CS 162 Introduction to Computer Science Chapter 17 C++ String Objects Herbert G. Mayer, PSU (Copied from Prof. Phillip Wong at PSU) Status 11/30/2014.

CS 162Introduction to Computer Science

Chapter 17C++ String Objects

Herbert G. Mayer, PSU(Copied from Prof. Phillip Wong at PSU)

Status 11/30/2014

Syllabus

String Data Simple String Operations Examples String I/O Functions String Library Functions String to Number Conversion Arrays of Strings

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String Data

C++ lacks the built-in type string Strings are approximated in C++ via arrays of

characters, i.e. arrays of type char Declaration: char stringname[ SIZE ]; Characters in the string occupy consecutive char

elements in the array

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The first '\0' (null) character found in a string literal marks the end of the string. It must be present for literals, inserted by your C++ compiler

When declaring a string, the array size should also be long enough to hold both the string and '\0’

Declaring a character array larger than the expected string length allows the array to accommodate a larger string at a later time

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Example:// Declare two stringschar str1[ 4 ];char str2[ 10 ];

// Initialize string one char at a timestr1[0]= 'C'; str1[1]= 'a'; str1[2]= 't';str1[3]= '\0';

// Use up only 5 out of 10 elementsstr2[0]= 'J'; str2[1]= 'a'; str2[2]= 'n'; str2[3]= 'e'; str2[4]= '\0';

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A string literal, A.K.A. constant is delimited by double quotation marks (e.g., "Hello, world!”)

A literal string automatically has a '\0' at the end of the string

A string variable can be declared and initialized with a string literalExample:char s[10] = "Hi, Bob!";char mystr[] = " Good night everybody! ”;char t1[4] = "Jim"; // OKchar t2[3] = "Jim"; // ERROR! Overflow

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Example:char s[10] = "Hi, Bob!";

s[0] 'H' 1218

s[1] 'i' 1219

s[2] ',' 1220

s[3] ' ' 1221

s[4] 'B' 1222

s[5] 'o' 1223

s[6] 'b' 1224

s[7] '!' 1225

s[8] '\0' 1226

s[9] 1227

s[0] s[1] s[2] s[3] s[4] s[5] s[6] s[7] s[8] s[9]

s → 'H' 'i' ',' ' ' 'B' 'o' 'b' '!' '\0'

s is the address of the first character in the string.s[j] is the j-th character in the string.s is equivalent to &s[0].

Declared array size = 10Valid array index range = 0 .. 9

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Example:char s[10] = "Hi, Bob!";

What happens if we do this: s[6] = '\0'; ?

The string stored in s is now “Hi, Bo” instead of “Hi, Bob!”

s[0] s[1] s[2] s[3] s[4] s[5] s[6] s[7] s[8] s[9]

s → 'H' 'i' ',' ' ' 'B' 'o' 'b' '!' '\0'

s[0] s[1] s[2] s[3] s[4] s[5] s[6] s[7] s[8] s[9]

s → 'H' 'i' ',' ' ' 'B' 'o' '\0' '!' '\0'

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To embed a quote " in a string, use \"Example:char str1[] = "Hello"; /* Hello */char str2[] = "\"Hello\""; /* "Hello" */

To embed a backslash in a string, use \\Example:char winpath[]="C:\\Windows”; // C:\Windows

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Examples of Simple String Operations

Assume these string declarations:char str1[ MAXSIZE ], str2[ MAXSIZE ];

Input a string (requires stdio.h)fgets( str1, MAXSIZE, stdin );

Print a stringprintf( "str1 = %s\n", str1 );

Copy a string (requires string.h)strcpy( str1, "Hello” );strcpy( str2, str1 );

Copies literal “Hello” to str1.

Copies contents of str1 to str2.

Stores input in str1.

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Example:

/* Simple string operations demonstration */#include <stdio.h>#include <string.h>

#define MAXLEN 100

int main( void ){ // main

char str1[ MAXLEN ], str2[ MAXLEN ];

printf( "What is your name? ” );fgets( str1, MAXLEN, stdin );printf( "Hello, %s!\n", str1 );return 0;

} //end main

Output:What is your name? JoeHello, Joe!

Why does the exclamation point show up on a separate line?fgets() stores the newline ('\n') that you typed as part of the string itself.

