System of two charges equal in magnitude but opposite in nature separated by certain distance is called an electric dipole.
Dipole moment is given as
P = qL, directed from -ve charge to +ve charge.
Electric potential due to electric dipole
Let us consider an electric dipole Ab of length L formed by charges +q and -q. Let p be a point at a distance r from centre of dipole where electric potential is to be found.
We know
V = V1+V2
V1 = 1/4πEo(q/PB)
V2 = 1/4πEo(q/AP)
if we consider dipole of very short length we can have AP = PN and PM = PB
also we know
PM = r-l/2cos(theta)
PN = r+l/2cos(theta)
Now
V = q/4πEo(1/r-l/2cos(theta) - 1/r+l/2cos(theta))
V = q/4πEo(lcos(theta)/r^2-l^2/4cos^2(theta))
since l^2/4cos^2(theta <<< r^2 we can have
V = pcos(theta)/4πEor^2
Sunday, June 13, 2010
Gauss Law:
The total lines of force(electric flux) passing through a surface is eqyal to 1/Eo times the total charge enclosed by that surface.
φ = 1/Eo * total charge
Application og gauss law
1) To find electric intnensity due to charged sphere
Let us consdier a charged sphere of radius E and p be a point at distance r from the centre of sphere
CASE I when P lies outside the sphere
To find the intensity at { drae a concentric sphere of radius r such that it will enclose charge on given sphere. Let it be q then according to gauss law,
Elelctrc flux, φ = 1/Eo * q
E *A = 1/Eo *q
E = 1/4πr^2 * q/Eo
E = 1/4πE0 *q/r^2
CASE II P lies on surface of sphere
IN this r = R
E = 1/4πE0 *q/R^2
CASE III P lies inside the sphere
In this case the total charge enclosed by a gaussian surface draws through P(sphere) will be zero.
φ = 1/Eo * q = 0
E *A =0
E = 0
II) Intensity due to a line charge
Let us consider a infinitely long conductor which contains the charge as linear charge density λ. Let p be a point at a distance r from the conductor where electric intensity is to be found. To find this let us draw a cylinder through P co-axial with the conductor of length L.
Then. from Gauss law, flux through this cylinder,
φ = 1/Eo * total charge
φ = 1/Eo * λ
E * A = 1/Eo λ L
E * 2πrl = 1/Eo λ l
E = λ/2πEor
III) Intensity due to charged plane sheet
Let us consider a plane charged sheet having surface charge density E. Let P be a point at distance r from plane sheet where intensity is to be found. To find the intensity, let us draw a cylinder enclosing P, perpendicular to plane sheet having surface area A. The form gauss law
φ = 1/Eo * total charge
φ = 1/Eo * 6A
E *A = 1/Eo * 6A
E = 6/Eo
φ = 1/Eo * total charge
Application og gauss law
1) To find electric intnensity due to charged sphere
Let us consdier a charged sphere of radius E and p be a point at distance r from the centre of sphere
CASE I when P lies outside the sphere
To find the intensity at { drae a concentric sphere of radius r such that it will enclose charge on given sphere. Let it be q then according to gauss law,
Elelctrc flux, φ = 1/Eo * q
E *A = 1/Eo *q
E = 1/4πr^2 * q/Eo
E = 1/4πE0 *q/r^2
CASE II P lies on surface of sphere
IN this r = R
E = 1/4πE0 *q/R^2
CASE III P lies inside the sphere
In this case the total charge enclosed by a gaussian surface draws through P(sphere) will be zero.
φ = 1/Eo * q = 0
E *A =0
E = 0
II) Intensity due to a line charge
Let us consider a infinitely long conductor which contains the charge as linear charge density λ. Let p be a point at a distance r from the conductor where electric intensity is to be found. To find this let us draw a cylinder through P co-axial with the conductor of length L.
