CS110: Data Types in C

Lecture 8V. Kamakoti

20th January 2008

Datatypes

• Types of Data you can use in C withoutexplicit definition by the user– Arithmetic

• Integer domain• Real domain

– Strings• Length one - character• Length greater than one - array of characters

– Sequence of characters

• Built-in Types in C• You can define your own data types• Construct called typedef

• Later in this course• User defined data type

• Creative Problem

Today’s Lecture Contents

Built-in Data types

• int• char• float• double• Sizes of each vary depending on

– Architecture and compiler

Variations

• int– signed, unsigned, long, short

• float– single and double precision

• Size of “short int” less than or equal to just“int” which is less than or equal to “long int”

• The builtin function sizeof() shall return thesize of the types.

Example Program• #include <stdio.h>• /* Program to print sizeof all datatypes */• main() {• printf("sizeof int is %d\n",sizeof(int));• printf("sizeof short int is %d\n",sizeof(short int));• printf("sizeof long int is %d\n",sizeof(long int));• printf("sizeof unsigned is %d\n",sizeof(unsigned));• printf("sizeof short unsigned is %d\n",sizeof(short unsigned));• printf("sizeof long unsigned is %d\n",sizeof(long unsigned));• printf("sizeof float is %d\n",sizeof(float));• printf("sizeof double is %d\n",sizeof(double));• }

Output

• Apple I Mac system - Intel Core 2 DuoArchitecture

• Compiler gcc-4.0.1 versionsizeof int is 4sizeof short int is 2sizeof long int is 4sizeof unsigned is 4sizeof short unsigned is 2sizeof long unsigned is 4sizeof float is 4sizeof double is 8

What does this mean?

• int i; // Allocated to memory 0x1000• It has 4 bytes

– B3,B2,B1,B0– B3 is most significant byte 0x1003– B0 is least significant byte stored in 0x1000– B1 is stored in 0x1001– B2 is stored in 0x1002

• This is called little-endian storage - Intelmachines use this

What does this mean?

• int i; // Allocated to memory 0x1000• It has 4 bytes

– B3,B2,B1,B0– B3 is most significant byte 0x1000– B0 is least significant byte stored in 0x1003– B1 is stored in 0x1002– B2 is stored in 0x1001

• This is called big-endian storage - SUNprocessors - Sparc use this

Constants in C• Remember the #define PI in the Area of the

circle problem?• That was a “real” constant

• 0, 1, 245, 623, 987 - are decimals• 0, 01, 0245, 0623 - are octals

• Begin with a “0”• So 125 is not equal to 0125• 0987 is illegal - why?

• 0x0, 0x1, 0xABCD - are hexadecimals• 0X0, 0X1, 0XABCD - are also correct way of

representing them.

Wrong ways for integer constants

• 12,245– no comma to enhance readability

• 36.0– no dots //Not integer constant

• 10 20 30– no space in between

• 123-45-6789– no hyphen to enhance readability

• 0900– not decimal

Wrong ways for octal constants

• 743– Should start with 0

• 05280– Illegal digit “8”

• 0777.77– No dots allowed for octal integers

Unsigned and Long IntegerConstants

• 50000U - decimal unsigned• 123456789L - decimal long• 123456789UL - decimal unsigned long• 0123456L - octal long• 0777777U - octal unsigned• 0x5000ABCU - hexa unsigned• 0XFFFFFUL - hexa unsigned long

Floating Point Constants

• 3 X 105

– 300000.– 3e5– 3e+5– 3E5– 3.0e+– .3e6– 0.3E6

• 3 X 105

– 30E4– 30.E+4– 300e3– ….

Floating Point Constants

• 5.026 X 10-17

– 5.026E-17– .5026E-16– 50.26E-18– .0005026E-13

Beware: Underflows and Overflows because of

Finite precision arithmetic

Character Constants

• ASCII code - introduced• ‘A’ is an example• ‘a1’ is wrong• sizeof(char) is 1 byte• Declared as char c;• Then assign as

– c = ‘A’ //Only single quotes

Escape Sequences

• ‘\n’ - Line feed• ‘\t’ - Tab• ‘\r’ carriage return• Where do you use?• Get an apostrophe ‘\’’• Double quote ‘\”’• ‘\0’ is the null character

ASCII values

char c; c = ‘A’; c = ‘\101’; //Octal of 65 c = ‘\x41’; //Hexa of 65 c = ‘\X41’;All are same. The ASCII value of

character c is decimal 65

Symbolic Constants

Those that are defined using#define

Remember the PI in your Area of thecircle program.

Statements

• Simple and compound statements• Expression and control statements• if (condition) S1 else S2 - Control• a = b+c; expression { pi = 3.141593; area = pi * radius * radius;} // A compound statement

Test Your Understanding

• Which of the following are not valid identifiers– record1– 1record– $tax– file_3– File_3– return– Name_and_address– Name-and-address

Test Your Understanding

• Which of the following are not valid identifiers– record1– 1record

– $tax– file_3– File_3– return– Name_and_address– Name-and-address

Test Your Understanding

• Which of the following is not a ANSI Clanguage keyword?– void– enum– goto– function

Test Your Understanding

• Which of the following is not a ANSI Clanguage keyword?– void– enum– goto– function

Test Your Understanding

• What will be the result of the followingprogram if the inputs are 2 and 3

main() { float a,b; scanf(“%d %d”,&a,&b); printf(“%f %f”,a,b);}

Test Your Understanding

main() { float a,b; scanf(“%d %d”,&a,&b); printf(“%f %f”,a,b);}

Ans: It compiles without any problem. It will begarbage. What does this tell?

Test Your Understandingmain() { float a,b; scanf(“%d %d”,&a,&b); printf(“%f %f”,a,b);}

Ans: The compiler only catches some type of bugs. Youhave to take much care. Just because your programcompiled without errors/warning, you are notguaranteed of correct results.

Test Your Understanding

• What will be the result of the followingprogram if the inputs are 2 and 3

main() { int a,b; scanf(“%d %d”,&a,&b); printf(“%f %f”,a,b);}

Test Your Understanding

• Junk will be printed for this too. main() { int a,b; scanf(“%d %d”,&a,&b); printf(“%f %f”,a,b);}

Test Your Understanding

• What will be the result of the followingprogram

main() { int a,b; a = 10; b = 20; printf(“%d ”,a,b);}

Test Your Understanding

• It shall print 10main() { int a,b; a = 10; b = 20; printf(“%d ”,a,b);}

Conclusion

• Compilers do not catch most of yourbugs

• Reasons are many– Attend the 5th semester Language

Translator BTech Core course in CSE Dept• You train yourself to be a CAREFUL

PROGRAMMER

Creative Problem

• Suppose you are using a program thatreads in a large English text file as inputand processes it. The program for somereason does not like a set of words andwhenever it sees the same in your textfile it halts and outputs “Error”. It is soangry that it does not say what the wordis and which line it occurred or anyother info.

Creative Problem

• You do not have a list of offendingwords. Devise a strategy to identify thefirst offending word in your text. Expressthe efficiency of your strategy in termsof the number of words in the input textfile.

Thank You

• SAARANG Time :-)– Do not go home - it is your festival– No lab this week– People who have pending work to be done

can visit labs on Monday and Tuesday tofinish backlogs.

– Lab exercises 1 and 2• Grades to be finalized this week.