A.Abhari CPS125 1

User_defined Structure Type• Database, record and structure• A structure type is a data type for a record composed of multiple componentsFor example a database of planets needs: Name: Jupiter Diameter: 142,980 km Moons: 16 Orbit time: 11.9 yr Rotation time: 9.925 hr

A.Abhari CPS125 2

User_defined Structure Type

• We need to use define a structure type:#define STRSIZE 10typedef struct {

char name[STRSIZE];double diameter;int moons;double orbit_time,

rotation_time;} planet_t; suffix _t is used to show it is a type

A.Abhari CPS125 3

User_defined Structure Type

• Also we need to define structure variables of the defined structure type:

planet_t previous_planet, current_planet;

blank_planet = {“”, 0, 0, 0, 0};

• In the program we use variables of the structure type:

current_planet = blank_planet;

A.Abhari CPS125 4

User_defined Structure Type

• Also we can define the a structure containing components that are structure. Such an structure is referred to as hierarchical structure:

For example: typedef struct {

double diameter;planet_t planets[9];char galaxy[STRSIZE];

} solar_sys_t;

A.Abhari CPS125 5

Manipulating Individual Components of a Structure Data Object

• Direct component selection (.) can be used to reference a component of a structure. For example:

strcpy(current_planet.name, “Jupiter”);

current_planet.diameter= 142980;

current_planet.moon=16;

current_planet.orbit_time=11.9;

current_planet.rotation_time=9.925;

A.Abhari CPS125 6

a[j], f(…), left highest++, -- (postfix) left++, --, !(prefix) right -, +, (unary) &, * right(type name) right*, /, % left+, - left<, >, <=, >= left==, != left&& left|| left=, +=, -=, *=, /=, %= right lowest

Operators Precedence

A.Abhari CPS125 7

Structure Type Data as Input and Output

Parameters

• We can assign one structure to another structure variable, however we can not apply the equality and inequality operators to a structure type as a unit.

• When a structure variable is passed to a function, if it is input its value is copied to the formal variable and if it is output we must apply the address of operator to pass it as a pointer to the variable.

A.Abhari CPS125 8

Function with a Structured Input Variable

/* * Displays with labels all components of a planet_t structure */voidprint_planet(planet_t pl) /* input - one planet structure */{ printf("%s\n", pl.name); printf(" Equatorial diameter: %.0f km\n", pl.diameter); printf(" Number of moons: %d\n", pl.moons); printf(" Time to complete one orbit of the sun: %.2f years\n", pl.orbit_time); printf(" Time to complete one rotation on axis: %.4f hours\n", pl.rotation_time);}

A.Abhari CPS125 9

Function Comparing Two Structured Values for Equality

#include <string.h>/* * Determines whether or not the components of planet_1

and planet_2 match */intplanet_equal(planet_t planet_1, /* input - planets to */ planet_t planet_2) /* compare */{ return (strcmp(planet_1.name, planet_2.name) == 0

&& planet_1.diameter == planet_2.diameter && planet_1.moons == planet_2.moons && planet_1.orbit_time == planet_2.orbit_time && planet_1.rotation_time == planet_2.rotation_time);}

A.Abhari CPS125 10

Function with a Structured Output Argument

/* * Fills a type planet_t structure with input data. Integer

returned as * function result is success/failure/EOF indicator. * 1 => successful input of one planet * 0 => error encountered * EOF => insufficient data before end of file * In case of error or EOF, value of type planet_t output

argument is * undefined. */

A.Abhari CPS125 11

intscan_planet(planet_t *plnp) /* output - address of planet_t

structure to fill */{ int result; result = scanf("%s%lf%d%lf%lf", (*plnp).name, &(*plnp).diameter, &(*plnp).moons, &(*plnp).orbit_time, &(*plnp).rotation_time); if (result == 5) result = 1; else if (result != EOF) result = 0; return (result);}

A.Abhari CPS125 12

Indirect Component Selection Operator

• In the indirect referencing &(*pnlp).moons parenthesizing was used to override the default operator precedence

• C provide a single operator that cobines indirection and component selection operator.

(*structp). component is equal to

strucpt -> component

A.Abhari CPS125 13

intscan_planet(planet_t *plnp) /* output - address of planet_t

structure to fill */{ int result; result = scanf("%s%lf%d%lf%lf", plnp->name,plnp->name,

&(plnp->diameter),&(plnp->diameter),&(plnp->moons),&(plnp->moons),&(plnp->orbit_time),&(plnp->orbit_time),&(plnp->rotation_time))&(plnp->rotation_time))

if (result == 5) result = 1; else if (result != EOF) result = 0; return (result);}

A.Abhari CPS125 14

Function get_planet Returning a Structured Result Type

/* * Gets and returns a planet_t structure */planet_tget_planet(void){ planet_t planet; scanf("%s%lf%d%lf%lf", planet.name, &planet.diameter, &planet.moons, &planet.orbit_time, &planet.rotation_time); return (planet);}

current_planet = get_planet() is same as scan_planet(&current_planet)

A.Abhari CPS125 15

A Function with Structure Result

Suppose we defined the following structure for time

typedef struct { int hour, minute, seconds; } time_t; And in the program time_now defined as a

variable of type time_t ;If time_now initialized by 21:58:32 and secs

initialized by 97 we want by calling time_now= new_time(time_now,secs);time_now becomes 22:00:09

A.Abhari CPS125 16

Function to Compute an Updated Time Value/* * Computes a new time represented as a time_t structure * and based on time of day and elapsed seconds. */time_tnew_time(time_t time_of_day, /* input - time to be updated */ int elapsed_secs) /* input - seconds since last update */{ int new_hr, new_min, new_sec; new_sec = time_of_day.second + elapsed_secs; time_of_day.second = new_sec % 60; new_min = time_of_day.minute + new_sec / 60; time_of_day.minute = new_min % 60; new_hr = time_of_day.hour + new_min / 60; time_of_day.hour = new_hr % 24; return (time_of_day);}

A.Abhari CPS125 17

Abstract Data Type

• A user-defined data type combined with a set of basic operations.