Then. from Gauss law, flux through this cylinder,
φ = 1/Eo * total charge
φ = 1/Eo * λ
E * A = 1/Eo λ L
E * 2πrl = 1/Eo λ l
E = λ/2πEor
III) Intensity due to charged plane sheet
Let us consider a plane charged sheet having surface charge density E. Let P be a point at distance r from plane sheet where intensity is to be found. To find the intensity, let us draw a cylinder enclosing P, perpendicular to plane sheet having surface area A. The form gauss law
φ = 1/Eo * total charge
φ = 1/Eo * 6A
E *A = 1/Eo * 6A
E = 6/Eo
Ultrasound
The sound wave which has the frequency greater then 20khz i.e beyond the audible region is called the untrasound. The sound wave which has frequency less than 20hz are known as infrasonic sound wave or infrasound.
Production of ultrasound
1) Piezo electric method
When a force is applied into a pair faces of quartz crystal then an electric potential will be setup into the opposite pair faces. Conversly, if an electric potential be applied on a pair of quartz crystal then there will be tension on the opposite faces. This effect in quartz crystal is known as piezo electric effect.
Now When an alternating potential be applied on a pair forces of a quartz crystal then opposite force will be in longitidunal vibration when freq. of applied potential becomes equal to natural frequency of quartz crystal then resonance will occur. In that case crystal will vibrate with maximum amplitude and a sound wave will be produced with ultrasound properties.
2) Magnetic striction method
In this method a ferromagnetic material is wounded closely by a coil and alternating current is passed through it. Then we may experience a longitudinal vibration in the ferromagnetic substance. When the frequency of the alternating current becomes equal to the natural frequency of ferromagnetc substance then resonance will occur and the material will vibrate with maximum amplitude. And a sound wave is generated with unltasound Properties.
Use of ultrasound
1) To find the depth of Seas
2) To locate distance object
3) To kill unwanted tissue(blood less surgery)
Characterstics of Sound
1) Pitch: Pitch is a characterstics of sound which directly related with frequency. Sound wave having high frequency is called high pitched sound and vice versa.
2) Intensity: Sound energy flowing per unit area per unit time is called intensity
3) Threshold frequency of hearing: The minimum value of sound intensity that can be heared by human being in general
4) Loudness: It is the magnitue of sensation of hearing and it is directly proportional to the logarithm of intensity of sound
5) Quality: Quality of sound depends on no. of harmonics presents in sound. HIgher no. of harmonic reveals high quality and vice versa.
Production of ultrasound
1) Piezo electric method
When a force is applied into a pair faces of quartz crystal then an electric potential will be setup into the opposite pair faces. Conversly, if an electric potential be applied on a pair of quartz crystal then there will be tension on the opposite faces. This effect in quartz crystal is known as piezo electric effect.
Now When an alternating potential be applied on a pair forces of a quartz crystal then opposite force will be in longitidunal vibration when freq. of applied potential becomes equal to natural frequency of quartz crystal then resonance will occur. In that case crystal will vibrate with maximum amplitude and a sound wave will be produced with ultrasound properties.
2) Magnetic striction method
In this method a ferromagnetic material is wounded closely by a coil and alternating current is passed through it. Then we may experience a longitudinal vibration in the ferromagnetic substance. When the frequency of the alternating current becomes equal to the natural frequency of ferromagnetc substance then resonance will occur and the material will vibrate with maximum amplitude. And a sound wave is generated with unltasound Properties.
Use of ultrasound
1) To find the depth of Seas
2) To locate distance object
3) To kill unwanted tissue(blood less surgery)
Characterstics of Sound
1) Pitch: Pitch is a characterstics of sound which directly related with frequency. Sound wave having high frequency is called high pitched sound and vice versa.
2) Intensity: Sound energy flowing per unit area per unit time is called intensity
3) Threshold frequency of hearing: The minimum value of sound intensity that can be heared by human being in general
4) Loudness: It is the magnitue of sensation of hearing and it is directly proportional to the logarithm of intensity of sound
5) Quality: Quality of sound depends on no. of harmonics presents in sound. HIgher no. of harmonic reveals high quality and vice versa.
