computers and programming the 3 rd lecture jiří Šebesta

Computers and programming

The 3rd lecture

Jiří Šebesta

TOPIC

1. Iteration statements II.2. Jump statements3. String terminated by NULL4. Functions stdio.h for characters and strings 5. Library string.h for strings

Iteration statements II. (1/4)

while(test) statement;

• while

• Meaning of while is identical with for (initialization and increment outside of the head)

Example: Ex23.c

float temp[31]={-1.1,-0.3,-7.6,-2.7,-1.2,-2.6,-3.3, … … 2.4};// day temperatures in January unsigned int day = 1;

while(temp[day-1]<=0) // pass the loop while temp<=0 day++; printf("%d. January temperature ovecomes 0, it was %3.1f deg. C!\n", day, temp[day-1]);

return 0;


• while


do statement; while(test);

• do - while

Example: Ex24.c

int code; // number for passworddo{

printf("Insert access code:\n");scanf("%d", &code); //number from stdin to

variable code}while(code != 12345); // test of proper password 12345printf("\n\nAccess allowed");printf("\n\nInsert <space>+ENTER for exit");do //exit after <space> insertion

c=getchar();while (c!=' ');return 0;


• do - while

Jump statements (1/3)

• Utilization:

– in bodies of loops (for, do, while);– in body of switch (switch).

• Interrupting a performed iteration cycle:– break: leaving the whole loop– continue: continuing by the next iteration

• continue, break


• break

Example: Ex25.c

char c; int code, test, n=3; // n is the number of trialsdo{

printf("Insert access code:\n");scanf("%d", &code); //number from stdin to

variable coden--; //decrementation of the number of trials test=(code == 12345); // test of proper password

”12345”if ((test==0)&&(n==0)) break; // if no proper

// code and 3 trials done - acces denied}while(!test); // in test is info about access,

// 0 means denied, 1 means allowed


• continue

Example: divider searching

Example: Ex26.c

int num; //input numberint test, n, m=0;printf("Input number:");scanf("%d", &num); //number from stdin to variable numfor(n=2; n<=100; n++) //loop for n from 2 to 100{

test = (num%n == 0); //test if n is dividerif(!test)

continue; //if not, next nprintf(“\n%d", n); //if yes, print nm++; //number of dividers – incr.

}printf(“Number of dividers: %d\n", m);

Strings terminated by null (1/8)

• String: an array of characters (in memory: 1 byte/character – ASCII coding)

• NTS (Null Terminated Strings) : the last character of string null – spec. notation v C/C++ ’\0’

• Array elements: pointers to characters


• String as an array of ASCII codes of characters in memory:

• ASCII codes in range 0 - 127 worldwide standardized

• ASCII codes in range 128 – 255 (unsigned) depend on actual character set (national characters, UTF-8, Windows-1250, ISO-8859-x) – Extended ASCII


int main( void){ unsigned char c;

printf("Actual character set: \n\n"); for(c=32; c<=255 && c>=32; c++)

printf("\n ASCII code %d = character %c", c, c);

scanf("%c", &c); return 0;}

Ex . All characters printing including Extended ASCII

Example: Ex27.c


• Declaration methods:• char strA[5] = {‘A’,’B’,’C’,’D’,’\0’};

– 4 characters + terminating character null

• char strB[5] = ”ABCD”;– 4 characters, null inserted automatically – the same principle as string in the funtion printf()

• char *strC = ”ABCD”;- pointer to the first character, the first character points to the second character etc. up to null (this declaration is not recommended – potential conflict in memory)


• Special characters:\b - backspace BS \f - form feed FF (also clear screen) \n - new line NL

\r - carriage return CR

\t - horizontal tab HT \v - vertical tab (not all versions) \“ - double quotes (not all versions) \' - single quote character '

\\ - backslash character \ \ddd - character ddd, where ddd is an ASCII code given in octal base

\xhhh - character hhh, where hhh is an ASCII code given in hexadecimal base


Example: Ex28.c

• Ex. Arithmetic manipulation with characters

#include <stdio.h>

int main(void){ char text[] = "Vjku\"oguucig\"ku\"ugetgv#"; unsigned int n;

for(n=0; text[n]!='\0'; n++)// all chars in string if(text[n]!=' ') // excluding space

text[n] -= 2; // character code shift printf("%s\n", text); scanf("%c", &c); return 0;}


#include "stdafx.h"#include "stdio.h"#include "string.h" //library for ops. with strings

int main(void){ char s_inp[6] = {'a','b','c','d','e','\0'}; // or char s_inp[6] = "abcde"; char s_out[6], c; strcpy(s_out, s_inp); //copying of string,

not possible s_out = s_inp; printf("%s\n", s_out); scanf("%c", &c); return 0;}

• Copying of strings (using strcpy form string.h)

