Programming for Engineers: Data and Cgozips.uakron.edu/~jla36/documents/peng2011/peng2011topic03.pdf · 21.01.2009 · A Sample Program Data Types ACLU #3 Programming Assignment Programming

A Sample ProgramData Types

ACLU #3Programming Assignment

Programming for Engineers:Data and C

19 January 200921 January 2009

Programming for Engineers: Data and C



rhodium.c

/* rhodium.c – your weight in rhodium */#include <stdio.h>#include <conio.h>

int main(void){

float weight; /* user weight */float value; /* rhodium equivalent */

printf(“Are you worth your weight in rhodium?∖n”);printf(“Let’s check it out.∖n”);printf(“Please enter your weight in pounds: ”);




rhodium.c

/* rhodium.c – your weight in rhodium */#include <stdio.h>#include <conio.h>

int main(void){

float weight; /* user weight */float value; /* rhodium equivalent */

printf(“Are you worth your weight in rhodium?∖n”);printf(“Let’s check it out.∖n”);printf(“Please enter your weight in pounds: ”);




rhodium.c (cont’d)

/* get input from the user */scanf(“%f”, &weight);/* assume rhodium is $7010 per ounce *//* 14.5833 converts pounds avd. to ounces troy */value = 7010.0 * weight * 14.5833;printf(“Your weight in rhodium is worth $%.2f.∖n”, value);printf(“You are easily worth that! If rhodium prices drop,∖n”);printf(“eat more to maintain your value.∖n”);

printf(”∖n∖nPress any key to continue...”);while(!kbhit());

return 0;} Programming for Engineers: Data and C



rhodium.c (cont’d)

/* get input from the user */scanf(“%f”, &weight);/* assume rhodium is $7010 per ounce *//* 14.5833 converts pounds avd. to ounces troy */value = 7010.0 * weight * 14.5833;printf(“Your weight in rhodium is worth $%.2f.∖n”, value);printf(“You are easily worth that! If rhodium prices drop,∖n”);printf(“eat more to maintain your value.∖n”);

printf(”∖n∖nPress any key to continue...”);while(!kbhit());

return 0;} Programming for Engineers: Data and C



Integer Typeschar Typefloat TypesBool TypeComplex and Imaginary Types

Data Types

Data Type Bytes Minimum Maximum C Versionint 4 −231 ≈ −2× 109 231 − 1 ≈ 2× 109 K&Rlong 4 −231 231 − 1 K&Rshort 2 −215 ≈ 32, 000 215 − 1 ≈ 32, 000 K&Runsigned 4 0 232 K&Rchar 1 −27 = 128 27 − 1 = 127 K&Rfloat 4 1.175× 10−38 1.175× 1037 K&Rdouble 8 2.225× 10−308 22.225× 10307 K&Rsigned 4 −231 231 − 1 K&Rvoid N. A. C90Bool N.A. C99Complex N.A. C99Imaginary N. A. C99





Integer vs. Floating Point

▶ Integer Type (char)0 0 0 0 0 1 1 1

0×27+0×26+0×25+0×24+0×23+1×22+1×21+1×20

= 7

▶ Floating Point0 1000 0000 1.100 1001 0000 1111 1101 0000

(−1)0 ×(1× 20 + 1× 2−1 + ⋅ ⋅ ⋅+ 0× 2−23

)×(2128−127

)= +3.14159





int Type

▶ Declarationint number;

▶ Initializationnumber=1527; //decimal constantnumber=02767; //octal constantnumber=0x5F7; //hexadecimal constant

▶ Displayprintf(“%d”,number); //decimalprintf(“%o”,number); //octalprintf(“%x”,number); //hexadecimal





Other int Types

▶ Short int▶ Short number

printf(“%hd”,number);▶ Unsigned short number

printf(“%hu”,number);▶ Signed short number

printf(“%hd”,number);

▶ Long int▶ Long number

printf(“%ld”,number);▶ Unsigned long number

printf(“%lu”,number);▶ Signed long number

printf(“%ld”,number);





Other int Types (cont’d)

▶ Long Long int

▶ Long long numberprintf(“%lld”,number);

▶ Unsigned long long numberprintf(“%llu”,number);

▶ Signed long long numberprintf(“%lld”,number);





Integer Mathematics

▶ Neither integer underflow nor integer overflow is detected orsignaled.

▶ Integer division discards the remainder. For example, 94 = 2

instead of 94 = 2.25.





char Type

▶ Declarationchar ch;

▶ Initializationchar=’A’; //good stylechar=65; //poor stylechar=0x41; //poor stylechar=0101; //poor style

▶ Readscanf(“%c”,&ch);

▶ Displayprintf(“%c”,ch);





Other Specifics for Integer Types

▶ Octal▶ Symbol are 0 to 7▶ Must start with zero▶ Example: 0732 = 7× 82 + 3× 81 + 2× 80 = 47410

▶ Hexadecimal▶ Symbol are 0 to 9, A to F , or a to f▶ Must start with 0x or 0X▶ Example: 0x732 = 7× 162 + 3× 161 + 2× 160 = 184210





Escape Sequences

Sequece Meaning∖a Alert or bell∖b Backspace∖f Form feed∖n New line∖r Carriage return∖t Horizontal tab∖v Verticle tab∖∖ Backslash∖’ Single quote∖” Double quote∖? Question mark∖Ooo Octal number, with o an octal digit of 0 to 7∖xhh Hexadecimal number, with h a hexadecimal digit 0 to F





float Types

▶ Declarationfloat number;

▶ Initializationnumber=3.14159;number = 0.314159e+1;

▶ Displayprintf(“%f”,number);//decimal fractionprintf(“%e”,number);//exp. notationprintf(“%Lf”,number);//long double ver.printf(“%Le”,number);//long double ver.





Special Floating Point Cases

▶ Overflows to +INF

▶ Underflow is soft to 0.000000e+00

▶ Domain errors to +NAN





Float vs. Double

▶ float0 1000 0000 1.100 1001 0000 1111 1101 0000

(−1)0 ×(1× 20 + 1× 2−1 + ⋅ ⋅ ⋅+ 0× 2−23

)×(2128−127

)= +3.14159

▶ double float0 100 0000 0000 1.100 1001 0000 1111 1101 0

1111 0000 0001 1011 1000 0110 0110 1110

(−1)0 ×(1× 20 + 1× 2−1 + ⋅ ⋅ ⋅

)×(21024−1023

)= +3.14159





Bool Type

▶ Bool type is a C99 addition

▶ Bool type is not supported in Turbo C++▶ Values

▶ true is represented by a 1.▶ false is represented by a 0.





Complex and Imaginary Types

▶ C99 supports complex and imaginary types▶ float complex▶ double complex▶ long double complex

▶ Turbo C++ does not support these types directly

▶ The imaginary type is similar to the complex type.




ACLU #3




Programming Assignment

Complete Chapter 3 Exercise 2Complete Chapter 3 Exercise 4Complete Chapter 3 Exercise 6

Due Date: 24 February 2011


Documents

Programming for Engineers: Data and Cgozips.uakron.edu/~jla36/documents/peng2011/peng2011topic03.pdf · 21.01.2009 · A Sample Program Data Types ACLU #3 Programming Assignment Programming