Beat and Beat frequency
When two sound of nearly equal frequency are sounded together then periodic rise and fall in sound intensity is heared. This phenomeno is called beat. For this phenomenon the difference between frequencies of two sound wave must not exceed 10hz
Beat frewuency
Let us consider two sound waves with frequency f1 and f2 such that f1-f2 <10 hz are sounded together then at the starting points can be given as
y1 = asinw1t
y2 = asinw2t
By the principal of superpostion the resultant wave will be,
y = y1+y2
y = asinw1t + asinw2t
y = A sin(pi)(f1+f2) * t
This is a periodic function with amplitude
A = 2acos(pi)(f1-f2) *t
CASE I Loud sound
Loud sound will be heard when value of A becomes maximum i.e when
cos(pi)(f1-f2)*t = 1
cos(pi)(f1-f2)*t = cpsn(pi)
f = n/(f1-f2)
loud sound will be heared at f = 0, 1/(f1-f2),2/(f1-f2)
Beat frequency = (f1-f2) hz
CASEII Soft Sound
Soft sound will be heared when amplitude is minimum i.e when
cos(pi)(f1-f2)*t = cos((2n+1)/2)*pi
f = 1/(f1-f2)
Beat frequency = (f1-f2)Hz
Beat frewuency
Let us consider two sound waves with frequency f1 and f2 such that f1-f2 <10 hz are sounded together then at the starting points can be given as
y1 = asinw1t
y2 = asinw2t
By the principal of superpostion the resultant wave will be,
y = y1+y2
y = asinw1t + asinw2t
y = A sin(pi)(f1+f2) * t
This is a periodic function with amplitude
A = 2acos(pi)(f1-f2) *t
CASE I Loud sound
Loud sound will be heard when value of A becomes maximum i.e when
cos(pi)(f1-f2)*t = 1
cos(pi)(f1-f2)*t = cpsn(pi)
f = n/(f1-f2)
loud sound will be heared at f = 0, 1/(f1-f2),2/(f1-f2)
Beat frequency = (f1-f2) hz
CASEII Soft Sound
Soft sound will be heared when amplitude is minimum i.e when
cos(pi)(f1-f2)*t = cos((2n+1)/2)*pi
f = 1/(f1-f2)
Beat frequency = (f1-f2)Hz
Wave
Wave is the disturbance( means a change in pressure, density or displacement of the particles of the medium about their equilibrium position) produced in the particles of the medium when an energy transfers through the medium.
Transverse wave
If the particle of the medium vibrate along the direction perpendicular to the direction of wave motionm then this tyoe if wave is called a transverse wave.
Longitudinal wave.
If the particle of the medium vibrate along the direction parallel to the direction of wave motion then this type of wave is called longitudinal wave.
Progressive wave
A wave in which the crest and trough( in transvers) or compression and refraction( in lonbgitudinal) travel is called progressive. They vary continiously.
Equation of Progressive wave
Let us consider a progressive wave originating from O travels along the positive x-axis. Let is have frequency f, wavelength λ amplitude a angular velocity V
Since the motion in the medium is simple harmonic motion, the displacement of the particle be given as,
y = a sinwt-----------1
Let us consider a point p at a distance x, from origin O where the wave lags. Let wave lags at P(i.e phase difference) by φ. Then the displacement equation will be
y = asin(wt- φ)-------------2
We know that at distance λ phase difference will be 2 so that the phase difference at distance x will be
φ = 2 x/ λ
equation 2 becomes
y = asin(wt-2 x/ λ)---------------------3
Again,
y = asin(wt-kx)---------------------4
Where k = 2 / λ = wave number
Again using w = 2 / λ we get
y = asin(2 / λ(vt-x))-------------5
These equation 3 4 and 5 are equation of progressive wave.
Particle velocity
The displacement of particle per unit time is called particle velocity we have the displacement of particle when a wave is travelling alon x-axis as
y = asin(wt-kx)---------------------1
Diffrentiating the equation wrt time we get
dy/dx = acos(wt-kx)w
dy/dt = awcos(wt-kx)------------2
Now
Differentiating equation wrt x we get
dy/dx = -akcos(wt-k)-----------3
we have w = 2(pi)f = 2(pi)V/ λ = kV
Equation 2 becomes
dy/dt = akvcos(wt-kx)-------------4
Now from equation 3 and 4 becomes
dy/dt = -v^2dy/dx-------------5
partical velocity = -(wave velocity) * slope of displacement curve at that point
Transverse wave
If the particle of the medium vibrate along the direction perpendicular to the direction of wave motionm then this tyoe if wave is called a transverse wave.
