cs 261 c basics page 2 1/14/2016 cs 261, wsu vancouver primitive types notes: 4 a is a constant; b...

CS 261

C Basics

05/03/23 CS 261, WSU Vancouver

Primitive Types

Notes: A is a constant; B is a variable


Primitive Operations


Common Statements

Notes: always put in the braces! remember the semicolons!

int sum(int a, int b) {return a+b;

}

abstraction

int length(char s[]) {declarationsstatements

}

x = x + 3;y = sum(3,5);z = length ("abc");A[i] = '+';*p = '+';

assignment & using functions

/* a comment */#include <stdio.h>#define MAXSIZE 512

pre-processor stuff

this is class standard for formatting

if ( ... ) {...

} else if ( ...) {...

} else {...

}

switch ( ... ) {case ... :

...break;

case ... :...break;

default:...

}

selection

*

* We don't often use these

repetition

while ( ... ) {...

}

for (i = 0; i < n; ++i) {...

}

*

do {...

} while ( ... )


int isEvenSumSquares(int n) {int sum, i;

i = 0; /* i = 0, 1, … */sum = 0; /* sum = 0*0 + 1*1 + … + (i-1)*(i-1) */while (i <= n) {

sum = sum + i*i;i = i+1;

}

if (sum % 2 == 0) {return 1;

} else {return 0;

}}

Example: Even Sum of Squares Complete the following function declaration so that the code will return

the parity of the sum of the squares 12 + 22 + 32 + … + n2. If the sum is even, return 1; otherwise, return 0. For example, if n = 5, the result is 0 since 1 + 4 + 9 + 16 + 25 = 55 is odd.

Notes:• clear logic always better• clear indentation essential


Example: Sum of Squares For the mathematically inclined, here is a shortcut:

int isEvenSumSquares(int n) {if ((n % 4 == 3) || (n % 4 == 0)) {

return 1;} else {

return 0;}

}

int isEvenSumSquares (int n) {return (n % 4 == 3) || (n % 4 == 0);

}


Example: Sum of Squares To use the routine, we could code a main routine:

#include <stdio.h>

int isEvenSumSquares(int n);

int main() {int result;result = isEvenSumSquares(20);printf ("result for %d is %d\n", 20, result);return 0;

}

main.c

int isEvenSumSquares(int n) {return (n % 4 == 3) || (n % 4 == 0);

}sum.c

% gcc -o sum sum.c main.c% sumresult for 20 is 1%

what files are created?


Fundamental Principle of Counting

In C, we count from zero

A is an array with4 elements:

The first elementis A[0]:

The second elementis A[1]:

remarks

int A[4];

A[0] = 21;

A[1] = 6;

A[2] = 4;

A[3] = 9;

example code

The last elementis A[3]:


int S[4];int i, n;

n = 4;

Example: Set to Squares Write a code fragment so that will set each element

of the array S[i] to i2, for i = 0, 1, …, n-1. For example, if n=4, the code would set

S[0] to 0, S[1] to 1, S[2] to 4, and S[3] to 9.

What goes here?


Fundamental Principle of Arrays

Arrays are pointers to the first element

Aallocates room onthe stack and makesA point to first element

Astores 21 where Apoints

A+1stores 6 at next intbeyond where A points

A+3A[i] is shorthand for*(A+i)

example code remarks

int A[4];

*A = 21;

*(A+1) = 6;

*(A+2) = 4;

*(A+3) = 9;


Can Point to Elements More Than Once

int A[4];int B[];

*(A+2) = 4;

B = A;

*(B+1) = 11;

A B

?

A B

?

A B

A B

wouldA = B;be legal?


Strings are arrays of characters where the last character is '\0'

Fundamental Principle of Strings

core dump!s now has anew value s[0] ==


char s[];

s = "abc";

s = "";

s = 0;

s is an arrayvariable s

?

s now has avalue "abc" s[0] == 'a'

s[2] == 'c's[3] == '\0's

s now has anew value s

s[0] == '\0'""


int containsSpace(char s[]) {int i = 0;

while (s[i] != '\0') {if (s[i] == ' ') return 1;++i;

}

return 0;}

Example: Contains Space Write a function that tests if a string contains a space (' '). Return 1 if a space appears, 0 otherwise. Correctly handle the case of a null string. Don't use any library routines.

Notes:• usually prefer only one "return"• but, this shortcut logic is an OK idiom


Whiteboard Exercise Write a function that tests if a string contains

exactly one '/'

int hasOneSlash (char s[]) {

}


Whiteboard Exercise Write a function that tests if two strings are

exactly the same length

int areSameLength(char s[], char t[]) {

}


Whiteboard Exercise Write a function that trims all spaces from the

beginning of a string

char[] leftTrim(char s[]) {

}


Example: Length

core dump!

length("abc")

length("012345")

length("")

length(0)


Let's write a function, "length", that returns the length of a string ...

Note: Library routine "strlen" does this

Value is 3

Value is 6

Value is 0

Value is ...


First Implementation of Length ...Using arrays and subscripts ...

function definition

i == _____

length ("abc")

ss[0] == 'a's[1] == 'b's[2] == 'c's[3] == '\0'

"abc"

int length(char s[]){

int i;

i = 0;while (s[i] != '\0') {

i = i+1;}

return i;}


char a[4];int n;

a[0] = 'e';n = 4;

doIt(a, n);

a[0] == ____n == ______

invocation:

Fundamental Principle of Invocation

Function parameters are initialized by assignment from invocation argumentsparameter1 = argument1, parameter2 = argument2, ...