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String I/O Functions

C has library functions to input and output strings Use #include <stdio.h> In the following table, char * is a pointer to a

character array

String I/O Functions

int printf( const char * fmt, … ) Prints formatted output to console

int sprintf( char * s, const char * fmt, … ) Prints formatted output to string s

int puts( const char * s ) Prints string to console

int scanf( const char * fmt, … ) Reads formatted input from console

int fgets( const char * s, FILE * stream ) Reads string from input stream

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printf()

int printf( const char * format, …);

Writes formatted output to the console: Use the %s format specifier for strings %s expects the address of a char array which contains a string

Example:char name[] = "Jane Doe";printf("[%s]\n", name ); → [Jane Doe]printf("[%s]\n", & name[0] ); → [Jane Doe]printf("[%s]\n", & name[5] ); → [Doe]

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sprintf()

int sprintf( char * s, const char * format, …);

Performs a formatted print and saves it into a string Works like printf except the output is stored in string s Nothing is printed to the console

Example:char str[50];int x = 4;sprintf( str, "x=%d y=%f", x, 1.5f );printf( "title = %s\n", str ); title = x=4 y=1.500000 <- and line feed

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puts()

int puts( const char * s );

Writes a string to the console A newline '\n' is automatically printed at the end of the string

Example:char name[] = "Jane Doe";puts( name );printf( "lives here.\n” );

Output:Jane Doelives here.

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scanf()

int scanf( const char * format, …);

Reads formatted input from the console Use the %s format specifier for strings %s expects the address of a char array Only reads single “words” at a time (whitespace delimited)

Example:char str[100]; /* String storage */scanf( "%s", str ); /* str is address */scanf( "%s", &str[0] ); // synonymous to str

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scanf() scanf skips whitespace( space, tab, newline) and then stores

in str all characters up to the next whitespace scanf automatically adds '\0' to the end of the array It does not check for char array overflow. The array must be

large enough to hold the expected string and the '\0’ Any unused text remains in the input streamExample:scanf( "%s", str1 );User types in: Hello, Portland !!!scanf( "%s", str2 );→ str1 holds Hello,→ str2 holds Portland // not the !!!

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fgets()

char * fgets( char * s, int m, FILE * stream );

Reads a string from the input streamFor input from the console, use stdin for the streamfgets reads input characters until one of these conditions is met:

m – 1 characters are read '\n' (newline) is read, will be stored if fits into space End-of-file is detected

If successful, returns pointer to sIf end-of-file, returns NULL pointer

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Example

#define MAXLEN 10000

char str[ MAXLEN ]; // String storage

if( fgets( str, MAXLEN, stdin ) ) {printf( "%s", str );

}else{printf( "End of file\n” );

} //end if

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If a newline is entered to signal the end of input, the '\n' character is also stored as part of string

fgets automatically terminates the string with '\0’

Example:fgets( str, 10, stdin );

Suppose the user types in: Jim May

printf( "[%s]", str ); → [Jim May]

str → 'J' 'i' 'm' ' ' 'M' 'a' 'y' '\n' '\0'

User types Enter key(newline)

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Is there a simple way to get rid of the extraneous '\n' when using fgets?

This is one approach that works:

fgets( s, MAXLEN, stdin );if( strchr( s,'\n') != NULL )

s[ strlen(s) - 1 ] = '\0';

Caveat: The if check is necessary because a Ctrl-D terminates input without adding an extra '\n’

Warning: There is also a function; gets(). DO NOT USE IT!!

gets() does not check for char array overflow

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String Library Functions

C has a library of string processing functions Use #include <string.h>

In the following table:

s is of type char * (pointer to char array)n is of type size_t (unsigned integer)cs and ct are of type const char *c is an int converted to char

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Some Common string lib functionssize_t strlen( cs ) return length of cs.

char * strcpy( s, ct ) copy string ct to string s, including ‘\0’; return s.

char * strncpy( s, ct, n ) copy at most n characters of string ct to s; return s. Pad with ‘\0’s if ct has fewer than n characters.

char * strcat( s, ct ) concatenate string ct to end of string s; return s.

char * strncat( s, ct, n ) concatenate at most n characters of string ct to string s, terminate s with ‘\0’; return s.

int strcmp( cs, ct ) compare string cs to string ct; return <0 if cs<ct;0 if cs==ct, or >0 if cs>ct.

int strncmp( cs, ct, n ) compare at most n characters of string cs to string ct; return <0 if cs<ct; 0 if cs==ct, or >0 if cs>ct.

char * strchr( cs, c ) return pointer to first occurrence of c in cs or NULL if not present

char * strrchr( cs, c ) return pointer to last occurrence of c in cs or NULL if not present

char * strstr( cs, ct ) return pointer to first occurrence of string ct in cs, or NULL if not present

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strlen()

size_t strlen( const char * s );

Determines the length of the string s. The function counts the number of characters in the array. The count starts at array index 0. It continues counting until the first '\0' is found.

('\0' itself is not included in the string length.)

Example:strlen("") → 0char x[] = "Cat"; strlen(x) → 3SL = strlen(" PSU Vikings") → 12

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strcpy()

char * strcpy( char * d, const char * s );

Copies contents of string s to string d The terminating '\0' is also copied to d. The contents of d are overwritten. The address of string d is returned. Strings cannot be copied using the = assignment operator. Use strcpy() instead.