Longitudinal wave.
If the particle of the medium vibrate along the direction parallel to the direction of wave motion then this type of wave is called longitudinal wave.
Progressive wave
A wave in which the crest and trough( in transvers) or compression and refraction( in lonbgitudinal) travel is called progressive. They vary continiously.
Equation of Progressive wave
Let us consider a progressive wave originating from O travels along the positive x-axis. Let is have frequency f, wavelength λ amplitude a angular velocity V
Since the motion in the medium is simple harmonic motion, the displacement of the particle be given as,
y = a sinwt-----------1
Let us consider a point p at a distance x, from origin O where the wave lags. Let wave lags at P(i.e phase difference) by φ. Then the displacement equation will be
y = asin(wt- φ)-------------2
We know that at distance λ phase difference will be 2 so that the phase difference at distance x will be
φ = 2 x/ λ
equation 2 becomes
y = asin(wt-2 x/ λ)---------------------3
Again,
y = asin(wt-kx)---------------------4
Where k = 2 / λ = wave number
Again using w = 2 / λ we get
y = asin(2 / λ(vt-x))-------------5
These equation 3 4 and 5 are equation of progressive wave.
Particle velocity
The displacement of particle per unit time is called particle velocity we have the displacement of particle when a wave is travelling alon x-axis as
y = asin(wt-kx)---------------------1
Diffrentiating the equation wrt time we get
dy/dx = acos(wt-kx)w
dy/dt = awcos(wt-kx)------------2
Now
Differentiating equation wrt x we get
dy/dx = -akcos(wt-k)-----------3
we have w = 2(pi)f = 2(pi)V/ λ = kV
Equation 2 becomes
dy/dt = akvcos(wt-kx)-------------4
Now from equation 3 and 4 becomes
dy/dt = -v^2dy/dx-------------5
partical velocity = -(wave velocity) * slope of displacement curve at that point
Simple Harmonic Motion(SHM)
If in an oscillatory motion the acceleration is directly proportional to displacement and is always directed towards the mean position then it is called SHM
If F be the force applied on the particle and x be the displacement then,
F α –x (-ve sign shows they are opposite in direction)
.-. F = -kx----------1 where k is force constants
If m be the mass of the particle then from Newtons second law of motion.
F = ma-------2
From 1 and 2
ma = -kx
a = -k/m*x
a = -ω^2x----4
This is the equation of simple harmonic motion
Motion of a helical(loded Spring)
Let us consider a negligible helical spring suspended through a rigid support. Now it will attach a mass in its free end then it will be elongated through same distance. let it be L. Then according to Hooke’s law, the restoring force is directly proportional to the elongation produces
F1 α L
F1 = -CL-------------1 (where –ve shows that the force is in the opposite direction of elongation.
Now let the mass be pulled down through a distance x and then realesed. The spring then starts to oscillate. Now elongation on spring will be l+x & restoring force is given as,
F2 = -C(l+x)--------------2
The resulting force is
F = F1 + F2 = C(L+x) + cl = -cx
.-. F = -Cx----------------3
Now
From Newtons second law of motion,
F=ma------------------4
From 3 and 4
ma = -Cx
a = -c/m * x------------5
equation 5 is similar to the equation of SHM. Hence loaded spring executes SHM.
Comparint the equation with SHM we can have
ω^2 = C/m
Now time period of oscillation T = 2 /√(c/m)
Mass attached in two spring
Let us consider a mass m be attached in between two strings s1 and s2 having force constant c1 and c2 respectively.
Let the mass and system be placed in a frictionless surface. Now suppose the mass be pulled toward the right direction through the distance x. Then clearly the spring S1 will be elongated by distance x and spring S2 will be compressed through distance x.