Arguments are passed by value

void doIt(char b[], int m) {

b[0] = 'f';m = m + 1;b = 0;

}

example code

"call-by-value"

b = a;m = n;

behind scenes


Fundamental Principle of Arrays, AgainArrays are just pointers to the first element

At this point:s[0] == 'a' == *(s+0)s[1] == 'b' == *(s+1)s[2] == 'c' == *(s+2)s[3] == '\0' == *(s+3)

At this point::i == _____

int length(char *s){int i;

i = 0;while (s[i] != '\0') {

i = i+1;}

return i;}

n = length("abc");

example function usageexample function definition

remarkswas:

char s[]


Some Versions of Length Using Pointers …

int length(char *s) {int i;

i = 0;while (s[i] != '\0') {

i = i+1;}return i;

}

v1int length(char *s) {

int i = 0;while (*(s+i) != '\0') {

++i;}return i;

}

v2

int length(char *s) {int i = 0;while (*s != '\0') {

++s; ++i;}return i;

}

v3 int length(char *s) {int i = 0;while (*s++ != '\0') {

++i;}return i;

}

v4


int length(char *s) {int i=0;

while (*s++ != '\0') ++i;

return i;

}

final implementation

Experienced C programmers can combine C facilities to writecompact but still clear code ...

String Length Again, Tersely ...

You are getting to right level if this code is clear 'char *s' same as 'char s[ ]' initializing declarations ++ operator, both prefix &

postfix call-by-value, so changes to

parameters don't change arguments

testing against '\0' (or 0) can be elided (but, I prefer to leave it explicit)

simulate on an example!


Array Declarations in Detail char A[4];

Allocates room for 4 characters and makes A a pointer to the first such character. A is a constant pointer that cannot be modified. A[i] is identical to *(A+i).

char *B; or char B[ ]; Allocates no storage; B has no defined value. B is a variable; its values point to a character. B[i] is identical to *(B+i).

B = A; Makes B point the same place as A. Thus B[i] are the same storage as A[i] for all i.

A = B; illegal; A is a constant (such as 253)

A

B ?

Note: there some subtle differences between char *B and char B[] involving initializing declarations


Let's write a routine, "copy", that copies one string over another

Note: Library routine "strcpy" does this

Whiteboard Exercise

char a[10];

copy(a, "1234");

copy(a+3, "xyz");

example usage remarks

Space set aside, but contents notpredictable

Contents now:a[0] == '1' a[1] == '2' a[2] == '3'a[3] == '4' a[4] == '\0'

Contents now:a[0] == '1' a[1] == '2' a[2] == '3'a[3] == 'x' a[4] == 'y' a[5] == 'z'a[6] == '\0'


Solution

void copy(char s[], char t[]) {

}

Want to copy all of t into s, like this:• s[0] = t[0]; s[1] = t[1]; s[2] = t[2]; …

Copying includes the '\0' at the end of t.We need to examine each char of t in turn.Let i be an index to do this.It will also be the index of next char in s.The logic will be:

• Set i to 0 (the index of the first char)• Repeat:

– Set s[i] to t[i].– If t[i] was '\0', quit this loop.– Otherwise, increment i and continue.

• Now we know t[i] is the '\0'.• So, all of t has been copied.• Return.

function logic function code


Some Useful Names & Functions


Let's write a function, "getline", that reads one line from stdin

Note: Library routine "fgets" does this

Another Example

#define MAXLINE 1000

char buffer[MAXLINE];

while (getline(buffer)) {... do something ...

}

example usage remarks

Define a constant

Set aside space

Process each line


1. Understand the Problem ... State the problem:

Read one line from stdin into s A line is terminated by '\n' or EOF (end of file) Fill s with all characters plus '\0'; exclude the '\n' (if any) Don't check if s is big enough Return 1 if a line was read, 0 if EOF detected right away

List representative test cases:

int getline(char s[]);


2. Code into C ... Write a description or draw a diagram that explains your approach:

read chars from stdin using getchar use c to remember the character just read use i to index next place in s to put a character

Outline the overall logic: set i to 0 repeat:

– read one char from stdin into c– if c is EOF, then quit this loop– if c is newline, then quit this loop– store c in s[i]– increment i

terminate s properly and return a value:– if c is EOF and i is 0, return 0– otherwise, set s[i] to '\0' and return 1

Convert the logic into code:


2. Code into C ... Convert the logic into code:

int getline(char s[]) {char c;int i = 0;

while ( c = getchar (), (c != EOF) && (c != '\n') ) {s[i] = c;++i;

}

if ((c == EOF) && (i == 0)) {return 0;

} else {s[i] = '\0';return 1;

}}

Notes:• what is EOF?• note = vs ==• note comma operation• note parens


3. Verify the solution ... Desk check the code by mentally simulating operation against test cases:

file with lines of varying length (0 chars, 1 char, 2 chars, …, 1000 chars)

file with varying number of lines (0 lines, 1 line, …)

file with last char as '\n' or not

Write a test routine, then compile and test:

#include <stdio.h>extern int getline (char s[]);

int main() {

char t[500];while (getline(t)) {

... write t plus '\n' to stdout ...}return 0;

}

... body of getline ...

When is outputdifferent from input?

cs 261 c basics page 2 1/14/2016 cs 261, wsu vancouver primitive types notes: 4 a is a constant; b...

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