Example: Ex29.c


char S[7], T[7], c; char R[7] = "ABCDEF"; int n;

for( n=0; n<6; n++) // last -> first S[5-n] = R[n]; //or S[5-n] = *(R+n); S[6] = '\0' ; printf("%s\n", S);

for( n=0; n<6; n++) // capital -> small T[n] = R[n]+32; //or *(T+n) = *(R+n)+32; T[n] = '\0' ; printf("%s\n", T); scanf("%c", &c);

• Changeover of text, small letter => capital letter

Example: Ex30.c

Funct. stdio.h for chars and strings (1/3)

int main(void){ char c,d; int n; puts("Is C/C++ …"); // unformated printing of

string to stdout followed by new line c=getchar(); // get char.from the standard input if (c=='y')

puts("Right, ..."); if (c=='n')

puts("I disagree, ..."); do c=getchar(); while(c!=' '); // wait for space return 0;}

• Functions puts() and getchar()

Example: Ex31.c


int main(void){ char c;

for(c='A';c<='Z';c++)// loop for all alphabet {

putchar(c); // printing character to the stdout - capital

putchar(c+32); // smallputchar('\n'); // new line

}

do c=getchar(); while(c!=' '); // wait for space return 0;}

• Function putchar()

Example: Ex32.c


int main(void){ char fname[20], sname[20], c;

printf("Insert your first name: "); gets(fname); // reading characters from stdin and

stores them as a string - first name printf("\nInsert your surname: "); gets(sname); // reading surname printf("\nYour whole name is: %s %s", fname, sname); scanf("%c", &c); return 0;}

• Function gets()

Example: Ex33.c

Library string.h for strings (1/4)

char S[20], T[20], c; char R[20] = "ABCDEF"; int n;

for(n=0; n<strlen(R); n++) // last -> first S[strlen(R)-n-1] = R[n]; S[n] = '\0'; printf("%s\n", S);

for( n=0; n<strlen(R); n++) // cap. -> small T[n] = R[n]+32; T[n] = '\0'; printf("%s\n", T);

scanf("%c", &c);

• Length of string – strlen()

Example: Ex34.c


char final[30], c;

char first[] = "wine"; char second[] = "women"; char third[] = "songs";

strcpy(final, first); strcat(final, second); // appends a copy of the source

string to the destination string strcat(final, third); printf("%s\n", final);

scanf("%c", &c); return 0;

• Concatenation of strings – strcat()

Example: Ex35.c


char number[11], c; char *ptr, digit = '8';

strcpy(number, "1487265098"); printf("Original string: %s\n", number);

ptr = strchr(number, digit); while( ptr) { *ptr = 'x' ; ptr = strchr(number, digit); } printf("Modified string: %s", number);

scanf("%c", &c);

• Character searching in string – strchr()

Example: Ex36.c


char txt1[51], txt2[5], txt3[5]="----",*ptr, c; int n;

printf("Insert text1: "); gets(txt1); printf("Insert text2 (searched): "); gets(txt2); n=strlen(txt2); ptr = strstr(txt1, txt2); while( ptr) { strncpy(ptr,txt3,n); ptr = strstr(txt1, txt2); } printf("Modified string: %s", txt1);

• String searching in string– strstr()+strncpy()

Example: Ex37.c

Strings – examples (1/2)

In the variable num, unknown number of integer and rational numbers is stored. Each number is ended by semicolon. Compute the total number of numbers and the number of integer numbers in the string .

char num[30] = "12;3.8;100;94.5;33;44;" , c; int n, total=0, rat=0;

printf("Numbers: %s\n\n", num); for (n=0; n<strlen(num); n++) { if(num[n] == ';') // total number of num. total++; if(num[n] == '.') // number of rational n. rat++; } printf("Total number of num.: %4d\n", total); printf("Integer numbers: %4d\n", total-rat); scanf("%c", &c);

Example: Ex38.c

Strings – examples (2/2)In the string num, unknown number of integer numbers is stored. Each number is ended by semicolon. Print the numbers to the column and justify them properly.

char num[30] ="12;4;100;95;33;44;"; char part[10], c; int m=0, n;

printf("Numbers: %s\n\n", num); for(n=0; n<strlen(num); n++) if(num[n] != ';') part[m++] = num[n]; else { part[m] = '\0'; printf("%6s\n", part); m = 0; }

Example: Ex39.c

TOPIC OF THE NEXT LECTURE

Pointers and arrays

THANK YOU FOR YOUR ATTENTION

computers and programming the 3 rd lecture jiří Šebesta

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