THerefore from Hook’s law restoring force acting on the mass will be
F = -C1x – C2x
F = -(C1+C2)x-------------1
This cause oscillation in mass from Newton’s second law of motion force acting on mass
F=ma-------------2
From 1 and 2
ma = -(C1+C2)x
a = -(C1+C2)/m *x--------3
This equation is similar to equation of SHM. Hence this executes SHM
Simple Pendulum
A point heavy mass suspended with mass less, weightless, extensible string through a rigid support. It is also called ideal pendulum.
Drawbacks of simple pendulum.
It is impossible to have heavy point mass.
It is impossible to have weightless string.
There will be relative motion between the bob and string at extreme position.
Length of pendulum, is difficult to measure exactly accurately.
Compound Pendulum
Compound pendulum is a rigid body of whatever shape capable to oscillate freely about a horizontal axis passing through it.
The point in which pendulum is suspended through rigid suppost is called point of suspension.
The distance between point of suspension and centre of gravity of the pendulum is called the length of pendulum.
Consider a compound pendulum having mass m and length l i.e SG = L. Let pendulum be displace through an angle ϴ. At this instant the weigh of pendulum mf act vertically downward direction. Then the restoring force that tends to bring pendulum torque acting on the body will be,
Ʈ = -mglsin ϴ------------1
Now,
If I be the moment of intertia and
α = a be angular accleration then the restoring torque will be given as
Ʈ = I * α = I * a--------------2
From equation 1 and 2
I * a = -mglsin ϴ
a = -mgl/I * ϴ --------------3
This equation 3 is also similar to equation of SHM. Hence compound also executes SHM.
Point of Suspension and point of oscillation are interchangeable. We have time period for a compound pendulum having length L as
T = 2 √((k^2/L + L)/g)
T = 2 π(√l1 + l)/g-------------1
such that k^2/L = L1
Now, let point of oscillation becomes point of suspension. Then length of pendulum
k^2/L = L1
Hence, time period in this case will be
T1 = 2 π(√k2/l1+l)/g
We have k^2 = LL1
T1 = 2π√LL1+L1/(l/2)
= 2 π√L1 + l)/g-------------2
If F be the force applied on the particle and x be the displacement then,
F α –x (-ve sign shows they are opposite in direction)
.-. F = -kx----------1 where k is force constants
If m be the mass of the particle then from Newtons second law of motion.
F = ma-------2
From 1 and 2
ma = -kx
a = -k/m*x
a = -ω^2x----4
This is the equation of simple harmonic motion
Motion of a helical(loded Spring)
Let us consider a negligible helical spring suspended through a rigid support. Now it will attach a mass in its free end then it will be elongated through same distance. let it be L. Then according to Hooke’s law, the restoring force is directly proportional to the elongation produces
F1 α L
F1 = -CL-------------1 (where –ve shows that the force is in the opposite direction of elongation.
Now let the mass be pulled down through a distance x and then realesed. The spring then starts to oscillate. Now elongation on spring will be l+x & restoring force is given as,
F2 = -C(l+x)--------------2
The resulting force is
F = F1 + F2 = C(L+x) + cl = -cx
.-. F = -Cx----------------3
Now
From Newtons second law of motion,
F=ma------------------4
From 3 and 4
ma = -Cx
a = -c/m * x------------5
equation 5 is similar to the equation of SHM. Hence loaded spring executes SHM.
Comparint the equation with SHM we can have
ω^2 = C/m
Now time period of oscillation T = 2 /√(c/m)
Mass attached in two spring
Let us consider a mass m be attached in between two strings s1 and s2 having force constant c1 and c2 respectively.
Let the mass and system be placed in a frictionless surface. Now suppose the mass be pulled toward the right direction through the distance x. Then clearly the spring S1 will be elongated by distance x and spring S2 will be compressed through distance x.
