsigned long^ c=299792458;printf(“%u\n”, (a + b + c) );return 0;}
Output:^300000000 Note:First,^ a^ is upcast to
unsigned int before to the first additionThen, the result is upcast to unsigned long^ before the second addition
Adeel Israr | COMSATS Institute of Information Technology, Islamabad
Example
-^ On the stack, an appropriate number of bytes are allocated to eachvariable^ #include // Function declarationsint main();// Function definitionsint main() {unsigned short a=42;unsigned int
b=207500;unsigned long^ c=299792458;printf(“%u\n”, (a + b + c) );return 0;}
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Wasted space?
-^ If an integer does not use all the bytes, the remaining bits are never-the-less allocated until the variable goes out of scope^ –^ In general-purpose computing, this is often not a problem^ –^ This is a critical issue, however, in embedded systems^ •^ More memory:^ –^ Costs more^ –^ Uses more power^ –^ Produces more heat
Determining the Adeel Israr | COMSATS Institute of Information Technology, Islamabad
size^ of a type
-^ We have said^
short,^ int^ and^ long
are 2, 4 and 8 bytes
-^ This is true on most every general-purpose computer • Unfortunately, the C++ specification doesn’t require this –^ Fortunately, the
sizeof^ operator gives you this information #include <stdio.h>int main();int main() { printf("An 'unsigned short' occupies %d bytes\n”, sizeof ( unsigned short );printf("An 'unsigned int' occupies %d bytes\n", sizeof ( unsigned int );printf("An 'unsigned long'^ occupies %d bytes\n" sizeof ( unsigned long ); return 0;}^ Output on :^ An 'unsigned short' occupies 2 bytesAn 'unsigned int'
occupies 4 bytes An 'unsigned long'
occupies 8 bytes
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Memory and initial values • Question: – What happens if the initial value cannot be stored? #include <stdio.h>int main();int main() {unsigned short c=299792458;printf("The speed of light is %ud m/s\n", c);return 0;}
Adeel Israr | COMSATS Institute of Information Technology, Islamabad
Memory and initial values • Fortunately, you get a warning: example.cpp: In function 'int main()':example.cpp:6:31:^ warning:^ narrowing
conversion^ of^ â299792458â
from^ 'int'^ to
'short unsigned int' inside { } [-Wnarrowing]unsigned short c=299792458;
^
warning: large integer implicitly truncated to unsignedtype [-Woverflow] • It still compiles and executes:^ The speed of light is 30794 m/s.
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Memory and arithmetic • What happens if the sum, difference or product of two integersexceeds what can be stored? #include <stdio.h>int main() {unsigned short smallest=0, largest=65535;printf("Smallest:\t%u\n",smallest);printf("Largest:\t%u\n",largest);smallest--;largest++;printf("Smallest:\t%u\n",smallest);printf("Largest:\t%u\n",largest);return 0;} Output:
Smallest:^0 Largest:^65535 Smallest minus 1: 65535Largest plus 1:
Adeel Israr | COMSATS Institute of Information Technology, Islamabad
Memory and arithmetic • Important: All unsigned integers arithmetic is performed:^16 modulo^2 for^ unsigned short^32 modulo^2 for^ unsigned int^64 modulo^2 for^ unsigned long • This is similar to all clock arithmetic being performed modulo 12
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Summary so far
-^ We have the following:^ –^ Unsigned integers are stored as either 1, 2, 4 or 8 bytes^ –^ The value is stored in the binary representation^ Type^^ –^ You should not memorize the exact ranges
Bytes^ Bits^ Range
Approximate Range unsigned char^
8 – 1^ 0, …, 255
unsigned short^
16 – 1^ 0, …, 65535
unsigned int^
32 – 1^ 0, …, 4.3 million unsigned long^
64 – 1^ 0, …, 1.8 trillion
Adeel Israr | COMSATS Institute of Information Technology, Islamabad
Signed types
-^ A better solution: •^ Note that^ –1 + 1 = 0, but also 11 + 1 = 0–5 + 2 = –3, but also 7 + 2 = 9, which we are equating to –
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Scientific notation
-^ Recall from secondary school scientific notation that allows us towrite numbers clearly and succinctly:^ Conventional notation
Scientific notation
–116.67408 × 10
8 2.99792458 × 10
0.0^000000000000000000000000
6.626070040 × 10
1.6021766208 × 10
0 8.3144598 × 10
3.14159265358979323 × 10
0
6.67408^ ×^10
Mantissa
Exponent
Base
Adeel Israr | COMSATS Institute of Information Technology, Islamabad
Scientific notation
-^ The number of decimal digits used is the precision:
Scientific notation
Precision –11^ 6.67408 × 10
8 2.99792458 × 10
–346.626070040 × 10
–191.6021766208 × 10
0 8.3144598 × 10
3.14159265358979323 × 10
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Scientific notation
-^ Without^ going
into^ detail,^ each
data^ type^ has^
an^ approximate
maximum precision it can store • There is generally only one situation where
float^ has acceptable
precision for engineering applications:^ –^ Computer graphics
Real WorldExample^ Ruler^ Screw gauge Data type
Approximatemaximum precision(decimal digits) float^
double^
Adeel Israr | COMSATS Institute of Information Technology, Islamabad
Scientific notation
-^ How could you store a floating-point number?^ –^ Store the exponent and mantissa separately, and assume a decimalpoint comes after the first digit^ Scientific notation
**Representation
float**^
double –11^ 6.67408 × 10
-11 6674080^
8 2.99792458 × 10
+08 2997925^
–34 6.626070040 × 10
–19 1.6021766208 × 10
0 8.3144598 × 10
+00 8314460^
+000 +3141592653589793 * In reality, these are stored in binary