Example:char dst[20], src[] = "Hello";strcpy( dst, "Siri” ); → dst holds Siristrcpy( dst, src ); → dst holds Hello

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strcat()

char * strcat( char * d, const char * s );

Concatenates contents of string s to string d The contents of s are appended to the end of whatever is

already in d The original '\0' in d is deleted before appending happens The address of the new string d is returned Strings cannot be concatenated using the + operator in C.

Use strcat() instead.Example:char dst[20] = "PSU", src[] = " rocks!";strcat( dst, src ); → dst holds PSU rocks!

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strcmp()

int strcmp( const char * s, const char * t );

Compares contents of string s to string t s and t are compared character by character Returns zero (0) if the strings are identical

positive number if s is lexically greater than t negative number if t is lexically greater than s

Strings cannot be compared using a relational operator. Use strcmp() instead

Example:char s1[5] = "PSU", s2[10] = "OSU";strcmp( s1, s2 ); → returns 1strcmp( s2, s1 ); → returns -1

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Example:char a[10], b[10]; /* Two strings */

strcpy( a, "Star” ); /* a: Star */strcpy( b, a ); /* b: Star */strcat( a, " Trek” ); /* a: Star Trek */strcat( b, " Wars” ); /* b: Star Wars */

if( strlen(a) > 0 && strlen(b) > 0 )if( strcmp( a, b ) == 0 )

printf( "You're kidding, right?\n” );else if( strcmp( a, b ) < 0 )

printf( "Trekker!\n” );

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strchr()

char * strchr( const char * s, int c );

Finds position of character c within string s. If c is found, returns pointer to first occurrence of c in s.

If c is not found, returns NULL pointer.

Example:char s[] = "PSU OSU";strchr( s, 'A’ ); → returns NULLp = strchr( s, 'U’ ); → returns pointer to first U

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strchr()

char * strstr( const char * s, const char * t );

Finds position of string t within string s. If t is found, returns pointer to first occurrence of t in s.

If t is not found, returns NULL pointer.

Example:char s[] = "PSU OSU";strstr( s, "A" ); → returns NULLp = strstr( s, "U O" ); → returns ptr to location

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String to Number Conversion

Use #include <stdlib.h>

Conversion functions (partial list)

Example:double x;char numstr[] = "12.75 HI";x = atof( numstr ); → x contains 12.75x = atof( "HI 12.75” ); → x contains 0.0printf( "%d\n", atoi( numstr ) ); → displays 12

double atof( const char * s ) Converts s to a number of type double

int atoi( const char * s ) Converts s to a number of type int

long atol( const char * s ) Converts s to a number of type long int

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Array of Strings

An array of strings (“string array”) is a 2-D array Each row represents a separate string

When declaring the array, the number of columns must be large enough to hold the largest expected string

An individual string within the array can be accessed by using just the row index

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Example:

/* 4 strings of up to 10 chars each */char s[4][10];strcpy( s[0], "Doe, Jane” );strcpy( s[3], "Rand, Bob” );

/* Assume user enters: Li, Joe */fgets( s[1], 10, stdin );

/* Assume user enters: Li, Joe */scanf( "%s", s[2] );

printf( "%s\n", s[0] ); /* Doe, Jane */printf( "%s\n", &s[0][5] ); /* Jane */printf( "%c\n", s[3][6] ); /* B */

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Example: (continued from previous page)

s is the name for the entire array of strings.

s[i] is the address of the i-th string in the array.

s[i][j] is j-th character of the i-th string.

s[i] is equivalent to &s[i][0].

s[3] → 'R' 'a' 'n' 'd' ',' ' ' 'B' o' 'b' '\0' length = 9

s[2] → 'L' 'i' ',' '\0' length = 3

s[1] → 'L' 'i' ',' ' ' 'J' 'o' 'e' '\n' '\0' length = 8

0 1 2 3 4 5 6 7 8 9

s[0] → 'D' 'o' 'e' ',' ' ' 'J' 'a' 'n' 'e' '\0' length = 9

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Example: Write a string length function using arrays#include <stdio.h>

/* Function returns the length of a string */int str_len( const char s[]){ // str_len

int k = 0; /* Array index for string */

while( s[k] != '\0') /* Look for string terminator */k++;

return k; /* k is also the length */} //end str_len

int main( void ){ // main

char a[] = "Hello!";

printf( "String length is %d.\n", str_len(a) );

return 0;} //end main

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Example: Write a string copy function using arrays

/* String copy: source: cs thus can be const, destination: st */char * str_cpy( char st[], const char cs[] ){

int k = 0; // Array index for string

while( cs[k] != '\0’ ) { /* Look for string terminator */st[k] = cs[k]; /* Copy single character at a time */k++;

} //end whilest[k] = '\0'; /* Add string terminator to destination */

return st; // st by itself is an address} //end str_cpy