THerefore from Hook’s law restoring force acting on the mass will be
F = -C1x – C2x
F = -(C1+C2)x-------------1
This cause oscillation in mass from Newton’s second law of motion force acting on mass
F=ma-------------2
From 1 and 2
ma = -(C1+C2)x
a = -(C1+C2)/m *x--------3
This equation is similar to equation of SHM. Hence this executes SHM
Simple Pendulum
A point heavy mass suspended with mass less, weightless, extensible string through a rigid support. It is also called ideal pendulum.
Drawbacks of simple pendulum.
It is impossible to have heavy point mass.
It is impossible to have weightless string.
There will be relative motion between the bob and string at extreme position.
Length of pendulum, is difficult to measure exactly accurately.
Compound Pendulum
Compound pendulum is a rigid body of whatever shape capable to oscillate freely about a horizontal axis passing through it.
The point in which pendulum is suspended through rigid suppost is called point of suspension.
The distance between point of suspension and centre of gravity of the pendulum is called the length of pendulum.
Consider a compound pendulum having mass m and length l i.e SG = L. Let pendulum be displace through an angle ϴ. At this instant the weigh of pendulum mf act vertically downward direction. Then the restoring force that tends to bring pendulum torque acting on the body will be,
Ʈ = -mglsin ϴ------------1
Now,
If I be the moment of intertia and
α = a be angular accleration then the restoring torque will be given as
Ʈ = I * α = I * a--------------2
From equation 1 and 2
I * a = -mglsin ϴ
a = -mgl/I * ϴ --------------3
This equation 3 is also similar to equation of SHM. Hence compound also executes SHM.
Point of Suspension and point of oscillation are interchangeable. We have time period for a compound pendulum having length L as
T = 2 √((k^2/L + L)/g)
T = 2 π(√l1 + l)/g-------------1
such that k^2/L = L1
Now, let point of oscillation becomes point of suspension. Then length of pendulum
k^2/L = L1
Hence, time period in this case will be
T1 = 2 π(√k2/l1+l)/g
We have k^2 = LL1
T1 = 2π√LL1+L1/(l/2)
= 2 π√L1 + l)/g-------------2
Wednesday, June 9, 2010
Structure
Array vs. Structure
1. An array is a collection of related data elements of same type. Structure can have elements of different data types.
2. An array is derived data type whereas a structure is a programmer-defined one.
3. Any array behaves like a built-in data type. All we have to do is to declare an array variable and use it. But in the case of a structure, first we have to design and declare a data structure before the variables of that type are declared anbd used.
Defining a structure
struct tag_name
{
data_type member1;
data_type member2;
};
1. The template is terminated with a semicolon
2. Wgile the entire difinition is considered as a statement, each member is declared independently for its name and type in a separate statement inside the template.
3. The tag name such as book_bank can be used to declare structure variables of the type, later in the program.
Declaring Structure Variables
struct book_bank, book1, book2,book3;
Elements included in declaring structure variable
1. The keyword struct
2. The structure tag name
3. List of variable names separated by commas
4. A terminating semicolon
Accessing Structure members
Members of a structure can be accesed by structure variable( member operator .) variable name
Structur Initialization
Rules for Initialization of Structure
1. We cannot initialize individual members inside the structure template
2. The order of values enclosed in braces must match the order of members in the structure difinition.
3. It is permitted to have a partial initialization. We can initialize only the first few members and leave the remaining blank. The uninitialized member should be only at the end of the list
4. The uninitialized members will be assigned default value as zero for intiger and #0 for character string
Unions
Major distinction between structure and union is in terms of storage. In structure, each member has its own storage location, whereas all the members of a union use the same location. This implies that, although a union may contain many member of different types, it can handle only one member at a time
1. An array is a collection of related data elements of same type. Structure can have elements of different data types.
2. An array is derived data type whereas a structure is a programmer-defined one.
3. Any array behaves like a built-in data type. All we have to do is to declare an array variable and use it. But in the case of a structure, first we have to design and declare a data structure before the variables of that type are declared anbd used.
Defining a structure
struct tag_name
{
data_type member1;
data_type member2;
};
1. The template is terminated with a semicolon
2. Wgile the entire difinition is considered as a statement, each member is declared independently for its name and type in a separate statement inside the template.