For example:

data type: planet_t

operations: scan_planet, print_planet, planet_equal

data type: complex_t (complex number a+bi)

operations: scan, print, add, subtract, multiply, divide, abs

A.Abhari CPS125 18

A User_Defined Type for Complex Numbers Partial Implementation of Type and Operators for Complex Numbers/* * Operators to process complex numbers */#include <stdio.h>#include <math.h>/* User-defined complex number type */typedef struct { double real, imag;} complex_t;

int scan_complex(complex_t *c);void print_complex(complex_t c);complex_t add_complex(complex_t c1, complex_t c2);complex_t subtract_complex(complex_t c1, complex_t c2);complex_t multiply_complex(complex_t c1, complex_t c2);complex_t divide_complex(complex_t c1, complex_t c2);complex_t abs_complex(complex_t c);

A.Abhari CPS125 19

/* Driver */intmain(void){ complex_t com1, com2; /* Gets two complex numbers */ printf("Enter the real and imaginary parts of a complex number\n"); printf("separated by a space> "); scan_complex(&com1); printf("Enter a second complex number> "); scan_complex(&com2); /* Forms and displays the sum */ printf("\n"); print_complex(com1); printf(" + "); print_complex(com2); printf(" = "); print_complex(add_complex(com1, com2));

A.Abhari CPS125 20

/* Forms and displays the difference */ printf("\n\n"); print_complex(com1); printf(" - "); print_complex(com2); printf(" = "); print_complex(subtract_complex(com1, com2)); /* Forms and displays the absolute value of the first number */ printf("\n\n|"); print_complex(com1); printf("| = "); print_complex(abs_complex(com1)); printf("\n"); return (0);}

A.Abhari CPS125 21

/* Returns sum of complex values c1 and c2*/complex_tadd_complex(complex_t c1, complex_t c2) /* input - values to add */{ complex_t csum; csum.real = c1.real + c2.real; csum.imag = c1.imag + c2.imag; return (csum);}

/*Returns difference c1 – c */complex_tsubtract_complex(complex_t c1, complex_t c2) /* input parameters */{ complex_t cdiff; cdiff.real = c1.real - c2.real; cdiff.imag = c1.imag - c2.imag; return (cdiff);}

A.Abhari CPS125 22

Parallel Arrays And Arrays of Structures

To represent student information we used parallel arrays:

int id[50];double gpa[50]Instead of using parallel arrays to represent the

student informationan array of structure can be used:typedef struct {

int id;double gpa;

} student_t;student_t stu_list[MAX_STU];

A.Abhari CPS125 23

Parallel Arrays And Arrays of Structures

If function scan_student is available we canuse for (i = 0; i < MAX_STU; ++i)

scan_student(&stu_list[i]);To fill the array andfor (i = 0; i < MAX_STU; ++i)

printf(“%d\t%.2f\n”, stu_list[i].id, stu_list[i].gpa);

To print the array contents

A.Abhari CPS125 24

Case Study

• Problem: Write a program that takes a measurement in one unit (e.g. 4.5 quarts) and converts it to another unit (e.g. liters).

• Structured data typeunit_t components:name /* e.g. “milligrams” */abbrev /* e.g. “mg” */class /* “liquid_volume”, “distance”, or “mass” */standard /* number of standard units equivalent to

this */

A.Abhari CPS125 25

Case Studytypedef struct { /* unit of measurement type */ char name[NAME_LEN]; /* character string such as "milligrams"

*/ char abbrev[ABBREV_LEN];/* shorter character string such as

"mg" */ char class[CLASS_LEN]; /* character string such as "pressure", "distance", "mass" */ double standard; /* number of standard units equivalent to this unit */} unit_t;int fscan_unit(FILE *filep, unit_t *unitp);void load_units(int unit_max, unit_t units[], int *unit_sizep);int search(const unit_t units[], const char *target, int n);double convert(double quantity, double old_stand, double new_stand);

Data File and Sample Run of Measurement Conversion ProgramData file units.dat:miles mi distance 1609.3kilometers km distance 1000yards yd distance 0.9144meters m distance 1quarts qt liquid_volume 0.94635

Sample run:Enter a conversion problem or q to quit.To convert 25 kilometers to miles, you would enter> 25 kilometers miles or, alternatively,> 25 km mi> 450 km milesAttempting conversion of 450.0000 km to miles . . .450.0000km = 279.6247 milesEnter a conversion problem or q to quit.> 2.5 qt lAttempting conversion of 2.5000 qt to l . . .2.5000qt = 2.3659 l

A.Abhari CPS125 27

Common Programming Errors

• Incorrect use of direct component selection ‘.’

• Indirect component selection ‘->’ can solve the problem of operator precedence of ‘.’

• There is no comparison operator on structure types and they can not directly used in printf and scanf