3. The tag name such as book_bank can be used to declare structure variables of the type, later in the program.
Declaring Structure Variables
struct book_bank, book1, book2,book3;
Elements included in declaring structure variable
1. The keyword struct
2. The structure tag name
3. List of variable names separated by commas
4. A terminating semicolon
Accessing Structure members
Members of a structure can be accesed by structure variable( member operator .) variable name
Structur Initialization
Rules for Initialization of Structure
1. We cannot initialize individual members inside the structure template
2. The order of values enclosed in braces must match the order of members in the structure difinition.
3. It is permitted to have a partial initialization. We can initialize only the first few members and leave the remaining blank. The uninitialized member should be only at the end of the list
4. The uninitialized members will be assigned default value as zero for intiger and #0 for character string
Unions
Major distinction between structure and union is in terms of storage. In structure, each member has its own storage location, whereas all the members of a union use the same location. This implies that, although a union may contain many member of different types, it can handle only one member at a time
Pointers
Why Pointers
1. Pointers are more efficient in handling arrays and data tables.
2. Pointers can be used to return multiple values from a function via function arguments.
3.Pointers permit references to functions and thereby facilitating passing of function s argument to another function.
4.The use of pointer arrays to character strings results in saving of data storage space in memory.
5.Pointers allow C to support dynamic memory managment.
6. Pointers reduce length and complexity of programs
7. They reduce the execution speed and thus reduce the program execution time.
Pointer Constant Pointer values Pointer Variable
Memory addresses within a computer are referred to as pointer constants. We can not change them; we can only use them to store data values. They are like house numbers.
We cannot save the value of a memory address directly. We can only obtain the value through the variable stored there using the address operator &. The value thus obtained is known as pointer value. The pointer value(i.e. the address of a variable) may change from one run of the program to another.
Once we have a pointer value, it can be stored into another variable. The variable that contains a pointer value is called pointer variable.
Rules of Pointer Operations
1. A pointer variable can be assigned the address of another variable.
2. A pointer variable can be assigned the values of another pointer variable.
3. A pointer variable can be initialized with NULL or zero value.
4. A pointer variable can be pre-fixed or post-fixed with increment or decrement operators
5. An integer value may be added or subtracted from a pointer variable.
6. When two pointers point to the same array, one pointer variable can b e subtracted from another.
7. When two pointers points to the objects of the same data types, they can be compared using relational operators.
8. A pointer variable can not be muktiplied by a constant
9. Two pointer variables can not be added
10. A value can not be assigned to an arbitary address
1. Pointers are more efficient in handling arrays and data tables.
2. Pointers can be used to return multiple values from a function via function arguments.
3.Pointers permit references to functions and thereby facilitating passing of function s argument to another function.
4.The use of pointer arrays to character strings results in saving of data storage space in memory.
5.Pointers allow C to support dynamic memory managment.
6. Pointers reduce length and complexity of programs
7. They reduce the execution speed and thus reduce the program execution time.
Pointer Constant Pointer values Pointer Variable
Memory addresses within a computer are referred to as pointer constants. We can not change them; we can only use them to store data values. They are like house numbers.
We cannot save the value of a memory address directly. We can only obtain the value through the variable stored there using the address operator &. The value thus obtained is known as pointer value. The pointer value(i.e. the address of a variable) may change from one run of the program to another.
Once we have a pointer value, it can be stored into another variable. The variable that contains a pointer value is called pointer variable.
Rules of Pointer Operations
1. A pointer variable can be assigned the address of another variable.
2. A pointer variable can be assigned the values of another pointer variable.
3. A pointer variable can be initialized with NULL or zero value.
4. A pointer variable can be pre-fixed or post-fixed with increment or decrement operators
5. An integer value may be added or subtracted from a pointer variable.
6. When two pointers point to the same array, one pointer variable can b e subtracted from another.
7. When two pointers points to the objects of the same data types, they can be compared using relational operators.
8. A pointer variable can not be muktiplied by a constant
9. Two pointer variables can not be added
10. A value can not be assigned to an arbitary address
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